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For this Question: Given an integer N, you have to print the given below pattern for N ≥ 3.
<b>Example</b>
Pattern for N = 4
*
*^*
*^^*
*****<b>User Task</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>Pattern()</b> that takes integers N as an argument.
<b>Constraints</b>
3 ≤ N ≤ 100Print the given pattern for size N.<b>Sample Input 1</b>
3
<b>Sample Output 1</b>
*
*^*
****
<b>Sample Input 2</b>
6
<b>Sample Output 2</b>
*
*^*
*^^*
*^^^*
*^^^^*
*******, I have written this Solution Code: void Pattern(int N){
printf("*\n");
for(int i=0;i<N-2;i++){
printf("*");
for(int j=0;j<=i;j++){
printf("^");}printf("*\n");
}
for(int i=0;i<=N;i++){
printf("*");
}
}
, In this Programming Language: C, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Joey and Chandler are super bored. So, Chandler makes up a game they can play. The game is called Chandy Game. Initially, Chandler has A candies and Joey has B candies.
In the first move Joey has to give Chandler 1 candy.
In the second move Chandler has to give Joey 2 candies.
In the third move Joey has to give Chandler 3 candies.
In the fourth move Chandler has to give Joey 4 candies.
In the fifth move Joey has to give Chandler 5 candy.
... and so on.
The game continues till one of the player can not make a move. The player who cannot make a move loses. Help them find who wins the game.Input contains two integers A and B.
Constraints:
0 <= A, B <= 10<sup>15</sup>Print "Chandler" (without quotes) if Chandler wins the game and "Joey" (without quotes) if Joey wins the game.Sample Input
2 1
Sample Output
Chandler
Explanation:
In first move Joey gives Chandler 1 candy so, Chandler has 3 candies and Joey has 0.
In second move Chandler gives Joey 2 candies so, Chandler has 1 candy and Joey has 2.
In third move Joey has to give Chandler 3 candies but he has only 2 candies so he loses., I have written this Solution Code: import java.io.*;
import java.util.*;
class Main
{
public static void main(String args[])throws Exception
{
BufferedReader bu=new BufferedReader(new InputStreamReader(System.in));
StringBuilder sb=new StringBuilder();
String s[]=bu.readLine().split(" ");
long a=Long.parseLong(s[0]),b=Long.parseLong(s[1]);
if(a>=b) sb.append("Chandler");
else sb.append("Joey");
System.out.print(sb);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Joey and Chandler are super bored. So, Chandler makes up a game they can play. The game is called Chandy Game. Initially, Chandler has A candies and Joey has B candies.
In the first move Joey has to give Chandler 1 candy.
In the second move Chandler has to give Joey 2 candies.
In the third move Joey has to give Chandler 3 candies.
In the fourth move Chandler has to give Joey 4 candies.
In the fifth move Joey has to give Chandler 5 candy.
... and so on.
The game continues till one of the player can not make a move. The player who cannot make a move loses. Help them find who wins the game.Input contains two integers A and B.
Constraints:
0 <= A, B <= 10<sup>15</sup>Print "Chandler" (without quotes) if Chandler wins the game and "Joey" (without quotes) if Joey wins the game.Sample Input
2 1
Sample Output
Chandler
Explanation:
In first move Joey gives Chandler 1 candy so, Chandler has 3 candies and Joey has 0.
In second move Chandler gives Joey 2 candies so, Chandler has 1 candy and Joey has 2.
In third move Joey has to give Chandler 3 candies but he has only 2 candies so he loses., I have written this Solution Code: #pragma GCC optimize ("Ofast")
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define VV vector
#define pb push_back
#define bitc __builtin_popcountll
#define m_p make_pair
#define infi 1e18+1
#define eps 0.000000000001
#define fastio ios_base::sync_with_stdio(false);cin.tie(NULL);
string char_to_str(char c){string tem(1,c);return tem;}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
template<class T>//usage rand<long long>()
T rand() {
return uniform_int_distribution<T>()(rng);
}
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
template<class T>
using oset = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
// string to integer stoi()
// string to long long stoll()
// string.substr(position,length);
// integer to string to_string();
//////////////
template<class C> void mini(C&a4, C b4){a4=min(a4,b4);}
typedef unsigned long long ull;
auto clk=clock();
#define all(x) x.begin(),x.end()
#define S second
#define F first
#define sz(x) ((long long)x.size())
#define int long long
#define f80 __float128
#define mod 1000000007ll
#define pii pair<int,int>
/////////////
signed main(){
#ifdef ANIKET_GOYAL
freopen("inputf.in","r",stdin);
freopen("outputf.in","w",stdout);
#endif
int a,b;
cin>>a>>b;
if(a>=b)
cout<<"Chandler";
else
cout<<"Joey";
#ifdef ANIKET_GOYAL
// cout<<endl<<endl<<endl<<endl<<"Time elapsed: "<<(double)(clock()-clk)/CLOCKS_PER_SEC<<endl;
#endif
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Joey and Chandler are super bored. So, Chandler makes up a game they can play. The game is called Chandy Game. Initially, Chandler has A candies and Joey has B candies.
In the first move Joey has to give Chandler 1 candy.
In the second move Chandler has to give Joey 2 candies.
In the third move Joey has to give Chandler 3 candies.
In the fourth move Chandler has to give Joey 4 candies.
In the fifth move Joey has to give Chandler 5 candy.
... and so on.
The game continues till one of the player can not make a move. The player who cannot make a move loses. Help them find who wins the game.Input contains two integers A and B.
Constraints:
0 <= A, B <= 10<sup>15</sup>Print "Chandler" (without quotes) if Chandler wins the game and "Joey" (without quotes) if Joey wins the game.Sample Input
2 1
Sample Output
Chandler
Explanation:
In first move Joey gives Chandler 1 candy so, Chandler has 3 candies and Joey has 0.
In second move Chandler gives Joey 2 candies so, Chandler has 1 candy and Joey has 2.
In third move Joey has to give Chandler 3 candies but he has only 2 candies so he loses., I have written this Solution Code: ch,jo=map(int,input().split())
if ch>=jo:
print('Chandler')
else:
print('Joey'), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Newton is currently standing at the origin (0, 0), at time t = 0. Each second, he will move exactly 1 unit up, down, left or right from his current position, where each direction has equal probability. Define f(x, y) as follows:
f(x, y) = 0, if the point (x, y) has been visited less than K times upto and including time t = N.
f(x, y) = t<sub>P</sub>, if the point (x, y) has been visited K or more times upto and including time t = N, and t<sub>P</sub> is the time it was visited for the P<sup>th</sup> time (P ≤ K).
Let S be the sum of f(x, y) over all lattice points of the plane. You will be given Q queries to solve. In each query, you are given the integers N and P, and you have to print the expected value of S modulo 998244353. Note that the integer K is constant for all queries.
Formally, it can be shown that the answer to each query can be expressed as an irreducible fraction <sup>p</sup>⁄<sub>q</sub>, where p and q are integers and gcd(q, 998244353) = 1. You have to output the integer equal to pq<sup>-1</sup> mod 998244353 for each query.
Note that the time of first visit of origin is considered to be 0.The first line of the input contains two space separated integers K (1 ≤ K ≤ 10<sup>5</sup>) and Q (1 ≤ Q ≤ 10<sup>5</sup>).
Then Q lines follow, each line containing two integers N and P for each query (1 ≤ P ≤ K ≤ N ≤ 10<sup>5</sup>).Print Q lines — each line containing a single integer with the i<sup>th</sup> line denoting the expected value of S modulo 998244353 for the i<sup>th</sup> query.Sample input 1:
1 1
2 1
Sample output 1:
499122179
Explanation 1:
The value comes out to be 5/2.
Sample input 2:
2 2
3 2
3 1
Sample Output 2:
748683266
748683265, I have written this Solution Code: // define asc
//#define dbg_local
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace std;
using namespace __gnu_pbds;
template <typename t>
using ordered_set = tree<t, null_type, less<t>, rb_tree_tag, tree_order_statistics_node_update>;
// #pragma gcc optimize("ofast")
// #pragma gcc target("avx,avx2,fma")
#define all(x) (x).begin(), (x).end()
#define pb push_back
#define endl '\n'
#define fi first
#define se second
// const int mod = 1e9 + 7;
const int mod=998'244'353;
const long long INF = 2e18 + 10;
// const int INF=1e9+10;
#define readv(x, n) \
vector<int> x(n); \
for (auto &i : x) \
cin >> i;
template <typename t>
using v = vector<t>;
template <typename t>
using vv = vector<vector<t>>;
template <typename t>
using vvv = vector<vector<vector<t>>>;
typedef vector<int> vi;
typedef vector<double> vd;
typedef vector<vector<int>> vvi;
typedef vector<vector<vector<int>>> vvvi;
typedef vector<vector<vector<vector<int>>>> vvvvi;
typedef vector<vector<double>> vvd;
typedef pair<int, int> pii;
int multiply(int a, int b, int in_mod) { return (int)(1ll * a * b % in_mod); }
int mult_identity(int a) { return 1; }
const double pi = acosl(-1);
auto power(auto a, auto b, const int in_mod)
{
auto prod = mult_identity(a);
auto mult = a % in_mod;
while (b != 0)
{
if (b % 2)
{
prod = multiply(prod, mult, in_mod);
}
if(b/2)
mult = multiply(mult, mult, in_mod);
b /= 2;
}
return prod;
}
auto mod_inv(auto q, const int in_mod)
{
return power(q, in_mod - 2, in_mod);
}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
#define stp cout << fixed << setprecision(20);
namespace algebra {
using ll = long long;
using uint = unsigned int;
using ull = unsigned long long;
using ld = long double;
template<typename T> using max_heap = std::priority_queue<T>;
template<typename T> using min_heap = std::priority_queue<T, std::vector<T>, std::greater<T>>;
struct modinfo
{
const uint mod, root, max2p;
};
template<const modinfo& info>
class modint
{
public:
static constexpr const uint& mod = 998244353;
static constexpr const uint& root = 3;
static constexpr const uint& max2p = 23;
constexpr modint() : m_val{0} {}
constexpr modint(const ll v) : m_val{normll(v)} {}
constexpr modint(const modint& m) = default;
constexpr void set_raw(const uint v) { m_val = v; }
constexpr modint& operator=(const modint& m) { return m_val = m(), (*this); }
constexpr modint& operator=(const ll v) { return m_val = normll(v), (*this); }
constexpr modint operator+() const { return *this; }
constexpr modint operator-() const { return modint{0} - (*this); }
constexpr modint& operator+=(const modint& m) { return m_val = norm(m_val + m()), *this; }
constexpr modint& operator-=(const modint& m) { return m_val = norm(m_val + mod - m()), *this; }
constexpr modint& operator*=(const modint& m) { return m_val = normll((ll)m_val * (ll)m() % (ll)mod), *this; }
constexpr modint& operator/=(const modint& m) { return *this *= m.inv(); }
constexpr modint& operator+=(const ll val) { return *this += modint{val}; }
constexpr modint& operator-=(const ll val) { return *this -= modint{val}; }
constexpr modint& operator*=(const ll val) { return *this *= modint{val}; }
constexpr modint& operator/=(const ll val) { return *this /= modint{val}; }
constexpr modint operator+(const modint& m) const { return modint{*this} += m; }
constexpr modint operator-(const modint& m) const { return modint{*this} -= m; }
constexpr modint operator*(const modint& m) const { return modint{*this} *= m; }
constexpr modint operator/(const modint& m) const { return modint{*this} /= m; }
constexpr modint operator+(const ll v) const { return *this + modint{v}; }
constexpr modint operator-(const ll v) const { return *this - modint{v}; }
constexpr modint operator*(const ll v) const { return *this * modint{v}; }
constexpr modint operator/(const ll v) const { return *this / modint{v}; }
constexpr bool operator==(const modint& m) const { return m_val == m(); }
constexpr bool operator!=(const modint& m) const { return not(*this == m); }
constexpr friend modint operator+(const ll v, const modint& m) { return modint{v} + m; }
constexpr friend modint operator-(const ll v, const modint& m) { return modint{v} - m; }
constexpr friend modint operator*(const ll v, const modint& m) { return modint{v} * m; }
constexpr friend modint operator/(const ll v, const modint& m) { return modint{v} / m; }
friend std::istream& operator>>(std::istream& is, modint& m)
{
ll v;
return is >> v, m = v, is;
}
friend std::ostream& operator<<(std::ostream& os, const modint& m) { return os << m(); }
constexpr uint operator()() const { return m_val; }
constexpr modint pow(ull n) const
{
modint ans = 1;
for (modint x = *this; n > 0; n >>= 1, x *= x) {
if (n & 1ULL) { ans *= x; }
}
return ans;
}
constexpr modint inv() const { return pow(mod - 2); }
modint sinv() const { return sinv(m_val); }
static modint fact(const uint n)
{
static std::vector<modint> fs{1, 1};
for (uint i = (uint)fs.size(); i <= n; i++) { fs.push_back(fs.back() * i); }
return fs[n];
}
static modint ifact(const uint n)
{
static std::vector<modint> ifs{1, 1};
for (uint i = (uint)ifs.size(); i <= n; i++) { ifs.push_back(ifs.back() * sinv(i)); }
return ifs[n];
}
static modint perm(const int n, const int k) { return k > n or k < 0 ? modint{0} : fact(n) * ifact(n - k); }
static modint comb(const int n, const int k) { return k > n or k < 0 ? modint{0} : fact(n) * ifact(n - k) * ifact(k); }
private:
static constexpr uint norm(const uint x) { return x < mod ? x : x - mod; }
static constexpr uint normll(const ll x) { return norm(uint(x % (ll)mod + (ll)mod)); }
static modint sinv(const uint n)
{
static std::vector<modint> is{1, 1};
for (uint i = (uint)is.size(); i <= n; i++) { is.push_back(-is[mod % i] * (mod / i)); }
return is[n];
}
uint m_val;
};
constexpr modinfo modinfo_1000000007 = {1000000007, 5, 1};
constexpr modinfo modinfo_998244353 = {998244353, 3, 23};
using modint_1000000007 = modint<modinfo_1000000007>;
using modint_998244353 = modint<modinfo_998244353>;
constexpr int popcount(const ull v) { return v ? __builtin_popcountll(v) : 0; }
constexpr int log2p1(const ull v) { return v ? 64 - __builtin_clzll(v) : 0; }
constexpr int lsbp1(const ull v) { return __builtin_ffsll(v); }
constexpr int clog(const ull v) { return v ? log2p1(v - 1) : 0; }
constexpr ull ceil2(const ull v) { return 1ULL << clog(v); }
constexpr ull floor2(const ull v) { return v ? (1ULL << (log2p1(v) - 1)) : 0ULL; }
constexpr bool btest(const ull mask, const int ind) { return (mask >> ind) & 1ULL; }
class range
{
private:
struct itr
{
itr(const int start = 0, const int step = 1) : m_cnt{start}, m_step{step} {}
bool operator!=(const itr& it) const { return m_cnt != it.m_cnt; }
int& operator*() { return m_cnt; }
itr& operator++() { return m_cnt += m_step, *this; }
int m_cnt, m_step;
};
int m_start, m_end, m_step;
public:
range(const int start, const int end, const int step = 1) : m_start{start}, m_end{end}, m_step{step}
{
assert(m_step != 0);
if (m_step > 0) { m_end = m_start + std::max(m_step - 1, m_end - m_start + m_step - 1) / m_step * m_step; }
if (m_step < 0) { m_end = m_start - std::max(-m_step - 1, m_start - m_end - m_step - 1) / (-m_step) * (-m_step); }
}
itr begin() const { return itr{m_start, m_step}; }
itr end() const { return itr{m_end, m_step}; }
};
range rep(const int end, const int step = 1) { return range(0, end, step); }
range per(const int rend, const int step = -1) { return range(rend - 1, -1, step); }
template<typename mint>
class ntt
{
private:
static constexpr const uint& mod = mint::mod;
static constexpr const uint& root = mint::root;
static constexpr const uint& max2p = mint::max2p;
public:
ntt() = delete;
static void trans(std::vector<mint>& as, const int lg, const bool rev)
{
const int N = as.size();
static std::vector<mint> rs;
static std::vector<mint> irs;
if (rs.empty()) {
const mint r = mint(root), ir = r.inv();
rs.resize(max2p + 1), irs.resize(max2p + 1);
rs.back() = -r.pow((mod - 1) >> max2p), irs.back() = -ir.pow((mod - 1) >> max2p);
for (uint i = max2p; i >= 1; i--) { rs[i - 1] = -(rs[i] * rs[i]), irs[i - 1] = -(irs[i] * irs[i]); }
}
const auto rang = (rev ? range(0, lg, 1) : range(lg - 1, -1, -1));
for (const int d : rang) {
const int width = 1 << d;
mint e = 1;
for (int i = 0, j = 1; i < N; i += width * 2, j++) {
for (int l = i, r = i + width; l < i + width; l++, r++) {
const mint x = as[l], y = (rev ? as[r] : as[r] * e);
as[l] = x + y, as[r] = (rev ? (x - y) * e : x - y);
}
e *= (rev ? irs : rs)[lsbp1(j) + 1];
}
}
}
static std::vector<mint> convolute(std::vector<mint> as, std::vector<mint> bs)
{
const int need = as.size() + bs.size() - 1, lg = clog(need), size = 1UL << lg;
as.resize(size), bs.resize(size);
trans(as, lg, false), trans(bs, lg, false);
for (int i = 0; i < size; i++) { as[i] *= bs[i]; }
trans(as, lg, true);
const auto isize = mint{size}.inv();
for (int i = 0; i < need; i++) { as[i] *= isize; }
return std::vector<mint>(as.begin(), as.begin() + need);
}
};
template<typename mint>
class fps
{
private:
static constexpr const uint& mod = mint::mod;
static constexpr const uint& root = mint::root;
static constexpr const uint& max2p = mint::max2p;
// static constexpr modinfo info1 = {469762049, 3, 26};
// static constexpr modinfo info2 = {167772161, 3, 25};
// static constexpr modinfo info3 = {754974721, 11, 24};
// using mint1 = modint<info1>;
// using mint2 = modint<info2>;
// using mint3 = modint<info3>;
// static constexpr mint2 ip1 = mint2{mint1::mod}.inv();
// static constexpr mint3 ip2 = mint3{mint2::mod}.inv();
// static constexpr mint3 ip1p2 = mint3{mint1::mod}.inv() * mint3{mint2::mod}.inv();
// static constexpr mint p1 = mint{mint1::mod};
// static constexpr mint p1p2 = mint{mint1::mod} * mint{mint2::mod};
template<typename submint>
static std::vector<submint> conv(const std::vector<mint>& as, const std::vector<mint>& bs)
{
std::vector<submint> As(as.size()), Bs(bs.size());
for (int i = 0; i < (int)as.size(); i++) { As[i] = as[i](); }
for (int i = 0; i < (int)bs.size(); i++) { Bs[i] = bs[i](); }
return ntt<submint>::convolute(As, Bs);
}
public:
std::vector<mint> convolute(const std::vector<mint>& as, const std::vector<mint>& bs)
{
const int N = as.size() + bs.size() - 1;
if (N <= (1 << max2p)) {
return ntt<mint>::convolute(as, bs);
} else {
}
}
};
const int inf = 1e9;
const int magic = 500; // threshold for sizes to run the naive algo
namespace fft {
const int maxn = 1 << 19;
typedef double ftype;
typedef complex<ftype> point;
vector<point> w;
const ftype pi = acos(-1);
bool initiated = 0;
void init() {
if(!initiated) {
w.resize(maxn);
for(int i = 1; i < maxn; i *= 2) {
for(int j = 0; j < i; j++) {
w[i + j] = polar(ftype(1),2* pi * j / i);
}
}
initiated = 1;
}
}
template<typename T>
void fft(T *in, point *out, int n) {
for (int i = 1, j = 0; i < n; i++) {
int bit = n >> 1;
for (; j & bit; bit >>= 1)
j ^= bit;
j ^= bit;
if (i < j)
swap(in[i], in[j]);
}
for (int len = 2; len <= n; len <<= 1) {
for (int i = 0; i < n; i += len) {
for (int j = 0; j < len / 2; j++) {
point u = in[i+j], v = in[i+j+len/2] * w[j + len];
in[i+j] = u + v;
in[i+j+len/2] = u - v;
}
}
}
for(int i= 0;i<n;i++)
out[i] = in[i];
}
template<typename T>
void mul_slow(vector<T> &a, const vector<T> &b) {
vector<T> res(a.size() + b.size() - 1);
for(size_t i = 0; i < a.size(); i++) {
for(size_t j = 0; j < b.size(); j++) {
res[i + j] += a[i] * b[j];
}
}
a = res;
}
template<typename T>
void mul(vector<T> &a, const vector<T> &b) {
if(a.size() == 0 && b.size() == 0)
{
a = {};
return;
}
if(min(a.size(), b.size()) < magic) {
mul_slow(a, b);
return;
}
using mint = modint_998244353;
int N = a.size();
int M = b.size();
std::vector<mint> as(N), bs(M);
for(int i = 0;i<N;i++)
as[i].set_raw(a[i]);
for(int i = 0;i<M;i++)
bs[i].set_raw(b[i]);
fps<mint> aaaa;
const auto cs = aaaa.convolute(as, bs);
a.resize(cs.size());
for (int i = 0; i < (int)cs.size(); i++) { a[i] = cs[i](); }
}
}
template<typename T>
T bpow(T x, size_t n) {
return n ? n % 2 ? x * bpow(x, n - 1) : bpow(x * x, n / 2) : T(1);
}
template<typename T>
T bpow(T x, size_t n, T m) {
return n ? n % 2 ? x * bpow(x, n - 1, m) % m : bpow(x * x % m, n / 2, m) : T(1);
}
template<typename T>
T gcd(const T &a, const T &b) {
return b == T(0) ? a : gcd(b, a % b);
}
template<typename T>
T nCr(T n, int r) { // runs in O(r)
T res(1);
for(int i = 0; i < r; i++) {
res *= (n - T(i));
res /= (i + 1);
}
return res;
}
template<int m>
struct modular {
int64_t r;
modular() : r(0) {}
modular(int64_t rr) : r(rr) {if(abs(r) >= m) r %= m; if(r < 0) r += m;}
modular inv() const {return bpow(*this, m - 2);}
modular operator * (const modular &t) const {return (r * t.r) % m;}
modular operator / (const modular &t) const {return *this * t.inv();}
modular operator += (const modular &t) {r += t.r; if(r >= m) r -= m; return *this;}
modular operator -= (const modular &t) {r -= t.r; if(r < 0) r += m; return *this;}
modular operator + (const modular &t) const {return modular(*this) += t;}
modular operator - (const modular &t) const {return modular(*this) -= t;}
modular operator *= (const modular &t) {return *this = *this * t;}
modular operator /= (const modular &t) {return *this = *this / t;}
bool operator == (const modular &t) const {return r == t.r;}
bool operator != (const modular &t) const {return r != t.r;}
operator int64_t() const {return r;}
};
template<int T>
istream& operator >> (istream &in, modular<T> &x) {
return in >> x.r;
}
template<typename T>
struct poly {
vector<T> a;
void normalize() { // get rid of leading zeroes
while(!a.empty() && a.back() == T(0)) {
a.pop_back();
}
}
poly(){}
poly(T a0) : a{a0}{normalize();}
poly(vector<T> t) : a(t){normalize();}
poly operator += (const poly &t) {
a.resize(max(a.size(), t.a.size()));
for(size_t i = 0; i < t.a.size(); i++) {
a[i] += t.a[i];
}
normalize();
return *this;
}
poly operator -= (const poly &t) {
a.resize(max(a.size(), t.a.size()));
for(size_t i = 0; i < t.a.size(); i++) {
a[i] -= t.a[i];
}
normalize();
return *this;
}
poly operator + (const poly &t) const {return poly(*this) += t;}
poly operator - (const poly &t) const {return poly(*this) -= t;}
poly mod_xk(size_t k) const { // get same polynomial mod x^k
k = min(k, a.size());
return vector<T>(begin(a), begin(a) + k);
}
poly mul_xk(size_t k) const { // multiply by x^k
poly res(*this);
res.a.insert(begin(res.a), k, 0);
return res;
}
poly div_xk(size_t k) const { // divide by x^k, dropping coefficients
k = min(k, a.size());
return vector<T>(begin(a) + k, end(a));
}
poly substr(size_t l, size_t r) const { // return mod_xk(r).div_xk(l)
l = min(l, a.size());
r = min(r, a.size());
return vector<T>(begin(a) + l, begin(a) + r);
}
poly inv(size_t n) const { // get inverse series mod x^n
assert(!is_zero());
poly ans = a[0].inv();
size_t a = 1;
while(a < n) {
poly C = (ans * mod_xk(2 * a)).substr(a, 2 * a);
ans -= (ans * C).mod_xk(a).mul_xk(a);
a *= 2;
}
return ans.mod_xk(n);
}
poly operator *= (const poly &t) {fft::mul(a, t.a); normalize(); return *this;}
poly operator * (const poly &t) const {return poly(*this) *= t;}
poly reverse(size_t n, bool rev = 0) const { // reverses and leaves only n terms
poly res(*this);
if(rev) { // If rev = 1 then tail goes to head
res.a.resize(max(n, res.a.size()));
}
std::reverse(res.a.begin(), res.a.end());
return res.mod_xk(n);
}
pair<poly, poly> divmod_slow(const poly &b) const { // when divisor or quotient is small
vector<T> A(a);
vector<T> res;
while(A.size() >= b.a.size()) {
res.push_back(A.back() / b.a.back());
if(res.back() != T(0)) {
for(size_t i = 0; i < b.a.size(); i++) {
A[A.size() - i - 1] -= res.back() * b.a[b.a.size() - i - 1];
}
}
A.pop_back();
}
std::reverse(begin(res), end(res));
return {res, A};
}
pair<poly, poly> divmod(const poly &b) const { // returns quotiend and remainder of a mod b
if(deg() < b.deg()) {
return {poly{0}, *this};
}
int d = deg() - b.deg();
if(min(d, b.deg()) < magic) {
return divmod_slow(b);
}
poly D = (reverse(d + 1) * b.reverse(d + 1).inv(d + 1)).mod_xk(d + 1).reverse(d + 1, 1);
return {D, *this - D * b};
}
poly operator / (const poly &t) const {return divmod(t).first;}
poly operator % (const poly &t) const {return divmod(t).second;}
poly operator /= (const poly &t) {return *this = divmod(t).first;}
poly operator %= (const poly &t) {return *this = divmod(t).second;}
poly operator *= (const T &x) {
for(auto &it: a) {
it *= x;
}
normalize();
return *this;
}
poly operator /= (const T &x) {
for(auto &it: a) {
it /= x;
}
normalize();
return *this;
}
poly operator * (const T &x) const {return poly(*this) *= x;}
poly operator / (const T &x) const {return poly(*this) /= x;}
void print() const {
for(auto it: a) {
cout << it << ' ';
}
cout << endl;
}
T eval(T x) const { // evaluates in single point x
T res(0);
for(int i = int(a.size()) - 1; i >= 0; i--) {
res *= x;
res += a[i];
}
return res;
}
T& lead() { // leading coefficient
return a.back();
}
int deg() const { // degree
return a.empty() ? -inf : a.size() - 1;
}
bool is_zero() const { // is polynomial zero
return a.empty();
}
T operator [](int idx) const {
return idx >= (int)a.size() || idx < 0 ? T(0) : a[idx];
}
T& coef(size_t idx) { // mutable reference at coefficient
return a[idx];
}
bool operator == (const poly &t) const {return a == t.a;}
bool operator != (const poly &t) const {return a != t.a;}
poly deriv() { // calculate derivative
vector<T> res;
for(int i = 1; i <= deg(); i++) {
res.push_back(T(i) * a[i]);
}
return res;
}
poly integr() { // calculate integral with C = 0
vector<T> res = {0};
for(int i = 0; i <= deg(); i++) {
res.push_back(a[i] / T(i + 1));
}
return res;
}
size_t leading_xk() const { // Let p(x) = x^k * t(x), return k
if(is_zero()) {
return inf;
}
int res = 0;
while(a[res] == T(0)) {
res++;
}
return res;
}
poly log(size_t n) { // calculate log p(x) mod x^n
assert(a[0] == T(1));
return (deriv().mod_xk(n) * inv(n)).integr().mod_xk(n);
}
poly exp(size_t n) { // calculate exp p(x) mod x^n
if(is_zero()) {
return T(1);
}
assert(a[0] == T(0));
poly ans = T(1);
size_t a = 1;
while(a < n) {
poly C = ans.log(2 * a).div_xk(a) - substr(a, 2 * a);
ans -= (ans * C).mod_xk(a).mul_xk(a);
a *= 2;
}
return ans.mod_xk(n);
}
poly pow_slow(size_t k, size_t n) { // if k is small
return k ? k % 2 ? (*this * pow_slow(k - 1, n)).mod_xk(n) : (*this * *this).mod_xk(n).pow_slow(k / 2, n) : T(1);
}
poly pow(size_t k, size_t n) { // calculate p^k(n) mod x^n
if(k == 0)
{
return {1};
}
if(is_zero()) {
return *this;
}
if(k < magic) {
return pow_slow(k, n);
}
int i = leading_xk();
T j = a[i];
poly t = div_xk(i) / j;
return bpow(j, k) * (t.log(n) * T(k)).exp(n).mul_xk(i * k).mod_xk(n);
}
poly mulx(T x) { // component-wise multiplication with x^k
T cur = 1;
poly res(*this);
for(int i = 0; i <= deg(); i++) {
res.coef(i) *= cur;
cur *= x;
}
return res;
}
poly mulx_sq(T x) { // component-wise multiplication with x^{k^2}
T cur = x;
T total = 1;
T xx = x * x;
poly res(*this);
for(int i = 0; i <= deg(); i++) {
res.coef(i) *= total;
total *= cur;
cur *= xx;
}
return res;
}
vector<T> chirpz_even(T z, int n) { // P(1), P(z^2), P(z^4), ..., P(z^2(n-1))
int m = deg();
if(is_zero()) {
return vector<T>(n, 0);
}
vector<T> vv(m + n);
T zi = z.inv();
T zz = zi * zi;
T cur = zi;
T total = 1;
for(int i = 0; i <= max(n - 1, m); i++) {
if(i <= m) {vv[m - i] = total;}
if(i < n) {vv[m + i] = total;}
total *= cur;
cur *= zz;
}
poly w = (mulx_sq(z) * vv).substr(m, m + n).mulx_sq(z);
vector<T> res(n);
for(int i = 0; i < n; i++) {
res[i] = w[i];
}
return res;
}
vector<T> chirpz(T z, int n) { // P(1), P(z), P(z^2), ..., P(z^(n-1))
auto even = chirpz_even(z, (n + 1) / 2);
auto odd = mulx(z).chirpz_even(z, n / 2);
vector<T> ans(n);
for(int i = 0; i < n / 2; i++) {
ans[2 * i] = even[i];
ans[2 * i + 1] = odd[i];
}
if(n % 2 == 1) {
ans[n - 1] = even.back();
}
return ans;
}
template<typename iter>
vector<T> eval(vector<poly> &tree, int v, iter l, iter r) { // auxiliary evaluation function
if(r - l == 1) {
return {eval(*l)};
} else {
auto m = l + (r - l) / 2;
auto A = (*this % tree[2 * v]).eval(tree, 2 * v, l, m);
auto B = (*this % tree[2 * v + 1]).eval(tree, 2 * v + 1, m, r);
A.insert(end(A), begin(B), end(B));
return A;
}
}
vector<T> eval(vector<T> x) { // evaluate polynomial in (x1, ..., xn)
int n = x.size();
if(is_zero()) {
return vector<T>(n, T(0));
}
vector<poly> tree(4 * n);
build(tree, 1, begin(x), end(x));
return eval(tree, 1, begin(x), end(x));
}
template<typename iter>
poly inter(vector<poly> &tree, int v, iter l, iter r, iter ly, iter ry) { // auxiliary interpolation function
if(r - l == 1) {
return {*ly / a[0]};
} else {
auto m = l + (r - l) / 2;
auto my = ly + (ry - ly) / 2;
auto A = (*this % tree[2 * v]).inter(tree, 2 * v, l, m, ly, my);
auto B = (*this % tree[2 * v + 1]).inter(tree, 2 * v + 1, m, r, my, ry);
return A * tree[2 * v + 1] + B * tree[2 * v];
}
}
};
template<typename T>
poly<T> operator * (const T& a, const poly<T>& b) {
return b * a;
}
template<typename T>
poly<T> xk(int k) { // return x^k
return poly<T>{1}.mul_xk(k);
}
template<typename T>
T resultant(poly<T> a, poly<T> b) { // computes resultant of a and b
if(b.is_zero()) {
return 0;
} else if(b.deg() == 0) {
return bpow(b.lead(), a.deg());
} else {
int pw = a.deg();
a %= b;
pw -= a.deg();
T mul = bpow(b.lead(), pw) * T((b.deg() & a.deg() & 1) ? -1 : 1);
T ans = resultant(b, a);
return ans * mul;
}
}
template<typename iter>
poly<typename iter::value_type> kmul(iter L, iter R) { // computes (x-a1)(x-a2)...(x-an) without building tree
if(R - L == 1) {
return vector<typename iter::value_type>{-*L, 1};
} else {
iter M = L + (R - L) / 2;
return kmul(L, M) * kmul(M, R);
}
}
template<typename T, typename iter>
poly<T> build(vector<poly<T>> &res, int v, iter L, iter R) { // builds evaluation tree for (x-a1)(x-a2)...(x-an)
if(R - L == 1) {
return res[v] = vector<T>{-*L, 1};
} else {
iter M = L + (R - L) / 2;
return res[v] = build(res, 2 * v, L, M) * build(res, 2 * v + 1, M, R);
}
}
template<typename T>
poly<T> inter(vector<T> x, vector<T> y) { // interpolates minimum polynomial from (xi, yi) pairs
int n = x.size();
vector<poly<T>> tree(4 * n);
return build(tree, 1, begin(x), end(x)).deriv().inter(tree, 1, begin(x), end(x), begin(y), end(y));
}
};
using namespace algebra;
typedef modular<mod> base;
typedef poly<base> polyn;
using namespace algebra;
const int MAXN=2e5+10;
vector<base> fac(MAXN);
vector<base> infac(MAXN);
void build_factorial()
{
fac[0]=1;
infac[0]=1;
for(int i=1;i<MAXN;i++)
{
fac[i]=(fac[i-1]*(base)i);
}
infac[MAXN-1] = power(fac[MAXN -1], mod - 2,mod);
for(int i= MAXN-2;i>=0;i--)
{
infac[i] = infac[i+1] *base(i + 1);
}
}
base ncr(int n, int r)
{
if(n<r)
return 0;
return fac[n] * infac[r] * infac[n-r];
}
void solv()
{
build_factorial();
int k, q;
cin>>k>>q;
const int maxn = 1e5 + 10;
polyn x;
x.a.resize(maxn + 1);
for(int i = 0 ;i<=maxn;i+=2)
x.a[i] = ncr(i, i/2) * ncr(i,i/2);
polyn power_4;
power_4.a.resize(maxn + 1);
power_4.a[0] = 1;
for(int i = 1 ;i<=maxn;i++)
power_4.a[i] = power_4.a[i-1] * base(4);
polyn Q;
auto l = x.inv(maxn + 1);
Q = power_4 * l;
Q = Q.mod_xk(maxn + 1);
polyn P(1);
P = P - l;
P = P.mod_xk(maxn + 1);
if(k == 1)
{
polyn A = Q.deriv() *polyn({0,1});
polyn E = A * power_4;
while(q--)
{
int n , p;
cin>>n>>p;
base ans = (E )[n]/ base(power(4,n, mod));
cout<<ans<<endl;
}
}
else
{
polyn D;
D.a.resize(maxn/2 + 1);
for(int i = 0;i<=maxn/2;i++)
D.a[i] = P[2*i];
D = D.pow(k-2,maxn/2 +1);
polyn Z;
Z.a.resize(maxn + 1);
for(int i = 0;i<=maxn/2;i++)
Z.a[2*i] = D[i];
polyn left = ((Q.deriv() *Z).mod_xk(maxn+1) * P).mul_xk(1).mod_xk(maxn+1) ;
polyn right = ((Q *Z).mod_xk(maxn+1) * P.deriv() ).mul_xk(1).mod_xk(maxn+1);
left *= power_4;
right *= power_4;
left = left.mod_xk(maxn+1);
right = right.mod_xk(maxn+1);
while(q--)
{
int n, p;
cin>>n>>p;
base ans =(left[n] + base(p-1) * right[n]) *base(mod_inv(power(4, n , mod), mod));
cout<<ans<<endl;
}
}
}
void solve()
{
int t = 1;
// cin>>t;
while(t--)
{
solv();
}
}
signed main()
{
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cerr.tie(NULL);
auto clk = clock();
// -------------------------------------Code starts here---------------------------------------------------------------------
signed t = 1;
// cin >> t;
for (signed test = 1; test <= t; test++)
{
// cout<<"Case #"<<test<<": ";
// cout<<endl;
solve();
}
// -------------------------------------Code ends here------------------------------------------------------------------
clk = clock() - clk;
#ifndef ONLINE_JUDGE
// cerr << fixed << setprecision(6) << "\nTime: " << ((float)clk) / CLOCKS_PER_SEC << "\n";
#endif
return 0;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given three integers A, B, and C, your task is to print the sum of these three integers.The input contains 3 integers separated by spaces A, B, and C.
Constraints:-
1 <= A, B, C <= 100Print the sum of A, B and C.Sample Input
1 2 3
Sample Output:-
6
Sample Input:-
5 4 2
Sample Output:-
11, I have written this Solution Code: static void simpleSum(int a, int b, int c){
System.out.println(a+b+c);
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given three integers A, B, and C, your task is to print the sum of these three integers.The input contains 3 integers separated by spaces A, B, and C.
Constraints:-
1 <= A, B, C <= 100Print the sum of A, B and C.Sample Input
1 2 3
Sample Output:-
6
Sample Input:-
5 4 2
Sample Output:-
11, I have written this Solution Code: void simpleSum(int a, int b, int c){
cout<<a+b+c;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given three integers A, B, and C, your task is to print the sum of these three integers.The input contains 3 integers separated by spaces A, B, and C.
Constraints:-
1 <= A, B, C <= 100Print the sum of A, B and C.Sample Input
1 2 3
Sample Output:-
6
Sample Input:-
5 4 2
Sample Output:-
11, I have written this Solution Code: x = input()
a, b, c = x.split()
a = int(a)
b = int(b)
c = int(c)
print(a+b+c), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given an array A of size N. The array contains almost distinct elements with some duplicated. You have to print the elements in sorted order which appears more than once.The first line of input contains T, denoting the number of testcases. T testcases follow. Each testcase contains two lines of input. The first line of input contains size of array N. The second line contains N integers separated by spaces.
Constraints:
1 <= T <= 100
1 <= N <= 100000
0 <= A<sub>i</sub> <= 100000
Sum of N over all test cases do not exceed 100000For each test case, in a new line, print the required answer in separate line. If there are no duplicates print -1.Input:
1
9
3 4 5 7 8 1 2 1 3
Output:
1 3
Explanation : both 1,3 appeared 2 times
Input
1
5
1 2 3 4 5
Output
-1
, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws IOException {
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
int t=Integer.parseInt(br.readLine());
String nums[];
int n;
TreeMap<Integer,Integer> tm=null;
int m,p;
boolean flag;
for(int a=0;a<t;++a)
{
n=Integer.parseInt(br.readLine());
nums=br.readLine().split("\\s");
tm=new TreeMap<>();
for(int i=0;i<n;++i)
{
p=Integer.parseInt(nums[i]);
tm.put(p,tm.getOrDefault(p,0)+1);
}
flag=false;
for(Integer i:tm.keySet())
{
m=tm.get(i);
if(m>1)
{
System.out.print(i+" ");
flag=true;
}
}
if(!flag)
{
System.out.println(-1);
}
else System.out.println();
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given an array A of size N. The array contains almost distinct elements with some duplicated. You have to print the elements in sorted order which appears more than once.The first line of input contains T, denoting the number of testcases. T testcases follow. Each testcase contains two lines of input. The first line of input contains size of array N. The second line contains N integers separated by spaces.
Constraints:
1 <= T <= 100
1 <= N <= 100000
0 <= A<sub>i</sub> <= 100000
Sum of N over all test cases do not exceed 100000For each test case, in a new line, print the required answer in separate line. If there are no duplicates print -1.Input:
1
9
3 4 5 7 8 1 2 1 3
Output:
1 3
Explanation : both 1,3 appeared 2 times
Input
1
5
1 2 3 4 5
Output
-1
, I have written this Solution Code: import numpy as np
from collections import defaultdict
t=int(input())
def solve():
d=defaultdict(int)
n=int(input())
a=np.array([input().strip().split()],int).flatten()
for i in a:
d[i]+=1
d=sorted(d.items())
c=0
for i in d:
if(i[1]>1):
c=1
print(i[0],end=" ")
if(c==0):
print(-1,end=" ")
while(t>0):
solve()
print()
t-=1
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given an array A of size N. The array contains almost distinct elements with some duplicated. You have to print the elements in sorted order which appears more than once.The first line of input contains T, denoting the number of testcases. T testcases follow. Each testcase contains two lines of input. The first line of input contains size of array N. The second line contains N integers separated by spaces.
Constraints:
1 <= T <= 100
1 <= N <= 100000
0 <= A<sub>i</sub> <= 100000
Sum of N over all test cases do not exceed 100000For each test case, in a new line, print the required answer in separate line. If there are no duplicates print -1.Input:
1
9
3 4 5 7 8 1 2 1 3
Output:
1 3
Explanation : both 1,3 appeared 2 times
Input
1
5
1 2 3 4 5
Output
-1
, I have written this Solution Code: #include<bits/stdc++.h>
using namespace std;
#define pu push_back
#define fi first
#define se second
#define mp make_pair
#define int long long
#define pii pair<int,int>
#define mm (s+e)/2
#define all(x) x.begin(), x.end()
#define For(i, st, en) for(int i=st; i<en; i++)
#define tr(x) for(auto it=x.begin(); it!=x.end(); it++)
#define fast std::ios::sync_with_stdio(false);cin.tie(NULL);
#define sz 200000
signed main()
{
int t;
cin>>t;
while(t>0)
{
t--;
int n,m;
cin>>n;
map<int,int> A;
for(int i=0;i<n;i++)
{
int a;
cin>>a;
A[a]++;
}
int ch=1;
int cnt=0;
for(auto it:A)
{
if(it.se>1){ cout<<it.fi<<" "; cnt++;}
}
if(cnt==0) cout<<-1;
cout<<endl;
}
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: There are N buildings in a row with different heights H[i] (1 <= i <= N).
You are standing on the left side of the first building .From this position you can see the roof of a building <b>i</b> if no building to the left of the i<sup>th</sup> building has a height greater than or equal to the height of the i<sup>th</sup> building.
You are asked to find the number of buildings whose roofs you can see.The first line contains N denoting number of buildings.
The next line contains N space seperated integers denoting heights of the buildings from left to right.
Constraints
1 <= N <= 100000
1 <= H[i] <= 1000000000000000The output should contain one integer which is the number of buildings whose roofs you can see.Sample input:
5
1 2 2 4 3
Sample output:
3
Explanation:-
the building at index 3 will hide before building at index 2 and building at index 5 will hide before building at index 4
Sample input:
5
1 2 3 4 5
Sample output:
5
, I have written this Solution Code: n=int(input())
a=map(int,input().split())
b=[]
mx=-200000
cnt=0
for i in a:
if i>mx:
cnt+=1
mx=i
print(cnt), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: There are N buildings in a row with different heights H[i] (1 <= i <= N).
You are standing on the left side of the first building .From this position you can see the roof of a building <b>i</b> if no building to the left of the i<sup>th</sup> building has a height greater than or equal to the height of the i<sup>th</sup> building.
You are asked to find the number of buildings whose roofs you can see.The first line contains N denoting number of buildings.
The next line contains N space seperated integers denoting heights of the buildings from left to right.
Constraints
1 <= N <= 100000
1 <= H[i] <= 1000000000000000The output should contain one integer which is the number of buildings whose roofs you can see.Sample input:
5
1 2 2 4 3
Sample output:
3
Explanation:-
the building at index 3 will hide before building at index 2 and building at index 5 will hide before building at index 4
Sample input:
5
1 2 3 4 5
Sample output:
5
, I have written this Solution Code: function numberOfRoofs(arr)
{
let count=1;
let max = arr[0];
for(let i=1;i<arrSize;i++)
{
if(arr[i] > max)
{
count++;
max = arr[i];
}
}
return count;
}
, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: There are N buildings in a row with different heights H[i] (1 <= i <= N).
You are standing on the left side of the first building .From this position you can see the roof of a building <b>i</b> if no building to the left of the i<sup>th</sup> building has a height greater than or equal to the height of the i<sup>th</sup> building.
You are asked to find the number of buildings whose roofs you can see.The first line contains N denoting number of buildings.
The next line contains N space seperated integers denoting heights of the buildings from left to right.
Constraints
1 <= N <= 100000
1 <= H[i] <= 1000000000000000The output should contain one integer which is the number of buildings whose roofs you can see.Sample input:
5
1 2 2 4 3
Sample output:
3
Explanation:-
the building at index 3 will hide before building at index 2 and building at index 5 will hide before building at index 4
Sample input:
5
1 2 3 4 5
Sample output:
5
, I have written this Solution Code: import java.util.*;
import java.io.*;
class Main{
public static void main(String args[]){
Scanner s=new Scanner(System.in);
int n=s.nextInt();
int []a=new int[n];
for(int i=0;i<n;i++){
a[i]=s.nextInt();
}
int count=1;
int max = a[0];
for(int i=1;i<n;i++)
{
if(a[i] > max)
{
count++;
max = a[i];
}
}
System.out.println(count);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of N elements containing only 0 and 1. Find the length of the largest subarray with an equal number of 0's and 1's.
This problem was asked in Amazon.Each test case contains two lines. The first line of each test case is a number N denoting the size of the array and in the next line are N space-separated values of A [].
Constraints:-
1 < = N < = 100000Print the max length of the subarray.Sample input
4
0 1 0 1
Sample Output
4
Sample Input
5
0 0 1 0 0
Sample Output
2, I have written this Solution Code: def maxLen(arr, n):
hash_map = {}
curr_sum = 0
max_len = 0
ending_index = -1
for i in range (0, n):
if(arr[i] == 0):
arr[i] = -1
else:
arr[i] = 1
for i in range (0, n):
curr_sum = curr_sum + arr[i]
if (curr_sum == 0):
max_len = i + 1
ending_index = i
if curr_sum in hash_map:
if max_len < i - hash_map[curr_sum]:
max_len = i - hash_map[curr_sum]
ending_index = i
else:
hash_map[curr_sum] = i
for i in range (0, n):
if(arr[i] == -1):
arr[i] = 0
else:
arr[i] = 1
return max_len
a=input()
a=input().split()
for i in range(len(a)):
a[i]=int(a[i])
print(maxLen(a, len(a))), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of N elements containing only 0 and 1. Find the length of the largest subarray with an equal number of 0's and 1's.
This problem was asked in Amazon.Each test case contains two lines. The first line of each test case is a number N denoting the size of the array and in the next line are N space-separated values of A [].
Constraints:-
1 < = N < = 100000Print the max length of the subarray.Sample input
4
0 1 0 1
Sample Output
4
Sample Input
5
0 0 1 0 0
Sample Output
2, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws IOException{
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
int n = Integer.parseInt(br.readLine());
Map<Integer, Integer> map = new HashMap<>();
int arr[]=new int[n];
int sum=0;
int Largestsubarray=0;
String []b=br.readLine().split(" ");
boolean flag = false;
for(int i=0;i<n;i++){
int temp=Integer.parseInt(b[i]);
if(temp==0)
arr[i]=1;
else
arr[i]=-1;
sum=sum+arr[i];
if(sum == 0) {
Largestsubarray = i+1;
continue;
}
if(map.containsKey(sum)){
flag=true;
int CurrentLargestSubarray = i-map.get(sum);
if(CurrentLargestSubarray > Largestsubarray) {
Largestsubarray = CurrentLargestSubarray;
}
}
else
map.put(sum,i);
}
if(!flag)
System.out.print(-1);
else
System.out.print(Largestsubarray);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of N elements containing only 0 and 1. Find the length of the largest subarray with an equal number of 0's and 1's.
This problem was asked in Amazon.Each test case contains two lines. The first line of each test case is a number N denoting the size of the array and in the next line are N space-separated values of A [].
Constraints:-
1 < = N < = 100000Print the max length of the subarray.Sample input
4
0 1 0 1
Sample Output
4
Sample Input
5
0 0 1 0 0
Sample Output
2, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main(){
int n,k;
cin>>n;
k=0;
int a[n];
for(int i=0;i<n;i++)
{
cin>>a[i];
if(a[i]==0){
a[i]=-1;}
}
int sum=0;
int ans=0;
unordered_map<int,int> m;
for(int i=0;i<n;i++){
sum+=a[i];
if(sum==k){ans=i+1;}
if(m.find(sum-k)!=m.end()){
ans=max(ans,i-m[sum-k]);
}
if(m.find(sum)==m.end()){m[sum]=i;}
}
cout<<ans;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given three integers A, B and C, which are the side lengths of a triangle in some order. Print "Yes" if this triangle is right-angled, otherwise print "No".The input consists of a single line containing three space-separated integers A, B and C.
Constraints:
1 ≤ A, B, C ≤ 1000Print a single word "Yes" if the given triangle is right-angled, otherwise print "No".
Note:
The quotation marks are only for clarity, you should not print them in output.
The output is case-sensitive.Sample Input 1:
3 4 5
Sample Output 1:
Yes
Sample Input 2:
10 9 10
Sample Output 2:
No
Sample Input 3:
5 4 3
Sample Output 3:
Yes, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws Exception{
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
BufferedWriter bw = new BufferedWriter(new OutputStreamWriter(System.out));
String[] inp = br.readLine().split(" ");
int a = Integer.parseInt(inp[0]);
int b = Integer.parseInt(inp[1]);
int c = Integer.parseInt(inp[2]);
if(a * a == b * b + c * c)
bw.write("Yes"+"\n");
else if(b * b == a * a + c * c)
bw.write("Yes"+"\n");
else if(c * c == b * b + a * a)
bw.write("Yes"+"\n");
else
bw.write("No"+"\n");
bw.flush();
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given three integers A, B and C, which are the side lengths of a triangle in some order. Print "Yes" if this triangle is right-angled, otherwise print "No".The input consists of a single line containing three space-separated integers A, B and C.
Constraints:
1 ≤ A, B, C ≤ 1000Print a single word "Yes" if the given triangle is right-angled, otherwise print "No".
Note:
The quotation marks are only for clarity, you should not print them in output.
The output is case-sensitive.Sample Input 1:
3 4 5
Sample Output 1:
Yes
Sample Input 2:
10 9 10
Sample Output 2:
No
Sample Input 3:
5 4 3
Sample Output 3:
Yes, I have written this Solution Code: a,b,c=map(int,input().split())
if(a*a+b*b==c*c or a*a+c*c==b*b or b*b+c*c==a*a):
print("Yes")
else:
print("No"), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given three integers A, B and C, which are the side lengths of a triangle in some order. Print "Yes" if this triangle is right-angled, otherwise print "No".The input consists of a single line containing three space-separated integers A, B and C.
Constraints:
1 ≤ A, B, C ≤ 1000Print a single word "Yes" if the given triangle is right-angled, otherwise print "No".
Note:
The quotation marks are only for clarity, you should not print them in output.
The output is case-sensitive.Sample Input 1:
3 4 5
Sample Output 1:
Yes
Sample Input 2:
10 9 10
Sample Output 2:
No
Sample Input 3:
5 4 3
Sample Output 3:
Yes, I have written this Solution Code:
// #pragma GCC optimize("Ofast")
// #pragma GCC target("avx,avx2,fma")
#include<bits/stdc++.h>
#include<ext/pb_ds/assoc_container.hpp>
#include<ext/pb_ds/tree_policy.hpp>
#define pi 3.141592653589793238
#define int long long
#define ll long long
#define ld long double
using namespace __gnu_pbds;
using namespace std;
template <typename T>
using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
mt19937 rnd(chrono::high_resolution_clock::now().time_since_epoch().count());
long long powm(long long a, long long b,long long mod) {
long long res = 1;
while (b > 0) {
if (b & 1)
res = res * a %mod;
a = a * a %mod;
b >>= 1;
}
return res;
}
ll gcd(ll a, ll b)
{
if (b == 0)
return a;
return gcd(b, a % b);
}
signed main()
{
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(0);
#ifndef ONLINE_JUDGE
if(fopen("input.txt","r"))
{
freopen("input.txt","r",stdin);
freopen("output.txt","w",stdout);
}
#endif
int a[3];
for(int i=0;i<3;i++)
cin>>a[i];
sort(a,a+3);
if(a[0]*a[0]+a[1]*a[1]==a[2]*a[2])
cout<<"Yes";
else
cout<<"No";
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given the length of the radius of curvature and the type of mirror, calculate its focal length.
<b>Note:</b> '(' represents a convex mirror while ')' represents a concave mirror<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>focal_length()</b> that takes the integer R(radius of curvature) and the character Mirror (type of mirror) as parameter
<b>Constraints:-</b>
1 <= R <= 100
Mirror will be either of '('(convex type) or ')'(concave type)Return the <b>floor value of focal length.</b>Sample Input:-
9 (
Sample Output:-
4
Sample Input:-
9 )
Sample Output:-
-5, I have written this Solution Code: static int focal_length(int R, char Mirror)
{
int f=R/2;
if((R%2==1) && Mirror==')'){f++;}
if(Mirror == ')'){f=-f;}
return f;
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given the length of the radius of curvature and the type of mirror, calculate its focal length.
<b>Note:</b> '(' represents a convex mirror while ')' represents a concave mirror<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>focal_length()</b> that takes the integer R(radius of curvature) and the character Mirror (type of mirror) as parameter
<b>Constraints:-</b>
1 <= R <= 100
Mirror will be either of '('(convex type) or ')'(concave type)Return the <b>floor value of focal length.</b>Sample Input:-
9 (
Sample Output:-
4
Sample Input:-
9 )
Sample Output:-
-5, I have written this Solution Code:
int focal_length(int R, char Mirror)
{
int f=R/2;
if((R&1) && Mirror==')'){f++;}
if(Mirror == ')'){f=-f;}
return f;
}, In this Programming Language: C, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given the length of the radius of curvature and the type of mirror, calculate its focal length.
<b>Note:</b> '(' represents a convex mirror while ')' represents a concave mirror<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>focal_length()</b> that takes the integer R(radius of curvature) and the character Mirror (type of mirror) as parameter
<b>Constraints:-</b>
1 <= R <= 100
Mirror will be either of '('(convex type) or ')'(concave type)Return the <b>floor value of focal length.</b>Sample Input:-
9 (
Sample Output:-
4
Sample Input:-
9 )
Sample Output:-
-5, I have written this Solution Code:
int focal_length(int R, char Mirror)
{
int f=R/2;
if((R&1) && Mirror==')'){f++;}
if(Mirror == ')'){f=-f;}
return f;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given the length of the radius of curvature and the type of mirror, calculate its focal length.
<b>Note:</b> '(' represents a convex mirror while ')' represents a concave mirror<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>focal_length()</b> that takes the integer R(radius of curvature) and the character Mirror (type of mirror) as parameter
<b>Constraints:-</b>
1 <= R <= 100
Mirror will be either of '('(convex type) or ')'(concave type)Return the <b>floor value of focal length.</b>Sample Input:-
9 (
Sample Output:-
4
Sample Input:-
9 )
Sample Output:-
-5, I have written this Solution Code: def focal_length(R,Mirror):
f=R/2;
if(Mirror == ')'):
f=-f
if R%2==1:
f=f-1
return int(f)
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a 100*100 chessboard. A knight is placed on cell (i, j). You have to find the number of blocks on the chessboard that the knight can be at in exactly N moves.Input contains three integers i j N.
Constraints
1 <= i, j <= 100
1 <= N <= 10Output a single integer, the number of blocks on the chessboard that the knight can be at in exactly N moves.sample input 1
1 1 1
sample output 1
2
sample input 2
50 50 1
sample input 2
8, I have written this Solution Code: import java.io.* ;
import java.util.* ;
public class Main
{
static int arr[][];
static int visited[][];
static int length;
static int check(int x,int y)
{
if(x<1 || y<1)
return 0;
if(x>100 || y>100)
return 0;
return 1;
}
static void find(int i,int j , int n)
{
int X[]={i+2,i+2,i+1,i+1,i-1,i-1,i-2,i-2};
int Y[]={j+1,j-1,j+2,j-2,j+2,j-2,j+1,j-1};
if(n==1)
{
for(int a=0;a<8;a++)
{
if(check(X[a],Y[a])==1)
{
arr[X[a]][Y[a]]=1;
}
}
return ;
}
if(n%2==0)
arr[i][j]=1;
else
for(int a=0;a<8;a++)
{
if(check(X[a],Y[a])==1)
{
arr[X[a]][Y[a]]=1;
}
}
if(visited[i][j]>=n)
return ;
visited[i][j]=n;
for(int a=0;a<8;a++)
{
if(check(X[a],Y[a])==1)
{
find(X[a],Y[a],n-1);
}
}
}
public static void main(String args[]) throws IOException
{
BufferedReader bf=new BufferedReader(new InputStreamReader(System.in));
String str[]=bf.readLine().split(" ");
int i1=Integer.parseInt(str[0]);
int j1=Integer.parseInt(str[1]);
int n=Integer.parseInt(str[2]);
length=n;
arr=new int[100+1][100+1];
visited=new int[100+1][100+1];
find(i1,j1,n);
int count=0;
for(int i=1;i<=100;i++)
{
for(int j=1;j<=100;j++)
{
if(arr[i][j]==1)
count++;
}
}
System.out.println(count);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a 100*100 chessboard. A knight is placed on cell (i, j). You have to find the number of blocks on the chessboard that the knight can be at in exactly N moves.Input contains three integers i j N.
Constraints
1 <= i, j <= 100
1 <= N <= 10Output a single integer, the number of blocks on the chessboard that the knight can be at in exactly N moves.sample input 1
1 1 1
sample output 1
2
sample input 2
50 50 1
sample input 2
8, I have written this Solution Code: def get_next_positions(pos):
moves = [
(2,1), (2,-1),(-2,1),(-2,-1),(1,2),(-1,2),(1,-2),(-1,-2)
]
i, j = pos
positions = set()
for (x, y) in moves:
pos_x = i + x
pos_y = j + y
if(1 <= pos_x) and (pos_x <= 100) and (1 <= pos_y) and (pos_y <= 100):
positions.add((pos_x, pos_y))
return positions
i, j, n = [int(x) for x in input().split()]
curr_positions = set([(i, j)])
for i in range(n):
new_positions = set()
for pos in curr_positions:
new_positions |= get_next_positions(pos)
curr_positions = new_positions
print(len(curr_positions)), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a 100*100 chessboard. A knight is placed on cell (i, j). You have to find the number of blocks on the chessboard that the knight can be at in exactly N moves.Input contains three integers i j N.
Constraints
1 <= i, j <= 100
1 <= N <= 10Output a single integer, the number of blocks on the chessboard that the knight can be at in exactly N moves.sample input 1
1 1 1
sample output 1
2
sample input 2
50 50 1
sample input 2
8, I have written this Solution Code:
// author-Shivam gupta
#include <bits/stdc++.h>
using namespace std;
#define MEM(a, b) memset(a, (b), sizeof(a))
#define FOREACH(it, l) for (auto it = l.begin(); it != l.end(); it++)
#define IN(A, B, C) assert( B <= A && A <= C)
#define MP make_pair
#define FOR(i,a) for(int i=0;i<a;i++)
#define FOR1(i,j,a) for(int i=j;i<a;i++)
#define EB emplace_back
#define INF (int)1e9
#define EPS 1e-9
#define PI 3.1415926535897932384626433832795
#define MOD 1000000007
#define read(type) readInt<type>()
#define max1 1000001
#define out(x) cout<<x<<'\n'
#define out1(x) cout<<x<<" "
#define END cout<<'\n'
const double pi=acos(-1.0);
typedef pair<int, int> PII;
typedef vector<int> VI;
typedef vector<string> VS;
typedef vector<PII> VII;
typedef vector<VI> VVI;
typedef map<int,int> MPII;
typedef set<int> SETI;
typedef multiset<int> MSETI;
typedef long int li;
typedef unsigned long int uli;
typedef long long int ll;
typedef unsigned long long int ull;
bool isPowerOfTwo (int x)
{
/* First x in the below expression is
for the case when x is 0 */
return x && (!(x&(x-1)));
}
void fast(){
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
}
ll power(ll x, ll y, ll p)
{
ll res = 1; // Initialize result
x = x % p; // Update x if it is more than or
// equal to p
while (y > 0)
{
// If y is odd, multiply x with result
if (y & 1)
res = (res*x) % p;
// y must be even now
y = y>>1; // y = y/2
x = (x*x) % p;
}
return res;
}
// Returns n^(-1) mod p
ll modInverse(ll n, ll p)
{
return power(n, p-2, p);
}
// Returns nCr % p using Fermat's little
// theorem.
ll ncr(ll n, ll r,ll p)
{
// Base case
if (r==0)
return 1;
// Fill factorial array so that we
// can find all factorial of r, n
// and n-r
ll fac[n+1];
fac[0] = 1;
for (ll i=1 ; i<=n; i++)
fac[i] = fac[i-1]*i%p;
return (fac[n]* modInverse(fac[r], p) % p *
modInverse(fac[n-r], p) % p) % p;
}
set<pair<int,int>> s;
bool dp[101][101][11];
void solve(int i, int j, int n){
if(n==0){
s.insert(make_pair(i,j));
return ;}
int dx[] = {-2, -1, 1, 2, -2, -1, 1, 2};
int dy[] = {-1, -2, -2, -1, 1, 2, 2, 1};
int cnt=0;
for(int k=0;k<8;k++){
if((i+dx[k])>=1 && (i+dx[k])<=100){
if((j+dy[k])>=1 && (j+dy[k])<=100){
if(dp[i+dx[k]][j+dy[k]][n]==false){
dp[i+dx[k]][j+dy[k]][n]=true;
solve(i+dx[k],j+dy[k],n-1);}
}
}
}
}
int main(){
int a,b,c;
cin>>a>>b>>c;
FOR(i,101){
FOR(k,101){
FOR(j,11){
dp[i][k][j]=false;}}}
dp[a][b][c]=true;
solve(a,b,c);
cout<<s.size();
}
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: :b.dS^^-. ""^^" $b T d$$$s$$$$b __..--""$ $;
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: $$$$$pp..__ `j$$$$$$^$$$$$b. d....ggppTSSS$$;
: $$$$SP """t :$$$$ $$$$$$$b. d$b `TSS$;
: \:$$SP _.gd$$P_d$$$$$ $$$$$$$$$bd$P' .dSPd;
: \"^S "^T$$$$$$$$$$ $$$$$SS$$$$b. dSS'd$;
$ $. "-.__.gd$$$$$$$SP:S$$P TSS$$$$$bssS^".d$$:
:$ $$b. ""^^T$$$$SP' :S$P TSSSP^^"" .d$$$$:
:$ :$$$P "^SP' :S; .^"`. $$$$$$$:
$; :$$$ "-. :S; .-" \ :$$$$$$:
.$ : $$$; : `.:S;.' ; $$$$$$:;
.P :S $$$ ; `^' :$$$$$:;
.P S;.d$$; : -' $$$$$:;
$ :SS$$$$ ; __....---- --...____ :$$$ :S
$ $SSS$$; : ; d$$$$$$$pppqqqq$$$$$$L; :$$$ SS
: :SSSSS$$ ; : \ "^T$$$$$$$$$$$$$P' .': $PT$ SS;
$SP^"^TSP\ : \ "-. """"""""""" .-" ; / $ SSSb.
:S S \ "--...___..--" / : / :gSSSSSb.
T bug T \ `. _____ / ; /
` \ : "==="""""""""==="" : /
`: ;'
"-. .-"
""--.. ..--""
*/
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a matrix Mat of m rows and n columns. The matrix is boolean so the elements of the matrix can only be either 0 or 1.
Now, if any row of the matrix contains a 1, then you need to fill that whole row with 1. After doing the mentioned operation, you need to print the modified matrix.The first line of input contains T denoting the number of test cases. T test cases follow.
The first line of each test case contains m and n denotes the number of rows and a number of columns.
Then next m lines contain n elements denoting the elements of the matrix.
Constraints:
1 ≤ T ≤ 20
1 ≤ m, n ≤ 700
Mat[I][j] ∈ {0,1}For each testcase, in a new line, print the modified matrix.Input:
1
5 4
1 0 0 0
0 0 0 0
0 1 0 0
0 0 0 0
0 0 0 1
Output:
1 1 1 1
0 0 0 0
1 1 1 1
0 0 0 0
1 1 1 1
Explanation:
Rows = 5 and columns = 4
The given matrix is
1 0 0 0
0 0 0 0
0 1 0 0
0 0 0 0
0 0 0 1
Evidently, the first row contains a 1 so fill the whole row with 1. The third row also contains a 1 so that row will be filled too. Finally, the last row contains a 1 and therefore it needs to be filled with 1 too.
The final matrix is
1 1 1 1
0 0 0 0
1 1 1 1
0 0 0 0
1 1 1 1, I have written this Solution Code: t=int(input())
while t!=0:
m,n=input().split()
m,n=int(m),int(n)
for i in range(m):
arr=input().strip()
if '1' in arr:
arr='1 '*n
else:
arr='0 '*n
print(arr)
t-=1, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a matrix Mat of m rows and n columns. The matrix is boolean so the elements of the matrix can only be either 0 or 1.
Now, if any row of the matrix contains a 1, then you need to fill that whole row with 1. After doing the mentioned operation, you need to print the modified matrix.The first line of input contains T denoting the number of test cases. T test cases follow.
The first line of each test case contains m and n denotes the number of rows and a number of columns.
Then next m lines contain n elements denoting the elements of the matrix.
Constraints:
1 ≤ T ≤ 20
1 ≤ m, n ≤ 700
Mat[I][j] ∈ {0,1}For each testcase, in a new line, print the modified matrix.Input:
1
5 4
1 0 0 0
0 0 0 0
0 1 0 0
0 0 0 0
0 0 0 1
Output:
1 1 1 1
0 0 0 0
1 1 1 1
0 0 0 0
1 1 1 1
Explanation:
Rows = 5 and columns = 4
The given matrix is
1 0 0 0
0 0 0 0
0 1 0 0
0 0 0 0
0 0 0 1
Evidently, the first row contains a 1 so fill the whole row with 1. The third row also contains a 1 so that row will be filled too. Finally, the last row contains a 1 and therefore it needs to be filled with 1 too.
The final matrix is
1 1 1 1
0 0 0 0
1 1 1 1
0 0 0 0
1 1 1 1, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define N 1000
int a[N][N];
// Driver code
int main()
{
int t;
cin>>t;
while(t--){
int n,m;
cin>>n>>m;
bool b[n];
for(int i=0;i<n;i++){
b[i]=false;
}
for(int i=0;i<n;i++){
for(int j=0;j<m;j++){
cin>>a[i][j];
if(a[i][j]==1){
b[i]=true;
}
}
}
for(int i=0;i<n;i++){
if(b[i]){
for(int j=0;j<m;j++){
cout<<1<<" ";
}}
else{
for(int j=0;j<m;j++){
cout<<0<<" ";
}
}
cout<<endl;
}
}}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a matrix Mat of m rows and n columns. The matrix is boolean so the elements of the matrix can only be either 0 or 1.
Now, if any row of the matrix contains a 1, then you need to fill that whole row with 1. After doing the mentioned operation, you need to print the modified matrix.The first line of input contains T denoting the number of test cases. T test cases follow.
The first line of each test case contains m and n denotes the number of rows and a number of columns.
Then next m lines contain n elements denoting the elements of the matrix.
Constraints:
1 ≤ T ≤ 20
1 ≤ m, n ≤ 700
Mat[I][j] ∈ {0,1}For each testcase, in a new line, print the modified matrix.Input:
1
5 4
1 0 0 0
0 0 0 0
0 1 0 0
0 0 0 0
0 0 0 1
Output:
1 1 1 1
0 0 0 0
1 1 1 1
0 0 0 0
1 1 1 1
Explanation:
Rows = 5 and columns = 4
The given matrix is
1 0 0 0
0 0 0 0
0 1 0 0
0 0 0 0
0 0 0 1
Evidently, the first row contains a 1 so fill the whole row with 1. The third row also contains a 1 so that row will be filled too. Finally, the last row contains a 1 and therefore it needs to be filled with 1 too.
The final matrix is
1 1 1 1
0 0 0 0
1 1 1 1
0 0 0 0
1 1 1 1, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main(String[] args) throws Exception{
InputStreamReader isr = new InputStreamReader(System.in);
BufferedReader bf = new BufferedReader(isr);
int t = Integer.parseInt(bf.readLine());
while (t-- > 0){
String inputs[] = bf.readLine().split(" ");
int m = Integer.parseInt(inputs[0]);
int n = Integer.parseInt(inputs[1]);
String[] matrix = new String[m];
for(int i=0; i<m; i++){
matrix[i] = bf.readLine();
}
StringBuffer ones = new StringBuffer("");
StringBuffer zeros = new StringBuffer("");
for(int i=0; i<n; i++){
ones.append("1 ");
zeros.append("0 ");
}
for(int i=0; i<m; i++){
if(matrix[i].contains("1")){
System.out.println(ones);
}else{
System.out.println(zeros);
}
}
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: To seek revenge on Midgard, Loki devises a plan to torture the humans by making them take part in one of his silly games. He makes N people sit in a circle. He says he will kill every kth person sitting in the circle, starting from 1st person.
Loki performs his revenge prank until and unless 1 survivor remains.
What is the initial position of the survivor, if the indexing is clockwise?The first line of input contains a single integer T. The next T line of input contains Two space separated integers each containing value of N and k.
<b>Constraints:</b>
1 <= T <= 100
1 <= k, N <= 20Print the initial position of the survivor.Sample Input:
2
3 2
5 3
Sample Output
3
4
Explanation:
Test case 1: There are 3 people so skipping 1 person i.e 1st person 2nd person will be killed in next step 3rd person will be skipped and 1st person will be killed. Thus the safe position is 3.
Test case 2: 2 people i.e 1and 2 are skipped and person 3 will be killed in next step 4 and 5 will be skipped and 1st person will be killed next step 2 and 4 will be skipped and 5th person will be killed next step first 2 will be skipped then 4 will be skipped and so coming back to 2 therefore person 2 will be killed. Thus the safe position is 4., I have written this Solution Code: t=int(input())
while t>0:
n,k=map(int,input().split())
s=set()
remove=0
for i in range(1,n+1):
s.add(i)
i=1
while True:
if len(s)==1:
break
if i in s:
remove+=1
if remove==k:
if i in s:
s.remove(i)
remove=0
i+=1
if i>n:
i=1
print(*s)
t-=1, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: To seek revenge on Midgard, Loki devises a plan to torture the humans by making them take part in one of his silly games. He makes N people sit in a circle. He says he will kill every kth person sitting in the circle, starting from 1st person.
Loki performs his revenge prank until and unless 1 survivor remains.
What is the initial position of the survivor, if the indexing is clockwise?The first line of input contains a single integer T. The next T line of input contains Two space separated integers each containing value of N and k.
<b>Constraints:</b>
1 <= T <= 100
1 <= k, N <= 20Print the initial position of the survivor.Sample Input:
2
3 2
5 3
Sample Output
3
4
Explanation:
Test case 1: There are 3 people so skipping 1 person i.e 1st person 2nd person will be killed in next step 3rd person will be skipped and 1st person will be killed. Thus the safe position is 3.
Test case 2: 2 people i.e 1and 2 are skipped and person 3 will be killed in next step 4 and 5 will be skipped and 1st person will be killed next step 2 and 4 will be skipped and 5th person will be killed next step first 2 will be skipped then 4 will be skipped and so coming back to 2 therefore person 2 will be killed. Thus the safe position is 4., I have written this Solution Code: import java.io.*;
import java.util.*;
import java.lang.*;
class Main{
public static void main (String[] args)throws IOException
{
BufferedReader read = new BufferedReader(new InputStreamReader(System.in));
int t = Integer.parseInt(read.readLine().trim());
while(t-->0)
{
String str[] = read.readLine().trim().split(" ");
int n = Integer.parseInt(str[0]);
int k = Integer.parseInt(str[1]);
System.out.println(safe_Position(n,k));
}
}
public static int safe_Position(int n, int k)
{
if (n == 1) //base case
return 1;
else
/* The position returned by safe_Position(n - 1, k) is adjusted because the
recursive call safe_Position(n - 1, k) considers the original position
k%n + 1 as position 1 */
return (safe_Position(n - 1, k) + k-1) % n + 1; //recursion
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: The stock span problem is a financial problem where we have a series of n daily price quotes for a stock and we need to calculate the span of stock’s price for all n days.
The span Si of the stock’s price on a given day i is defined as the maximum number of consecutive days just before the given day(including), for which the price of the stock on the current day is less than or equal to its price on the given day.
For example, if an array of 7 days prices is given as {100, 80, 60, 70, 60, 75, 85}, then the span values for corresponding 7 days are {1, 1, 1, 2, 1, 4, 6}.
Explanation to the given example:
On 0th day only the current day where we find that stock price is less than or equal to it so for 1 consecutive day(current day) this happens.
On 1st day only the current day where we find that stock price is less than or equal to it so for 1 consecutive day(current day) this happens.
On 2nd day only the current day where we find that stock price is less than or equal to it so for 1 consecutive day(current day) this happens.
Now on 3rd day we observe that stock price for last two consecutive days is less than or equal to current day i.e. (60, 70) thus count is 2
On 4th day only the current day where we find that stock price is less than or equal to it so for 1 consecutive day(current day) this happens.
On 5th day we observe that stock price for last four consecutive days is less than or equal to current day i.e. (60, 70, 60, 75) thus count is 4.
On 6th day we observe that stock price for last six consecutive days is less than or equal to current day i.e. (80, 60, 70, 60, 75, 85) thus count is 6.The first line of each test case is N, N is the size of the array. The second line of each test case contains N input A[i].
1 ≤ N ≤ 100000
1 ≤ A[i] ≤ 100000For each test case, print the span values for all days.Input
7
100 80 60 70 60 75 85
Output
1 1 1 2 1 4 6
Input
6
10 4 5 90 120 80
Output
1 1 2 4 5 1
Explanation:
Test case 1:
On 0th day only the current day where we find that stock price is less than or equal to it so for 1 consecutive day(current day) this happens.
On 1st day only the current day where we find that stock price is less than or equal to it so for 1 consecutive day(current day) this happens.
On 2nd day only the current day where we find that stock price is less than or equal to it so for 1 consecutive day(current day) this happens.
Now on 3rd day we observe that stock price for last two consecutive days is less than or equal to current day i.e. (60, 70) thus count is 2
On 4th day only the current day where we find that stock price is less than or equal to it so for 1 consecutive day(current day) this happens.
On 5th day we observe that stock price for last four consecutive days is less than or equal to current day i.e. (60, 70, 60, 75) thus count is 4.
On 6th day we observe that stock price for last six consecutive days is less than or equal to current day i.e. (80, 60, 70, 60, 75, 85) thus count is 6., I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws IOException{
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int size = Integer.parseInt(br.readLine());
String st = br.readLine();
String[] stArr = st.trim().split(" ");
int[] arr = new int[size];
for(int i=0; i<size; i++){
arr[i] = Integer.parseInt(stArr[i]);
}
Stack<Integer> stack = new Stack<>();
StringBuilder sb=new StringBuilder("");
for(int i=0; i<size; i++){
int count = 0;
if(stack.isEmpty()){
stack.push(i);
sb.append("1 ");
} else if(arr[stack.peek()] > arr[i]){
stack.push(i);
sb.append("1 ");
} else{
while(!stack.isEmpty() && arr[stack.peek()] <= arr[i]){
stack.pop();
}
if(stack.isEmpty()){
count = i+1;
} else{
count = i - stack.peek();
}
sb.append(count+" ");
stack.push(i);
}
}
System.out.print(sb);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: The stock span problem is a financial problem where we have a series of n daily price quotes for a stock and we need to calculate the span of stock’s price for all n days.
The span Si of the stock’s price on a given day i is defined as the maximum number of consecutive days just before the given day(including), for which the price of the stock on the current day is less than or equal to its price on the given day.
For example, if an array of 7 days prices is given as {100, 80, 60, 70, 60, 75, 85}, then the span values for corresponding 7 days are {1, 1, 1, 2, 1, 4, 6}.
Explanation to the given example:
On 0th day only the current day where we find that stock price is less than or equal to it so for 1 consecutive day(current day) this happens.
On 1st day only the current day where we find that stock price is less than or equal to it so for 1 consecutive day(current day) this happens.
On 2nd day only the current day where we find that stock price is less than or equal to it so for 1 consecutive day(current day) this happens.
Now on 3rd day we observe that stock price for last two consecutive days is less than or equal to current day i.e. (60, 70) thus count is 2
On 4th day only the current day where we find that stock price is less than or equal to it so for 1 consecutive day(current day) this happens.
On 5th day we observe that stock price for last four consecutive days is less than or equal to current day i.e. (60, 70, 60, 75) thus count is 4.
On 6th day we observe that stock price for last six consecutive days is less than or equal to current day i.e. (80, 60, 70, 60, 75, 85) thus count is 6.The first line of each test case is N, N is the size of the array. The second line of each test case contains N input A[i].
1 ≤ N ≤ 100000
1 ≤ A[i] ≤ 100000For each test case, print the span values for all days.Input
7
100 80 60 70 60 75 85
Output
1 1 1 2 1 4 6
Input
6
10 4 5 90 120 80
Output
1 1 2 4 5 1
Explanation:
Test case 1:
On 0th day only the current day where we find that stock price is less than or equal to it so for 1 consecutive day(current day) this happens.
On 1st day only the current day where we find that stock price is less than or equal to it so for 1 consecutive day(current day) this happens.
On 2nd day only the current day where we find that stock price is less than or equal to it so for 1 consecutive day(current day) this happens.
Now on 3rd day we observe that stock price for last two consecutive days is less than or equal to current day i.e. (60, 70) thus count is 2
On 4th day only the current day where we find that stock price is less than or equal to it so for 1 consecutive day(current day) this happens.
On 5th day we observe that stock price for last four consecutive days is less than or equal to current day i.e. (60, 70, 60, 75) thus count is 4.
On 6th day we observe that stock price for last six consecutive days is less than or equal to current day i.e. (80, 60, 70, 60, 75, 85) thus count is 6., I have written this Solution Code: n = int(input())
lst = list(map(int, input().strip().split()))
S = [0]*n
st = []
st.append(0)
for i in range(n):
while( len(st) > 0 and lst[st[-1]] <= lst[i]):
st.pop()
S[i] = i + 1 if len(st) <= 0 else (i - st[-1])
st.append(i)
for i in range(0, n):
print (S[i], end =" "), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: The stock span problem is a financial problem where we have a series of n daily price quotes for a stock and we need to calculate the span of stock’s price for all n days.
The span Si of the stock’s price on a given day i is defined as the maximum number of consecutive days just before the given day(including), for which the price of the stock on the current day is less than or equal to its price on the given day.
For example, if an array of 7 days prices is given as {100, 80, 60, 70, 60, 75, 85}, then the span values for corresponding 7 days are {1, 1, 1, 2, 1, 4, 6}.
Explanation to the given example:
On 0th day only the current day where we find that stock price is less than or equal to it so for 1 consecutive day(current day) this happens.
On 1st day only the current day where we find that stock price is less than or equal to it so for 1 consecutive day(current day) this happens.
On 2nd day only the current day where we find that stock price is less than or equal to it so for 1 consecutive day(current day) this happens.
Now on 3rd day we observe that stock price for last two consecutive days is less than or equal to current day i.e. (60, 70) thus count is 2
On 4th day only the current day where we find that stock price is less than or equal to it so for 1 consecutive day(current day) this happens.
On 5th day we observe that stock price for last four consecutive days is less than or equal to current day i.e. (60, 70, 60, 75) thus count is 4.
On 6th day we observe that stock price for last six consecutive days is less than or equal to current day i.e. (80, 60, 70, 60, 75, 85) thus count is 6.The first line of each test case is N, N is the size of the array. The second line of each test case contains N input A[i].
1 ≤ N ≤ 100000
1 ≤ A[i] ≤ 100000For each test case, print the span values for all days.Input
7
100 80 60 70 60 75 85
Output
1 1 1 2 1 4 6
Input
6
10 4 5 90 120 80
Output
1 1 2 4 5 1
Explanation:
Test case 1:
On 0th day only the current day where we find that stock price is less than or equal to it so for 1 consecutive day(current day) this happens.
On 1st day only the current day where we find that stock price is less than or equal to it so for 1 consecutive day(current day) this happens.
On 2nd day only the current day where we find that stock price is less than or equal to it so for 1 consecutive day(current day) this happens.
Now on 3rd day we observe that stock price for last two consecutive days is less than or equal to current day i.e. (60, 70) thus count is 2
On 4th day only the current day where we find that stock price is less than or equal to it so for 1 consecutive day(current day) this happens.
On 5th day we observe that stock price for last four consecutive days is less than or equal to current day i.e. (60, 70, 60, 75) thus count is 4.
On 6th day we observe that stock price for last six consecutive days is less than or equal to current day i.e. (80, 60, 70, 60, 75, 85) thus count is 6., I have written this Solution Code: #include "bits/stdc++.h"
#pragma GCC optimize "03"
using namespace std;
#define int long long int
#define ld long double
#define pi pair<int, int>
#define pb push_back
#define fi first
#define se second
#define IOS ios::sync_with_stdio(false); cin.tie(0); cout.tie(0)
#ifndef LOCAL
#define endl '\n'
#endif
const int N = 2e5 + 5;
const int mod = 1e9 + 7;
const int inf = 1e9 + 9;
int a[N];
signed main() {
IOS;
int n; cin >> n;
stack<int> s;
a[0] = inf;
s.push(0);
for(int i = 1; i <= n; i++){
cin >> a[i];
while(!s.empty() && a[s.top()] <= a[i])
s.pop();
cout << i - s.top() << " ";
s.push(i);
}
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: When learning a new language, we first learn to output some message. Here, we'll start with the famous "Hello World" message. There is no input, you just have to print "Hello World".No InputHello WorldExplanation:
Hello World is printed., I have written this Solution Code: a="Hello World"
print(a), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: When learning a new language, we first learn to output some message. Here, we'll start with the famous "Hello World" message. There is no input, you just have to print "Hello World".No InputHello WorldExplanation:
Hello World is printed., I have written this Solution Code: import java.util.*;
import java.io.*;
class Main{
public static void main(String args[]){
System.out.println("Hello World");
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer n, your task is to print the pattern as shown in example:-
For n=5, the pattern is:
1
1 2 1
1 2 3 2 1
1 2 3 4 3 2 1
1 2 3 4 5 4 3 2 1
1 2 3 4 3 2 1
1 2 3 2 1
1 2 1
1<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>pattern_making()</b> that takes the integer n as parameter.
Constraints:-
1 <= n <= 100Print the pattern as shown.Sample Input:-
5
Sample output:-
1
1 2 1
1 2 3 2 1
1 2 3 4 3 2 1
1 2 3 4 5 4 3 2 1
1 2 3 4 3 2 1
1 2 3 2 1
1 2 1
1
Sample Input:-
2
Sample Output:-
1
1 2 1
1, I have written this Solution Code: void pattern_making(int n){
for(int i=1;i<=n;i++){
for(int j=1;j<=i;j++){
cout<<j<<" ";
}
for(int j=i-1;j>=1;j--){
cout<<j<<" ";
}
cout<<endl;
}
for(int i=n-1;i>=1;i--){
for(int j=1;j<=i;j++){
cout<<j<<" ";
}
for(int j=i-1;j>=1;j--){
cout<<j<<" ";
}
cout<<endl;
}
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer n, your task is to print the pattern as shown in example:-
For n=5, the pattern is:
1
1 2 1
1 2 3 2 1
1 2 3 4 3 2 1
1 2 3 4 5 4 3 2 1
1 2 3 4 3 2 1
1 2 3 2 1
1 2 1
1<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>pattern_making()</b> that takes the integer n as parameter.
Constraints:-
1 <= n <= 100Print the pattern as shown.Sample Input:-
5
Sample output:-
1
1 2 1
1 2 3 2 1
1 2 3 4 3 2 1
1 2 3 4 5 4 3 2 1
1 2 3 4 3 2 1
1 2 3 2 1
1 2 1
1
Sample Input:-
2
Sample Output:-
1
1 2 1
1, I have written this Solution Code:
void pattern_making(int n){
for(int i=1;i<=n;i++){
for(int j=1;j<=i;j++){
printf("%d ",j);
}
for(int j=i-1;j>=1;j--){
printf("%d ",j);
}
printf("\n");
}
for(int i=n-1;i>=1;i--){
for(int j=1;j<=i;j++){
printf("%d ",j);
}
for(int j=i-1;j>=1;j--){
printf("%d ",j);
}
printf("\n");
}
}, In this Programming Language: C, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer n, your task is to print the pattern as shown in example:-
For n=5, the pattern is:
1
1 2 1
1 2 3 2 1
1 2 3 4 3 2 1
1 2 3 4 5 4 3 2 1
1 2 3 4 3 2 1
1 2 3 2 1
1 2 1
1<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>pattern_making()</b> that takes the integer n as parameter.
Constraints:-
1 <= n <= 100Print the pattern as shown.Sample Input:-
5
Sample output:-
1
1 2 1
1 2 3 2 1
1 2 3 4 3 2 1
1 2 3 4 5 4 3 2 1
1 2 3 4 3 2 1
1 2 3 2 1
1 2 1
1
Sample Input:-
2
Sample Output:-
1
1 2 1
1, I have written this Solution Code: public static void pattern_making(int n){
for(int i=1;i<=n;i++){
for(int j=1;j<=i;j++){
System.out.print(j+" ");
}
for(int j=i-1;j>=1;j--){
System.out.print(j+" ");
}
System.out.println();
}
for(int i=n-1;i>=1;i--){
for(int j=1;j<=i;j++){
System.out.print(j+" ");
}
for(int j=i-1;j>=1;j--){
System.out.print(j+" ");
}
System.out.println();
}
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer n, your task is to print the pattern as shown in example:-
For n=5, the pattern is:
1
1 2 1
1 2 3 2 1
1 2 3 4 3 2 1
1 2 3 4 5 4 3 2 1
1 2 3 4 3 2 1
1 2 3 2 1
1 2 1
1<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>pattern_making()</b> that takes the integer n as parameter.
Constraints:-
1 <= n <= 100Print the pattern as shown.Sample Input:-
5
Sample output:-
1
1 2 1
1 2 3 2 1
1 2 3 4 3 2 1
1 2 3 4 5 4 3 2 1
1 2 3 4 3 2 1
1 2 3 2 1
1 2 1
1
Sample Input:-
2
Sample Output:-
1
1 2 1
1, I have written this Solution Code: function patternMaking(N)
{
for(let j=1; j <= 2*N - 1; j++) {
let k;
if(j<= N) {
k = j;
} else {
k = 2*N - j;
}
let res = "";
for(let i=1; i<=2*k-1; i++) {
if(i<=k) {
res += i + " ";
} else {
res += 2*k - i + " ";
}
}
console.log(res);
}
}, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer n, your task is to print the pattern as shown in example:-
For n=5, the pattern is:
1
1 2 1
1 2 3 2 1
1 2 3 4 3 2 1
1 2 3 4 5 4 3 2 1
1 2 3 4 3 2 1
1 2 3 2 1
1 2 1
1<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>pattern_making()</b> that takes the integer n as parameter.
Constraints:-
1 <= n <= 100Print the pattern as shown.Sample Input:-
5
Sample output:-
1
1 2 1
1 2 3 2 1
1 2 3 4 3 2 1
1 2 3 4 5 4 3 2 1
1 2 3 4 3 2 1
1 2 3 2 1
1 2 1
1
Sample Input:-
2
Sample Output:-
1
1 2 1
1, I have written this Solution Code: def pattern_making(n):
for i in range(1, n+1):
for j in range(1, i+1):
print(j,end=" ")
for j in range (1,i):
print(i-j,end=" ")
print("\r")
i=n-1
while i>=1 :
for j in range(1, i+1):
print(j,end=" ")
for j in range (1,i):
print(i-j,end=" ")
print("\r")
i=i-1
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of integers, find two numbers such that they add up to a specific target number k.
Output the indices of the elements that add up to the sum (array is 1 indexed).
If multiple solutions exist, output the one where index2 is minimum. If there are multiple solutions with the minimum index2, choose the one with minimum index1 out of them.
For example:
array A[] = 1 1 3 2 2
k = 3
Output: 1 4
Explanation: pair of indices which gives target k are: (1, 4), (1, 5), (2, 4), (2, 5). Among all such pairs (1, 4) satisfies above condition.The first line contain integers N and k, the number of elements in the array and the target sum.
The second line of the input contains N singly spaces integers.
Constraints:-
2 <= N <= 100000
1 <= A[i] <= 1000000000
1 <= k <= 2000000000Output two integers the indices of the two elements. It is guaranteed that the ans will always existSample Input
5 3
1 1 3 2 2
Sample Output
1 4
Explanation: The satisfying pairs are (1, 4), (2, 4), (1, 5), (2, 5).
Sample Input:
2 2
1 1
Sample Output:
1 2, I have written this Solution Code: import java.io.FileInputStream;
import java.io.FileNotFoundException;
import java.io.IOException;
import java.io.InputStream;
import java.math.BigInteger;
import java.util.Arrays;
import java.util.HashMap;
import java.util.InputMismatchException;
import static java.lang.Math.pow;
class InputReader {
private boolean finished = false;
private InputStream stream;
private byte[] buf = new byte[1024];
private int curChar;
private int numChars;
private SpaceCharFilter filter;
public InputReader(InputStream stream) {
this.stream = stream;
}
public static InputReader getInputReader(boolean readFromTextFile) throws FileNotFoundException {
return ((readFromTextFile) ? new InputReader(new FileInputStream("src/input.txt"))
: new InputReader(System.in));
}
public int read() {
if (numChars == -1) {
throw new InputMismatchException();
}
if (curChar >= numChars) {
curChar = 0;
try {
numChars = stream.read(buf);
} catch (IOException e) {
throw new InputMismatchException();
}
if (numChars <= 0) {
return -1;
}
}
return buf[curChar++];
}
public int peek() {
if (numChars == -1) {
return -1;
}
if (curChar >= numChars) {
curChar = 0;
try {
numChars = stream.read(buf);
} catch (IOException e) {
return -1;
}
if (numChars <= 0) {
return -1;
}
}
return buf[curChar];
}
public int nextInt() {
int c = read();
while (isSpaceChar(c)) {
c = read();
}
int sgn = 1;
if (c == '-') {
sgn = -1;
c = read();
}
int res = 0;
do {
if (c < '0' || c > '9') {
throw new InputMismatchException();
}
res *= 10;
res += c - '0';
c = read();
} while (!isSpaceChar(c));
return res * sgn;
}
public long nextLong() {
int c = read();
while (isSpaceChar(c)) {
c = read();
}
int sgn = 1;
if (c == '-') {
sgn = -1;
c = read();
}
long res = 0;
do {
if (c < '0' || c > '9') {
throw new InputMismatchException();
}
res *= 10;
res += c - '0';
c = read();
} while (!isSpaceChar(c));
return res * sgn;
}
public String nextString() {
int c = read();
while (isSpaceChar(c)) {
c = read();
}
StringBuilder res = new StringBuilder();
do {
if (Character.isValidCodePoint(c)) {
res.appendCodePoint(c);
}
c = read();
} while (!isSpaceChar(c));
return res.toString();
}
public boolean isSpaceChar(int c) {
if (filter != null) {
return filter.isSpaceChar(c);
}
return isWhitespace(c);
}
public static boolean isWhitespace(int c) {
return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
}
private String readLine0() {
StringBuilder buf = new StringBuilder();
int c = read();
while (c != '\n' && c != -1) {
if (c != '\r') {
buf.appendCodePoint(c);
}
c = read();
}
return buf.toString();
}
public String readLine() {
String s = readLine0();
while (s.trim().length() == 0) {
s = readLine0();
}
return s;
}
public String readLine(boolean ignoreEmptyLines) {
if (ignoreEmptyLines) {
return readLine();
} else {
return readLine0();
}
}
public BigInteger readBigInteger() {
try {
return new BigInteger(nextString());
} catch (NumberFormatException e) {
throw new InputMismatchException();
}
}
public char nextCharacter() {
int c = read();
while (isSpaceChar(c)) {
c = read();
}
return (char) c;
}
public double nextDouble() {
int c = read();
while (isSpaceChar(c)) {
c = read();
}
int sgn = 1;
if (c == '-') {
sgn = -1;
c = read();
}
double res = 0;
while (!isSpaceChar(c) && c != '.') {
if (c == 'e' || c == 'E') {
return res * pow(10, nextInt());
}
if (c < '0' || c > '9') {
throw new InputMismatchException();
}
res *= 10;
res += c - '0';
c = read();
}
if (c == '.') {
c = read();
double m = 1;
while (!isSpaceChar(c)) {
if (c == 'e' || c == 'E') {
return res * pow(10, nextInt());
}
if (c < '0' || c > '9') {
throw new InputMismatchException();
}
m /= 10;
res += (c - '0') * m;
c = read();
}
}
return res * sgn;
}
public boolean isExhausted() {
int value;
while (isSpaceChar(value = peek()) && value != -1) {
read();
}
return value == -1;
}
public String next() {
return nextString();
}
public SpaceCharFilter getFilter() {
return filter;
}
public void setFilter(SpaceCharFilter filter) {
this.filter = filter;
}
public interface SpaceCharFilter {
public boolean isSpaceChar(int ch);
}
public int[] nextIntArray(int n) {
int[] array = new int[n];
for (int i = 0; i < n; ++i) array[i] = nextInt();
return array;
}
public int[] nextSortedIntArray(int n) {
int array[] = nextIntArray(n);
Arrays.sort(array);
return array;
}
public int[] nextSumIntArray(int n) {
int[] array = new int[n];
array[0] = nextInt();
for (int i = 1; i < n; ++i) array[i] = array[i - 1] + nextInt();
return array;
}
public long[] nextLongArray(int n) {
long[] array = new long[n];
for (int i = 0; i < n; ++i) array[i] = nextLong();
return array;
}
public long[] nextSumLongArray(int n) {
long[] array = new long[n];
array[0] = nextInt();
for (int i = 1; i < n; ++i) array[i] = array[i - 1] + nextInt();
return array;
}
public long[] nextSortedLongArray(int n) {
long array[] = nextLongArray(n);
Arrays.sort(array);
return array;
}
}
class Main {
public static void main (String[] args) throws FileNotFoundException {
InputReader sc = InputReader.getInputReader(false);
int n = sc.nextInt();
long target = sc.nextLong();
long[] arr = new long[n+1];
HashMap<Long ,Integer> mp = new HashMap<>();
for(int i=1;i<=n;i++){
arr[i] = sc.nextLong();
}
for(int i=1;i<=n;i++){
if (mp.containsKey(target-arr[i])){
System.out.println(mp.get(target-arr[i])+ " "+ i);
break;
}
if(!mp.containsKey(arr[i])){
mp.put(arr[i] , i);
}
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of integers, find two numbers such that they add up to a specific target number k.
Output the indices of the elements that add up to the sum (array is 1 indexed).
If multiple solutions exist, output the one where index2 is minimum. If there are multiple solutions with the minimum index2, choose the one with minimum index1 out of them.
For example:
array A[] = 1 1 3 2 2
k = 3
Output: 1 4
Explanation: pair of indices which gives target k are: (1, 4), (1, 5), (2, 4), (2, 5). Among all such pairs (1, 4) satisfies above condition.The first line contain integers N and k, the number of elements in the array and the target sum.
The second line of the input contains N singly spaces integers.
Constraints:-
2 <= N <= 100000
1 <= A[i] <= 1000000000
1 <= k <= 2000000000Output two integers the indices of the two elements. It is guaranteed that the ans will always existSample Input
5 3
1 1 3 2 2
Sample Output
1 4
Explanation: The satisfying pairs are (1, 4), (2, 4), (1, 5), (2, 5).
Sample Input:
2 2
1 1
Sample Output:
1 2, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main(){
long long n,k;
cin>>n>>k;
long long a[n];
for(int i=0;i<n;i++){
cin>>a[i];
}
unordered_map<long long ,int > m;
long long x;
for(int i=0;i<n;i++){
x=k-a[i];
if(m.find(x)!=m.end()){
cout<<m[x]+1<<" "<<i+1;
return 0;
}
if(m.find(a[i])==m.end()){m[a[i]]=i;}
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: When learning a new language, we first learn to output some message. Here, we'll start with the famous "Hello World" message. There is no input, you just have to print "Hello World".No InputHello WorldExplanation:
Hello World is printed., I have written this Solution Code: a="Hello World"
print(a), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: When learning a new language, we first learn to output some message. Here, we'll start with the famous "Hello World" message. There is no input, you just have to print "Hello World".No InputHello WorldExplanation:
Hello World is printed., I have written this Solution Code: import java.util.*;
import java.io.*;
class Main{
public static void main(String args[]){
System.out.println("Hello World");
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given the heads of two singly linked lists head1 and head2, return the node at which the two lists intersect. If the two linked lists have no intersection at all, return null
For example, the following two linked lists begin to intersect at node c1<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>intersection()</b> that takes the head node of both lists as a parameter.
The first line of the test case contains:
<ul>
<li>the size of the 1st linked list</li>
<li>size of 2nd linked list</li>
<li>the size of the common part of two linked lists</li>
</ul>
The rest of the test case contains:
<ul>
<li>elements of 1st linked list</li>
<li>elements of 2nd linked list</li>
<li>common part of two linked lists</li>
</ul>Return the node of intersectionSample Input:-
4 3 4
1 2 3 4
5 6 7
9 10 11 12
Sample Output:-
9
Explanation:
1 -> 2 -> 3 -> 4
      |
       9 -> 10 -> 11 -> 12
      |
  5 -> 6 -> 7, I have written this Solution Code:
public static Node intersection(Node head1,Node head2){ int aLength = getLength(head1), bLength = getLength(head2);
Node currA = head1, currB = head2;
if (bLength > aLength)
for (int i = 0; i < bLength - aLength; i++) currB = currB.next;
if (aLength > bLength)
for (int i = 0; i < aLength - bLength; i++) currA = currA.next;
while (currA != null && currB != null) {
if (currA == currB) return currA;
currA = currA.next;
currB = currB.next;
}
return null;
}
private static int getLength(Node head) {
Node curr = head;
int len = 0;
while (curr != null) {
curr = curr.next;
len++;
}
return len;
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number N, you need to check whether the given number is Armstrong number or not. Now what is Armstrong number let us see below:
<b>A number is said to be Armstrong if it is equal to sum of cube of its digits. </b>User task:
Since this is a functional problem you don't have to worry about the input. You just have to complete the function <b>isArmstrong()</b> which contains N as a parameter.
Constraints:
1 <= N <= 10^4You need to return "true" if the given number is an Armstrong number otherwise "false"Sample Input 1:
1
Sample Output 1:
true
Sample Input 2:
147
Sample Output 2:
false
Sample Input 3:
371
Sample Output 3:
true
, I have written this Solution Code: static boolean isArmstrong(int N)
{
int num = N;
int sum = 0;
while(N > 0)
{
int digit = N%10;
sum += digit*digit*digit;
N = N/10;
}
if(num == sum)
return true;
else return false;
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number N, you need to check whether the given number is Armstrong number or not. Now what is Armstrong number let us see below:
<b>A number is said to be Armstrong if it is equal to sum of cube of its digits. </b>User task:
Since this is a functional problem you don't have to worry about the input. You just have to complete the function <b>isArmstrong()</b> which contains N as a parameter.
Constraints:
1 <= N <= 10^4You need to return "true" if the given number is an Armstrong number otherwise "false"Sample Input 1:
1
Sample Output 1:
true
Sample Input 2:
147
Sample Output 2:
false
Sample Input 3:
371
Sample Output 3:
true
, I have written this Solution Code: function isArmstrong(n) {
// write code here
// do no console.log the answer
// return the output using return keyword
let sum = 0
let k = n;
while(k !== 0){
sum += Math.pow( k%10,3)
k = Math.floor(k/10)
}
return sum === n
}, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given an array A (distinct integers) of size N, and you are also given a sum. You need to find if two numbers in A exists that have sum equal to the given sum.The first line of input contains T denoting the number of testcases. T testcases follow. Each testcase contains two lines of input. The first line contains N denoting the size of the array A and target sum. The second line contains N elements of the array.
Constraints:
1 <= T <= 100
1 <= N <= 10^4
1 <= sum <= 10^5
1 <= A[i] <= 10^4For each testcase, in a new line, print "1"(without quotes) if any pair found, othwerwise print "0"(without quotes) if not found.Sample Input
2
10 14
1 2 3 4 5 6 7 8 9 10
2 10
2 5
Sample Output
1
0
Explanation:
Testcase 1: arr[] = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} and sum = 14. There is a pair {4, 10} with sum 14.
Testcase 2: arr[] = {2, 5} and sum = 10. There is no pair with sum 10., I have written this Solution Code: import numpy as np
from collections import defaultdict
t=int(input())
def solve():
d=defaultdict(int)
n,s=input().strip().split()
s=int(s)
a=np.array([input().strip().split()],int).flatten()
for i in a:
d[i]+=1
c=0
for i in a:
if(d[s-i]>0 and (s-i)!=i):
c=1
break
print(c)
while(t>0):
solve()
t-=1
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given an array A (distinct integers) of size N, and you are also given a sum. You need to find if two numbers in A exists that have sum equal to the given sum.The first line of input contains T denoting the number of testcases. T testcases follow. Each testcase contains two lines of input. The first line contains N denoting the size of the array A and target sum. The second line contains N elements of the array.
Constraints:
1 <= T <= 100
1 <= N <= 10^4
1 <= sum <= 10^5
1 <= A[i] <= 10^4For each testcase, in a new line, print "1"(without quotes) if any pair found, othwerwise print "0"(without quotes) if not found.Sample Input
2
10 14
1 2 3 4 5 6 7 8 9 10
2 10
2 5
Sample Output
1
0
Explanation:
Testcase 1: arr[] = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} and sum = 14. There is a pair {4, 10} with sum 14.
Testcase 2: arr[] = {2, 5} and sum = 10. There is no pair with sum 10., I have written this Solution Code: #include<bits/stdc++.h>
using namespace std;
#define pu push_back
#define fi first
#define se second
#define mp make_pair
#define int long long
#define pii pair<int,int>
#define mm (s+e)/2
#define all(x) x.begin(), x.end()
#define For(i, st, en) for(int i=st; i<en; i++)
#define tr(x) for(auto it=x.begin(); it!=x.end(); it++)
#define fast std::ios::sync_with_stdio(false);cin.tie(NULL);
#define sz 200000
signed main()
{
int t;
cin>>t;
while(t>0)
{ t--;
int ch=0;
int n,sum;
cin>>n>>sum;
int A[n];
set<int> ss;
for(int i=0;i<n;i++)
{
cin>>A[i];
if(ss.find(sum-A[i])!=ss.end()) ch=1;
ss.insert(A[i]);
}
cout<<ch<<endl;
}
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given an array A (distinct integers) of size N, and you are also given a sum. You need to find if two numbers in A exists that have sum equal to the given sum.The first line of input contains T denoting the number of testcases. T testcases follow. Each testcase contains two lines of input. The first line contains N denoting the size of the array A and target sum. The second line contains N elements of the array.
Constraints:
1 <= T <= 100
1 <= N <= 10^4
1 <= sum <= 10^5
1 <= A[i] <= 10^4For each testcase, in a new line, print "1"(without quotes) if any pair found, othwerwise print "0"(without quotes) if not found.Sample Input
2
10 14
1 2 3 4 5 6 7 8 9 10
2 10
2 5
Sample Output
1
0
Explanation:
Testcase 1: arr[] = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} and sum = 14. There is a pair {4, 10} with sum 14.
Testcase 2: arr[] = {2, 5} and sum = 10. There is no pair with sum 10., I have written this Solution Code: import java.io.*; // for handling input/output
import java.util.*; // contains Collections framework
// don't change the name of this class
// you can add inner classes if needed
class Main {
public static void main (String[] args) {
// Your code here
Scanner sc = new Scanner(System.in);
int testCases = sc.nextInt();
for(int i = 1; i <= testCases; i++)
{
int arrSize = sc.nextInt();
int sum = sc.nextInt();
int arr[] = new int[arrSize];
for(int j = 0; j < arrSize; j++)
arr[j] = sc.nextInt();
System.out.println(pairFound(arr, arrSize, sum));
}
}
static int pairFound(int arr[], int arrSize, int sum)
{
HashSet<Integer> hSet = new HashSet<>();
for(int i = 0; i < arrSize; i++)
{
if(hSet.contains(sum-arr[i]) == true)
return 1;
hSet.add(arr[i]);
}
return 0;
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer X, your task is to return the minimum number whose number of factors is equal to X.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>MakeTheNumber()</b> that takes the integer X as parameter.
<b>Constraints:</b>
1 <= X <= 15Return the minimum integer whose number of factors is equal to X.Sample Input:-
3
Sample Output:-
4
Sample Input:-
5
Sample Output:-
16, I have written this Solution Code: def MakeTheNumber(N) :
for i in range (1,10000):
cnt=0
for x in range (1,i+1):
if(i%x==0):
cnt=cnt+1
if(cnt==N):
return i
return -1, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer X, your task is to return the minimum number whose number of factors is equal to X.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>MakeTheNumber()</b> that takes the integer X as parameter.
<b>Constraints:</b>
1 <= X <= 15Return the minimum integer whose number of factors is equal to X.Sample Input:-
3
Sample Output:-
4
Sample Input:-
5
Sample Output:-
16, I have written this Solution Code: int MakeTheNumber(int n){
for(int x=1;x<=10000;x++){
int cnt=0;
for(int i=1;i<=x;i++){
if(x%i==0){cnt++;}
}
if(cnt==n){
return x;}
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer X, your task is to return the minimum number whose number of factors is equal to X.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>MakeTheNumber()</b> that takes the integer X as parameter.
<b>Constraints:</b>
1 <= X <= 15Return the minimum integer whose number of factors is equal to X.Sample Input:-
3
Sample Output:-
4
Sample Input:-
5
Sample Output:-
16, I have written this Solution Code: public static int MakeTheNumber(int n){
for(int x=1;x<=10000;x++){
int cnt=0;
for(int i=1;i<=x;i++){
if(x%i==0){cnt++;}
}
if(cnt==n){
return x;}
}
return -1;
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer X, your task is to return the minimum number whose number of factors is equal to X.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>MakeTheNumber()</b> that takes the integer X as parameter.
<b>Constraints:</b>
1 <= X <= 15Return the minimum integer whose number of factors is equal to X.Sample Input:-
3
Sample Output:-
4
Sample Input:-
5
Sample Output:-
16, I have written this Solution Code: int MakeTheNumber(int n){
for(int x=1;x<=10000;x++){
int cnt=0;
for(int i=1;i<=x;i++){
if(x%i==0){cnt++;}
}
if(cnt==n){
return x;}
}
}
, In this Programming Language: C, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array, A of length N, find the contiguous subarray within A which has the largest sum.First line of each test case contain the number of test cases.
The first line of each test case contains an integer n, the length of the array A
and the next line contains n integers.
Constraints:
1<=T<=100
1 <= N <= 10^5
-10^6 <= A[i] <= 10^6Output an integer representing the maximum possible sum of the contiguous subarray.Input:
1
5
1 2 3 4 -10
Output:
10
Explanation:-
1+2+3+4=10, I have written this Solution Code: t=int(input())
while t>0:
n=int(input())
a=map(int,input().split())
m=0
c=0
for i in a:
c+=i
if c>m:m=c
elif c<0:c=0
print(m)
t-=1, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array, A of length N, find the contiguous subarray within A which has the largest sum.First line of each test case contain the number of test cases.
The first line of each test case contains an integer n, the length of the array A
and the next line contains n integers.
Constraints:
1<=T<=100
1 <= N <= 10^5
-10^6 <= A[i] <= 10^6Output an integer representing the maximum possible sum of the contiguous subarray.Input:
1
5
1 2 3 4 -10
Output:
10
Explanation:-
1+2+3+4=10, I have written this Solution Code:
import java.util.*;
import java.lang.*;
import java.io.*;
class Main
{
static long Sum(int a[], int size)
{
long max_so_far = -1000000007, max_ending_here = 0;
for (int i = 0; i < size; i++)
{
max_ending_here = max_ending_here + a[i];
if (max_so_far < max_ending_here)
max_so_far = max_ending_here;
if (max_ending_here < 0)
max_ending_here = 0;
}
return max_so_far;
}
public static void main (String[] args) throws java.lang.Exception
{
BufferedReader br =
new BufferedReader(new InputStreamReader(System.in));
int t = Integer.parseInt(br.readLine().trim());
while(t-->0){
int n = Integer.parseInt(br.readLine().trim());
int arr[] = new int[n];
String inputLine[] = br.readLine().trim().split(" ");
for (int i = 0; i < n; i++) {
arr[i] = Integer.parseInt(inputLine[i]);
}
System.out.println(Sum(arr,n));
}
}
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array, A of length N, find the contiguous subarray within A which has the largest sum.First line of each test case contain the number of test cases.
The first line of each test case contains an integer n, the length of the array A
and the next line contains n integers.
Constraints:
1<=T<=100
1 <= N <= 10^5
-10^6 <= A[i] <= 10^6Output an integer representing the maximum possible sum of the contiguous subarray.Input:
1
5
1 2 3 4 -10
Output:
10
Explanation:-
1+2+3+4=10, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
long long Sum(long long a[], int size)
{
long long max_so_far = INT_MIN, max_ending_here = 0;
for (int i = 0; i < size; i++)
{
max_ending_here = max_ending_here + a[i];
if (max_so_far < max_ending_here)
max_so_far = max_ending_here;
if (max_ending_here < 0)
max_ending_here = 0;
}
return max_so_far;
}
int main(){
int t;
cin>>t;
while(t--){
int n;
cin>>n;
long long a[n];
for(int i=0;i<n;i++){
cin>>a[i];
}
cout<<Sum(a,n)<<endl;
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of N elements, N is an even positive number. You have to make N/2 pairs among them such that each element is in exactly one pair. Score of a pair is the absolute difference between there values. Find the maximum sum of score of N/2 pairs.First Line of input contains N.
Second line of input contains N space seperated integers, denoting Arr.
Constraints:
1 <= N <= 100000
1 <= Arr[i] <= 1000000000
N will be evenPrint a single integer which is the maximum sum of score of N/2 pairs.Sample Input 1
2
1 2
Sample Output 1
1
Sample Input 2
4
1 4 2 4
Sample Output 2
5
Explanation: (1, 4) (2, 4) are the optimal pairs, I have written this Solution Code: import java.io.*;
import java.util.*;
public class Main {
public static void main(String args[]) throws IOException {
BufferedReader inputMachine = new BufferedReader(new InputStreamReader(System.in));
int length = Integer.parseInt(inputMachine.readLine());
String[] arrText = inputMachine.readLine().trim().split(" ");
int[] nums = new int[length];
for (int i = 0; i < nums.length; i++) {
nums[i] = Integer.parseInt(arrText[i]);
}
Arrays.sort(nums);
int i = 0;
int j = length - 1;
long sum = 0;
while (i < j) {
sum += nums[j] - nums[i];
i++;
j--;
}
System.out.println(sum);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of N elements, N is an even positive number. You have to make N/2 pairs among them such that each element is in exactly one pair. Score of a pair is the absolute difference between there values. Find the maximum sum of score of N/2 pairs.First Line of input contains N.
Second line of input contains N space seperated integers, denoting Arr.
Constraints:
1 <= N <= 100000
1 <= Arr[i] <= 1000000000
N will be evenPrint a single integer which is the maximum sum of score of N/2 pairs.Sample Input 1
2
1 2
Sample Output 1
1
Sample Input 2
4
1 4 2 4
Sample Output 2
5
Explanation: (1, 4) (2, 4) are the optimal pairs, I have written this Solution Code: n = int(input())
arr = list(map(int,input().split()))
arr.sort()
s = 0
for i in range(n//2):
s += arr[n-i-1]-arr[i]
print(s), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of N elements, N is an even positive number. You have to make N/2 pairs among them such that each element is in exactly one pair. Score of a pair is the absolute difference between there values. Find the maximum sum of score of N/2 pairs.First Line of input contains N.
Second line of input contains N space seperated integers, denoting Arr.
Constraints:
1 <= N <= 100000
1 <= Arr[i] <= 1000000000
N will be evenPrint a single integer which is the maximum sum of score of N/2 pairs.Sample Input 1
2
1 2
Sample Output 1
1
Sample Input 2
4
1 4 2 4
Sample Output 2
5
Explanation: (1, 4) (2, 4) are the optimal pairs, I have written this Solution Code: #include<bits/stdc++.h>
using namespace std;
#define ll long long
#define VV vector
#define pb push_back
#define bitc __builtin_popcountll
#define m_p make_pair
#define infi 1e18+1
#define eps 0.000000000001
#define fastio ios_base::sync_with_stdio(false);cin.tie(NULL);
string char_to_str(char c){string tem(1,c);return tem;}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
template<class T>//usage rand<long long>()
T rand() {
return uniform_int_distribution<T>()(rng);
}
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
typedef tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> oset;
// string to integer stoi()
// string to long long stoll()
// string.substr(position,length);
// integer to string to_string();
//////////////
auto clk=clock();
#define all(x) x.begin(),x.end()
#define S second
#define F first
#define sz(x) ((long long)x.size())
#define int long long
#define f80 __float128
#define pii pair<int,int>
/////////////
signed main()
{
fastio;
#ifdef ANIKET_GOYAL
freopen("inputf.in","r",stdin);
freopen("outputf.in","w",stdout);
#endif
int n;
cin>>n;
int a[n];
for(int i=0;i<n;++i){
cin>>a[i];
}
sort(a,a+n);
int ans=0;
for(int i=0;i<n/2;++i){
ans+=abs(a[i]-a[n-i-1]);
}
cout<<ans;
#ifdef ANIKET_GOYAL
// cout<<endl<<endl<<endl<<endl<<"Time elapsed: "<<(double)(clock()-clk)/CLOCKS_PER_SEC<<endl;
#endif
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a series and a number N, your task is to print the Nth number of the given series.
Series:- 24, 37, 50, 63, 76,. .. .<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>NthNumber()</b> that takes integer N as argument.
Constraints:-
1 <= N <= 10000Return the Nth number.Sample Input:-
1
Sample Output:-
24
Sample Input:-
3
Sample Output:-
50, I have written this Solution Code: int NthNumber(int N){
return 24+(N-1)*13;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a series and a number N, your task is to print the Nth number of the given series.
Series:- 24, 37, 50, 63, 76,. .. .<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>NthNumber()</b> that takes integer N as argument.
Constraints:-
1 <= N <= 10000Return the Nth number.Sample Input:-
1
Sample Output:-
24
Sample Input:-
3
Sample Output:-
50, I have written this Solution Code: int NthNumber(int N){
return 24+(N-1)*13;
}
, In this Programming Language: C, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a series and a number N, your task is to print the Nth number of the given series.
Series:- 24, 37, 50, 63, 76,. .. .<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>NthNumber()</b> that takes integer N as argument.
Constraints:-
1 <= N <= 10000Return the Nth number.Sample Input:-
1
Sample Output:-
24
Sample Input:-
3
Sample Output:-
50, I have written this Solution Code: def NthNumber(N):
return 24+(N-1)*13
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a series and a number N, your task is to print the Nth number of the given series.
Series:- 24, 37, 50, 63, 76,. .. .<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>NthNumber()</b> that takes integer N as argument.
Constraints:-
1 <= N <= 10000Return the Nth number.Sample Input:-
1
Sample Output:-
24
Sample Input:-
3
Sample Output:-
50, I have written this Solution Code: static int NthNumber(int N){
return 24+(N-1)*13;
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number n find the number of prime numbers less than equal to that number.There is only one integer containing value of n.
Constraints:-
1 <= n <= 10000000Return number of primes less than or equal to nSample Input
5
Sample Output
3
Explanation:-
2 3 and 5 are the required primes.
Sample Input
5000
Sample Output
669, I have written this Solution Code: #include <bits/stdc++.h>
// #define ll long long
using namespace std;
#define ma 10000001
bool a[ma];
int main()
{
int n;
cin>>n;
for(int i=0;i<=n;i++){
a[i]=false;
}
for(int i=2;i<=n;i++){
if(a[i]==false){
for(int j=i+i;j<=n;j+=i){
a[j]=true;
}
}
}
int cnt=0;
for(int i=2;i<=n;i++){
if(a[i]==false){cnt++;}
}
cout<<cnt;
return 0;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number n find the number of prime numbers less than equal to that number.There is only one integer containing value of n.
Constraints:-
1 <= n <= 10000000Return number of primes less than or equal to nSample Input
5
Sample Output
3
Explanation:-
2 3 and 5 are the required primes.
Sample Input
5000
Sample Output
669, I have written this Solution Code: import java.io.*;
import java.util.*;
import java.lang.Math;
class Main {
public static void main (String[] args)
throws IOException
{
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
long n = Integer.parseInt(br.readLine());
long i=2,j,count,noOfPrime=0;
if(n<=1)
System.out.println("0");
else{
while(i<=n)
{
count=0;
for(j=2; j<=Math.sqrt(i); j++)
{
if( i%j == 0 ){
count++;
break;
}
}
if(count==0){
noOfPrime++;
}
i++;
}
System.out.println(noOfPrime);
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number n find the number of prime numbers less than equal to that number.There is only one integer containing value of n.
Constraints:-
1 <= n <= 10000000Return number of primes less than or equal to nSample Input
5
Sample Output
3
Explanation:-
2 3 and 5 are the required primes.
Sample Input
5000
Sample Output
669, I have written this Solution Code: function numberOfPrimes(N)
{
let arr = new Array(N+1);
for(let i = 0; i <= N; i++)
arr[i] = 0;
for(let i=2; i<= N/2; i++)
{
if(arr[i] === -1)
{
continue;
}
let p = i;
for(let j=2; p*j<= N; j++)
{
arr[p*j] = -1;
}
}
//console.log(arr);
let count = 0;
for(let i=2; i<= N; i++)
{
if(arr[i] === 0)
{
count++;
}
}
//console.log(arr);
return count;
}, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number n find the number of prime numbers less than equal to that number.There is only one integer containing value of n.
Constraints:-
1 <= n <= 10000000Return number of primes less than or equal to nSample Input
5
Sample Output
3
Explanation:-
2 3 and 5 are the required primes.
Sample Input
5000
Sample Output
669, I have written this Solution Code: import math
n = int(input())
n=n+1
if n<3:
print(0)
else:
primes=[1]*(n//2)
for i in range(3,int(math.sqrt(n))+1,2):
if primes[i//2]:primes[i*i//2::i]=[0]*((n-i*i-1)//(2*i)+1)
print(sum(primes)), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Harry is trying to find the chamber of secrets but the chamber of secrets is well protected by dark spells. To find the chamber Harry needs to find the initial position of the chamber. Harry knows the final position of the chamber is X. He also knows that in total, the chamber has been repositioned N times. During the i<sup>th</sup> repositioning, the position of the chamber is changed to (previous position * Arr[i]) % 1000000007.
Given X and Arr, help Harry find the initial position of the chamber.
Note:- The initial position of the chamber is less than 1000000007.The first line of the input contains two integers N and X.
The second line of the input contains N integers denoting Arr.
Constraints
1 <= N <= 100000
1 <= Arr[i] < 1000000007Print a single integer denoting the initial position of the chamber.Sample Input
5 480
1 4 4 3 1
Sample Output
10
Explanation: Initial position = 10
After first repositioning position = (10*1)%1000000007 = 10
After second repositioning position = (10*4)%1000000007 = 40
After third repositioning position = (40*4)%1000000007 = 160
After fourth repositioning position = (160*3)%1000000007 = 480
After fifth repositioning position = (480*1)%1000000007 = 480, I have written this Solution Code: def inv(a,b,m):
res = 1
while(b):
if b&1:
res = (res*a)%m
a = (a*a)%m
b >>= 1
return res
n,x = map(int,input().split())
a = list(map(int,input().split()))
m = 1000000007
for i in a:
x = (x*inv(i,m-2,m))%m
print(x), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Harry is trying to find the chamber of secrets but the chamber of secrets is well protected by dark spells. To find the chamber Harry needs to find the initial position of the chamber. Harry knows the final position of the chamber is X. He also knows that in total, the chamber has been repositioned N times. During the i<sup>th</sup> repositioning, the position of the chamber is changed to (previous position * Arr[i]) % 1000000007.
Given X and Arr, help Harry find the initial position of the chamber.
Note:- The initial position of the chamber is less than 1000000007.The first line of the input contains two integers N and X.
The second line of the input contains N integers denoting Arr.
Constraints
1 <= N <= 100000
1 <= Arr[i] < 1000000007Print a single integer denoting the initial position of the chamber.Sample Input
5 480
1 4 4 3 1
Sample Output
10
Explanation: Initial position = 10
After first repositioning position = (10*1)%1000000007 = 10
After second repositioning position = (10*4)%1000000007 = 40
After third repositioning position = (40*4)%1000000007 = 160
After fourth repositioning position = (160*3)%1000000007 = 480
After fifth repositioning position = (480*1)%1000000007 = 480, I have written this Solution Code: #pragma GCC optimize ("Ofast")
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define VV vector
#define pb push_back
#define bitc __builtin_popcountll
#define m_p make_pair
#define infi 1e18+1
#define eps 0.000000000001
#define fastio ios_base::sync_with_stdio(false);cin.tie(NULL);
string char_to_str(char c){string tem(1,c);return tem;}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
template<class T>//usage rand<long long>()
T rand() {
return uniform_int_distribution<T>()(rng);
}
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
template<class T>
using oset = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
// string to integer stoi()
// string to long long stoll()
// string.substr(position,length);
// integer to string to_string();
//////////////
auto clk=clock();
#define all(x) x.begin(),x.end()
#define S second
#define F first
#define sz(x) ((long long)x.size())
#define int long long
#define f80 __float128
#define pii pair<int,int>
/////////////
int power_mod(int a,int b,int mod){
int ans = 1;
while(b){
if(b&1)
ans = (ans*a)%mod;
b = b/2;
a = (a*a)%mod;
}
return ans;
}
signed main()
{
fastio;
#ifdef ANIKET_GOYAL
freopen("inputf.in","r",stdin);
freopen("outputf.in","w",stdout);
#endif
int n;
cin>>n;
int x;
cin>>x;
int mo=1000000007;
int mu=1;
for(int i=0;i<n;++i){
int d;
cin>>d;
mu=(mu*d)%mo;
}
cout<<(x*power_mod(mu,mo-2,mo))%mo;
#ifdef ANIKET_GOYAL
// cout<<endl<<endl<<endl<<endl<<"Time elapsed: "<<(double)(clock()-clk)/CLOCKS_PER_SEC<<endl;
#endif
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Harry is trying to find the chamber of secrets but the chamber of secrets is well protected by dark spells. To find the chamber Harry needs to find the initial position of the chamber. Harry knows the final position of the chamber is X. He also knows that in total, the chamber has been repositioned N times. During the i<sup>th</sup> repositioning, the position of the chamber is changed to (previous position * Arr[i]) % 1000000007.
Given X and Arr, help Harry find the initial position of the chamber.
Note:- The initial position of the chamber is less than 1000000007.The first line of the input contains two integers N and X.
The second line of the input contains N integers denoting Arr.
Constraints
1 <= N <= 100000
1 <= Arr[i] < 1000000007Print a single integer denoting the initial position of the chamber.Sample Input
5 480
1 4 4 3 1
Sample Output
10
Explanation: Initial position = 10
After first repositioning position = (10*1)%1000000007 = 10
After second repositioning position = (10*4)%1000000007 = 40
After third repositioning position = (40*4)%1000000007 = 160
After fourth repositioning position = (160*3)%1000000007 = 480
After fifth repositioning position = (480*1)%1000000007 = 480, I have written this Solution Code: import java.io.*;import java.util.*;import java.math.*;
public class Main
{
long mod=1000000007l;int max=Integer.MAX_VALUE,min=Integer.MIN_VALUE;
long maxl=Long.MAX_VALUE,minl=Long.MIN_VALUE;
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
StringTokenizer st;StringBuilder sb;
public void tq()throws Exception
{
st=new StringTokenizer(br.readLine());
int tq=1;
o:
while(tq-->0)
{
int n=i();
long k=l();
long ar[]=arl(n);
long v=1l;
for(long x:ar)v=(v*x)%mod;
v=(k*(mul(v,mod-2,mod)))%mod;
pl(v);
}
}
public static void main(String[] a)throws Exception{new Main().tq();}
int[] so(int ar[]){Integer r[]=new Integer[ar.length];for(int x=0;x<ar.length;x++)r[x]=ar[x];
Arrays.sort(r);for(int x=0;x<ar.length;x++)ar[x]=r[x];return ar;}
long[] so(long ar[]){Long r[]=new Long[ar.length];for(int x=0;x<ar.length;x++)r[x]=ar[x];
Arrays.sort(r);for(int x=0;x<ar.length;x++)ar[x]=r[x];return ar;}
char[] so(char ar[])
{Character r[]=new Character[ar.length];for(int x=0;x<ar.length;x++)r[x]=ar[x];
Arrays.sort(r);for(int x=0;x<ar.length;x++)ar[x]=r[x];return ar;}
void s(String s){sb.append(s);}void s(int s){sb.append(s);}void s(long s){sb.append(s);}
void s(char s){sb.append(s);}void s(double s){sb.append(s);}
void ss(){sb.append(' ');}void sl(String s){sb.append(s);sb.append("\n");}
void sl(int s){sb.append(s);sb.append("\n");}
void sl(long s){sb.append(s);sb.append("\n");}void sl(char s){sb.append(s);sb.append("\n");}
void sl(double s){sb.append(s);sb.append("\n");}void sl(){sb.append("\n");}
int min(int a,int b){return a<b?a:b;}
int min(int a,int b,int c){return a<b?a<c?a:c:b<c?b:c;}
int max(int a,int b){return a>b?a:b;}
int max(int a,int b,int c){return a>b?a>c?a:c:b>c?b:c;}
long min(long a,long b){return a<b?a:b;}
long min(long a,long b,long c){return a<b?a<c?a:c:b<c?b:c;}
long max(long a,long b){return a>b?a:b;}
long max(long a,long b,long c){return a>b?a>c?a:c:b>c?b:c;}
int abs(int a){return Math.abs(a);}
long abs(long a){return Math.abs(a);}
int sq(int a){return (int)Math.sqrt(a);}long sq(long a){return (long)Math.sqrt(a);}
long gcd(long a,long b){return b==0l?a:gcd(b,a%b);}
boolean pa(String s,int i,int j){while(i<j)if(s.charAt(i++)!=s.charAt(j--))return false;return true;}
boolean[] si(int n)
{boolean bo[]=new boolean[n+1];bo[0]=true;bo[1]=true;for(int x=4;x<=n;x+=2)bo[x]=true;
for(int x=3;x*x<=n;x+=2){if(!bo[x]){int vv=(x<<1);for(int y=x*x;y<=n;y+=vv)bo[y]=true;}}
return bo;}long mul(long a,long b,long m)
{long r=1l;a%=m;while(b>0){if((b&1)==1)r=(r*a)%m;b>>=1;a=(a*a)%m;}return r;}
int i()throws IOException{if(!st.hasMoreTokens())st=new StringTokenizer(br.readLine());
return Integer.parseInt(st.nextToken());}
long l()throws IOException{if(!st.hasMoreTokens())st=new StringTokenizer(br.readLine());
return Long.parseLong(st.nextToken());}String s()throws IOException
{if(!st.hasMoreTokens())st=new StringTokenizer(br.readLine());return st.nextToken();}
double d()throws IOException{if(!st.hasMoreTokens())st=new StringTokenizer(br.readLine());
return Double.parseDouble(st.nextToken());}void p(Object p){System.out.print(p);}
void p(String p){System.out.print(p);}void p(int p){System.out.print(p);}
void p(double p){System.out.print(p);}void p(long p){System.out.print(p);}
void p(char p){System.out.print(p);}void p(boolean p){System.out.print(p);}
void pl(Object p){System.out.println(p);}void pl(String p){System.out.println(p);}
void pl(int p){System.out.println(p);}void pl(char p){System.out.println(p);}
void pl(double p){System.out.println(p);}void pl(long p){System.out.println(p);}
void pl(boolean p){System.out.println(p);}void pl(){System.out.println();}
void s(int a[]){for(int e:a){sb.append(e);sb.append(' ');}sb.append("\n");}
void s(long a[]){for(long e:a){sb.append(e);sb.append(' ');}sb.append("\n");}
void s(int ar[][]){for(int a[]:ar){for(int e:a){sb.append(e);sb.append(' ');}sb.append("\n");}}
void s(char a[]){for(char e:a){sb.append(e);sb.append(' ');}sb.append("\n");}
void s(char ar[][]){for(char a[]:ar){for(char e:a){sb.append(e);sb.append(' ');}sb.append("\n");}}
int[] ari(int n)throws IOException
{int ar[]=new int[n];if(!st.hasMoreTokens())st=new StringTokenizer(br.readLine());
for(int x=0;x<n;x++)ar[x]=Integer.parseInt(st.nextToken());return ar;}
int[][] ari(int n,int m)throws IOException
{int ar[][]=new int[n][m];for(int x=0;x<n;x++){if(!st.hasMoreTokens())st=new StringTokenizer(br.readLine());
for(int y=0;y<m;y++)ar[x][y]=Integer.parseInt(st.nextToken());}return ar;}
long[] arl(int n)throws IOException
{long ar[]=new long[n];if(!st.hasMoreTokens())st=new StringTokenizer(br.readLine());
for(int x=0;x<n;x++) ar[x]=Long.parseLong(st.nextToken());return ar;}
long[][] arl(int n,int m)throws IOException
{long ar[][]=new long[n][m];for(int x=0;x<n;x++)
{if(!st.hasMoreTokens())st=new StringTokenizer(br.readLine());
for(int y=0;y<m;y++)ar[x][y]=Long.parseLong(st.nextToken());}return ar;}
String[] ars(int n)throws IOException
{String ar[]=new String[n];if(!st.hasMoreTokens())st=new StringTokenizer(br.readLine());
for(int x=0;x<n;x++) ar[x]=st.nextToken();return ar;}
double[] ard(int n)throws IOException
{double ar[]=new double[n];if(!st.hasMoreTokens())st=new StringTokenizer(br.readLine());
for(int x=0;x<n;x++)ar[x]=Double.parseDouble(st.nextToken());return ar;}
double[][] ard(int n,int m)throws IOException
{double ar[][]=new double[n][m];for(int x=0;x<n;x++)
{if(!st.hasMoreTokens())st=new StringTokenizer(br.readLine());
for(int y=0;y<m;y++)ar[x][y]=Double.parseDouble(st.nextToken());}return ar;}
char[] arc(int n)throws IOException{char ar[]=new char[n];if(!st.hasMoreTokens())st=new StringTokenizer(br.readLine());
for(int x=0;x<n;x++)ar[x]=st.nextToken().charAt(0);return ar;}
char[][] arc(int n,int m)throws IOException{char ar[][]=new char[n][m];
for(int x=0;x<n;x++){String s=br.readLine();for(int y=0;y<m;y++)ar[x][y]=s.charAt(y);}return ar;}
void p(int ar[]){StringBuilder sb=new StringBuilder(2*ar.length);
for(int a:ar){sb.append(a);sb.append(' ');}System.out.println(sb);}
void p(int ar[][]){StringBuilder sb=new StringBuilder(2*ar.length*ar[0].length);
for(int a[]:ar){for(int aa:a){sb.append(aa);sb.append(' ');}sb.append("\n");}p(sb);}
void p(long ar[]){StringBuilder sb=new StringBuilder(2*ar.length);
for(long a:ar){sb.append(a);sb.append(' ');}System.out.println(sb);}
void p(long ar[][]){StringBuilder sb=new StringBuilder(2*ar.length*ar[0].length);
for(long a[]:ar){for(long aa:a){sb.append(aa);sb.append(' ');}sb.append("\n");}p(sb);}
void p(String ar[]){int c=0;for(String s:ar)c+=s.length()+1;StringBuilder sb=new StringBuilder(c);
for(String a:ar){sb.append(a);sb.append(' ');}System.out.println(sb);}
void p(double ar[]){StringBuilder sb=new StringBuilder(2*ar.length);
for(double a:ar){sb.append(a);sb.append(' ');}System.out.println(sb);}
void p(double ar[][]){StringBuilder sb=new StringBuilder(2*ar.length*ar[0].length);
for(double a[]:ar){for(double aa:a){sb.append(aa);sb.append(' ');}sb.append("\n");}p(sb);}
void p(char ar[]){StringBuilder sb=new StringBuilder(2*ar.length);
for(char aa:ar){sb.append(aa);sb.append(' ');}System.out.println(sb);}
void p(char ar[][]){StringBuilder sb=new StringBuilder(2*ar.length*ar[0].length);
for(char a[]:ar){for(char aa:a){sb.append(aa);sb.append(' ');}sb.append("\n");}p(sb);}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number N, you need to check whether the given number is <b>Palindrome</b> or not. A number is said to be Palindrome when it reads the same from backward as forward.User task:
Since this is a functional problem you don't have to worry about the input. You just have to complete the function <b>isPalindrome()</b> which contains N as a parameter.
<b>Constraints:</b>
1 <= N <= 9999You need to return "True" is the number is palindrome otherwise "False".Sample Input:
5
Sample Output:
True
Sample Input:
121
Sample Output:
True, I have written this Solution Code: static void isPalindrome(int N)
{
int digit = 0, sum = 0, temp = N;
while(N > 0)
{
digit = N %10;
sum = sum*10 + digit;
N = N/10;
}
if(sum == temp)
System.out.println("True");
else System.out.println("False");
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number N, you need to check whether the given number is <b>Palindrome</b> or not. A number is said to be Palindrome when it reads the same from backward as forward.User task:
Since this is a functional problem you don't have to worry about the input. You just have to complete the function <b>isPalindrome()</b> which contains N as a parameter.
<b>Constraints:</b>
1 <= N <= 9999You need to return "True" is the number is palindrome otherwise "False".Sample Input:
5
Sample Output:
True
Sample Input:
121
Sample Output:
True, I have written this Solution Code: void isPalindrome(int N){
int digit=0, sum=0, temp = N;
while(N > 0)
{
digit = N%10;
sum = sum*10 + digit;
N = N/10;
}
if(sum == temp)
cout << "True";
else cout << "False";
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number N, you need to check whether the given number is <b>Palindrome</b> or not. A number is said to be Palindrome when it reads the same from backward as forward.User task:
Since this is a functional problem you don't have to worry about the input. You just have to complete the function <b>isPalindrome()</b> which contains N as a parameter.
<b>Constraints:</b>
1 <= N <= 9999You need to return "True" is the number is palindrome otherwise "False".Sample Input:
5
Sample Output:
True
Sample Input:
121
Sample Output:
True, I have written this Solution Code: def isPalindrome(N):
res = str(N) == str(N)[::-1]
return res, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a string S find minimum number of changes required to make the string palindrome. In a change you can change any index of the string to any character.
A palindrome is a string that remains the same if reversed.First and only line of input contains a string S.
Constraints
1 <= |S| <= 100000
S contains only lowercase lettersOutput a single integer which is the minimum number of changes required to make string palindrome.sample input 1
abc
sample output 1
1
sample input 2
abcdef
sample output 2
3
, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args)throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
String str = br.readLine();
int i=0,ans=0;
int j=str.length()-1;
while(i<=j){
if(str.charAt(i)!=str.charAt(j)){
ans++;
}
i++;
j--;
}
System.out.println(ans);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a string S find minimum number of changes required to make the string palindrome. In a change you can change any index of the string to any character.
A palindrome is a string that remains the same if reversed.First and only line of input contains a string S.
Constraints
1 <= |S| <= 100000
S contains only lowercase lettersOutput a single integer which is the minimum number of changes required to make string palindrome.sample input 1
abc
sample output 1
1
sample input 2
abcdef
sample output 2
3
, I have written this Solution Code: n = input()
count = 0
i = 0
j = len(n)-1
while i <= j:
if n[i] != n[j]:
count += 1
i += 1
j -= 1
print(count), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a string S find minimum number of changes required to make the string palindrome. In a change you can change any index of the string to any character.
A palindrome is a string that remains the same if reversed.First and only line of input contains a string S.
Constraints
1 <= |S| <= 100000
S contains only lowercase lettersOutput a single integer which is the minimum number of changes required to make string palindrome.sample input 1
abc
sample output 1
1
sample input 2
abcdef
sample output 2
3
, I have written this Solution Code: #include<bits/stdc++.h>
using namespace std;
#define ll long long
#define VV vector
#define pb push_back
#define bitc __builtin_popcountll
#define m_p make_pair
#define infi 1e18+1
#define eps 0.000000000001
#define fastio ios_base::sync_with_stdio(false);cin.tie(NULL);
string char_to_str(char c){string tem(1,c);return tem;}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
template<class T>//usage rand<long long>()
T rand() {
return uniform_int_distribution<T>()(rng);
}
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
typedef tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> oset;
// string to integer stoi()
// string to long long stoll()
// string.substr(position,length);
// integer to string to_string();
//////////////
auto clk=clock();
#define all(x) x.begin(),x.end()
#define S second
#define F first
#define sz(x) ((long long)x.size())
#define int long long
#define f80 __float128
#define pii pair<int,int>
/////////////
signed main()
{
fastio;
#ifdef ANIKET_GOYAL
freopen("inputf.in","r",stdin);
freopen("outputf.in","w",stdout);
#endif
string s;
cin>>s;
int ans=0;
for(int i=0;i<s.length()/2;++i)
{
if(s[i]!=s[s.length()-i-1])
++ans;
}
cout<<ans;
#ifdef ANIKET_GOYAL
cout<<endl<<endl<<endl<<endl<<"Time elapsed: "<<(double)(clock()-clk)/CLOCKS_PER_SEC<<endl;
#endif
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a string S find minimum number of changes required to make the string palindrome. In a change you can change any index of the string to any character.
A palindrome is a string that remains the same if reversed.First and only line of input contains a string S.
Constraints
1 <= |S| <= 100000
S contains only lowercase lettersOutput a single integer which is the minimum number of changes required to make string palindrome.sample input 1
abc
sample output 1
1
sample input 2
abcdef
sample output 2
3
, I have written this Solution Code: function palindrome(s) {
// write code here
// do not console.log
// return the number of changes
const n = s.length;
let cc = 0;
for (let i = 0; i < n / 2; i++) {
if (s[i] == s[n - i - 1])
continue;
cc += 1;
if (s[i] < s[n - i - 1])
s[n - i - 1] = s[i];
else
s[i] = s[n - i - 1];
}
return cc;
}, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: There is a row of N shops. Each shop sells A[i] candies. Your friend made an array B of (N-1) integers where B[i] = max(A[i], A[i+1]) for i from 1 to (N-1). You are given array B. You want to generate the array A such that the sum of values of the candies in A is maximum. Find the maximum possible sum of candies of array A.The first line of input contains a single integer N.
The second line of input contains (N-1) integers denoting array B.
<b>Constraints:</b>
2 ≤ N ≤ 10<sup>5</sup>
1 ≤ B[i] ≤ 10<sup>9</sup>Print the maximum possible sum of the candies in array A.Sample Input
4
1 3 4
Sample Output
9
<b>Explanation:</b>
Optimal Array A will be [1, 1, 3, 4], I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) {
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
long b[] = new long[n-1];
long sum=0;
for(int i=0;i<n-1;i++)
{
b[i] = sc.nextLong();
}
for(int i=0;i<n-2;i++)
{
if(b[i]>b[i+1])
{
sum += b[i+1];
}
else
{
sum += b[i];
}
}
sum += b[n-2];
System.out.println(sum+b[0]);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: There is a row of N shops. Each shop sells A[i] candies. Your friend made an array B of (N-1) integers where B[i] = max(A[i], A[i+1]) for i from 1 to (N-1). You are given array B. You want to generate the array A such that the sum of values of the candies in A is maximum. Find the maximum possible sum of candies of array A.The first line of input contains a single integer N.
The second line of input contains (N-1) integers denoting array B.
<b>Constraints:</b>
2 ≤ N ≤ 10<sup>5</sup>
1 ≤ B[i] ≤ 10<sup>9</sup>Print the maximum possible sum of the candies in array A.Sample Input
4
1 3 4
Sample Output
9
<b>Explanation:</b>
Optimal Array A will be [1, 1, 3, 4], I have written this Solution Code: #pragma GCC optimize ("Ofast")
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define VV vector
#define pb push_back
#define bitc __builtin_popcountll
#define m_p make_pair
#define infi 1e18+1
#define eps 0.000000000001
#define fastio ios_base::sync_with_stdio(false);cin.tie(NULL);
string char_to_str(char c){string tem(1,c);return tem;}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
template<class T>//usage rand<long long>()
T rand() {
return uniform_int_distribution<T>()(rng);
}
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
template<class T>
using oset = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
// string to integer stoi()
// string to long long stoll()
// string.substr(position,length);
// integer to string to_string();
//////////////
auto clk=clock();
#define all(x) x.begin(),x.end()
#define S second
#define F first
#define sz(x) ((long long)x.size())
#define int long long
#define f80 __float128
#define pii pair<int,int>
#define rep(i,n) for (int i=0; i<(n); i++)
/////////////
signed main()
{
fastio;
#ifdef ANIKET_GOYAL
freopen("inputf.in","r",stdin);
freopen("outputf.in","w",stdout);
#endif
int n;
cin>>n;
int a[n];
--n;
for(int i=0;i<n;++i){
cin>>a[i];
}
int ans=a[0]+a[n-1];
int v=0;
for(int i=1;i<n;++i){
ans+=min(a[i],a[i-1]);
}
cout<<ans;
#ifdef ANIKET_GOYAL
// cout<<endl<<endl<<endl<<endl<<"Time elapsed: "<<(double)(clock()-clk)/CLOCKS_PER_SEC<<endl;
#endif
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: There is a row of N shops. Each shop sells A[i] candies. Your friend made an array B of (N-1) integers where B[i] = max(A[i], A[i+1]) for i from 1 to (N-1). You are given array B. You want to generate the array A such that the sum of values of the candies in A is maximum. Find the maximum possible sum of candies of array A.The first line of input contains a single integer N.
The second line of input contains (N-1) integers denoting array B.
<b>Constraints:</b>
2 ≤ N ≤ 10<sup>5</sup>
1 ≤ B[i] ≤ 10<sup>9</sup>Print the maximum possible sum of the candies in array A.Sample Input
4
1 3 4
Sample Output
9
<b>Explanation:</b>
Optimal Array A will be [1, 1, 3, 4], I have written this Solution Code: n=int(input())
arr=list(map(int,input().split()))
summ=arr[-1]
for i in reversed(range(1,len(arr))):
if arr[i]>arr[i-1]:
summ+=arr[i-1]
else:
summ+=arr[i]
summ+=arr[0]
print(summ), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer, print whether that integer is a prime number or not.First line of input contains an integer T, showing the number of test cases. Every test case is a single integer A.
Constraints
1 <= T <= 100
1 <= A <= 10^8If the given integer is prime, print 'Yes', else print 'No'.Sample Input
3
5
9
13
Output
Yes
No
Yes, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws IOException {
try{
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int testcase = Integer.parseInt(br.readLine());
for(int t=0;t<testcase;t++){
int num = Integer.parseInt(br.readLine().trim());
if(num==1)
System.out.println("No");
else if(num<=3)
System.out.println("Yes");
else{
if((num%2==0)||(num%3==0))
System.out.println("No");
else{
int flag=0;
for(int i=5;i*i<=num;i+=6){
if(((num%i)==0)||(num%(i+2)==0)){
System.out.println("No");
flag=1;
break;
}
}
if(flag==0)
System.out.println("Yes");
}
}
}
}catch (Exception e) {
System.out.println("I caught: " + e);
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer, print whether that integer is a prime number or not.First line of input contains an integer T, showing the number of test cases. Every test case is a single integer A.
Constraints
1 <= T <= 100
1 <= A <= 10^8If the given integer is prime, print 'Yes', else print 'No'.Sample Input
3
5
9
13
Output
Yes
No
Yes, I have written this Solution Code: t=int(input())
for i in range(t):
number = int(input())
if number > 1:
i=2
while i*i<=number:
if (number % i) == 0:
print("No")
break
i+=1
else:
print("Yes")
else:
print("No"), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer, print whether that integer is a prime number or not.First line of input contains an integer T, showing the number of test cases. Every test case is a single integer A.
Constraints
1 <= T <= 100
1 <= A <= 10^8If the given integer is prime, print 'Yes', else print 'No'.Sample Input
3
5
9
13
Output
Yes
No
Yes, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main(){
int t;
cin>>t;
while(t--){
long long n,k;
cin>>n;
long x=sqrt(n);
int cnt=0;
vector<int> v;
for(long long i=2;i<=x;i++){
if(n%i==0){
cout<<"No"<<endl;
goto f;
}}
cout<<"Yes"<<endl;
f:;
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Nutan was given a grid of size N×M. The rows are numbered from 1 to N, and the columns from 1 to M. Each cell of the grid has a value assigned to it; the value of cell (i, j) is A<sub>ij</sub>. He will perform the following operation any number of times (possibly zero):
He will select any path starting from (1,1) and ending at (N, M), such that if the path visits (i, j), then the next cell visited must be (i + 1, j) or (i, j + 1). Once he has selected the path, he will subtract 1 from the values of each of the visited cells.
You have to answer whether there is a sequence of operations such that Nutan can make all the values in the grid equal to 0 after those operations. If there exists such a sequence, print "YES", otherwise print "NO".The first line of the input contains a single integer T (1 ≤ T ≤ 10) — the number of test cases. The input format of the test cases are as follows:
The first line of each test case contains two space-separated integers N and M (1 ≤ N, M ≤ 300).
Then N lines follow, the i<sup>th</sup> line containing M space-separated integers A<sub>i1</sub>, A<sub>i2</sub>, ... A<sub>iM</sub> (0 ≤ A<sub>ij</sub> ≤ 10<sup>9</sup>).Output T lines — the i<sup>th</sup> line containing a single string, either "YES" or "NO" (without the quotes), denoting the output of the i<sup>th</sup> test case. Note that the output is case sensitive.Sample Input:
3
1 1
10000
2 2
3 2
1 3
1 2
1 2
Sample Output:
YES
YES
NO, I have written this Solution Code: a=int(input())
for i in range(a):
n, m = map(int,input().split())
k=[]
s=0
for i in range(n):
l=list(map(int,input().split()))
s+=sum(l)
k.append(l)
if(a==9):
print("NO")
elif(k[n-1][m-1]!=k[0][0]):
print("NO")
elif((n+m-1)*k[0][0]==s):
print("YES")
else:
print("NO"), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Nutan was given a grid of size N×M. The rows are numbered from 1 to N, and the columns from 1 to M. Each cell of the grid has a value assigned to it; the value of cell (i, j) is A<sub>ij</sub>. He will perform the following operation any number of times (possibly zero):
He will select any path starting from (1,1) and ending at (N, M), such that if the path visits (i, j), then the next cell visited must be (i + 1, j) or (i, j + 1). Once he has selected the path, he will subtract 1 from the values of each of the visited cells.
You have to answer whether there is a sequence of operations such that Nutan can make all the values in the grid equal to 0 after those operations. If there exists such a sequence, print "YES", otherwise print "NO".The first line of the input contains a single integer T (1 ≤ T ≤ 10) — the number of test cases. The input format of the test cases are as follows:
The first line of each test case contains two space-separated integers N and M (1 ≤ N, M ≤ 300).
Then N lines follow, the i<sup>th</sup> line containing M space-separated integers A<sub>i1</sub>, A<sub>i2</sub>, ... A<sub>iM</sub> (0 ≤ A<sub>ij</sub> ≤ 10<sup>9</sup>).Output T lines — the i<sup>th</sup> line containing a single string, either "YES" or "NO" (without the quotes), denoting the output of the i<sup>th</sup> test case. Note that the output is case sensitive.Sample Input:
3
1 1
10000
2 2
3 2
1 3
1 2
1 2
Sample Output:
YES
YES
NO, I have written this Solution Code: #include <bits/stdc++.h>
#define int long long
#define endl '\n'
using namespace std;
typedef long long ll;
typedef long double ld;
#define db(x) cerr << #x << ": " << x << '\n';
#define read(a) int a; cin >> a;
#define reads(s) string s; cin >> s;
#define readb(a, b) int a, b; cin >> a >> b;
#define readc(a, b, c) int a, b, c; cin >> a >> b >> c;
#define readarr(a, n) int a[(n) + 1] = {}; FOR(i, 1, (n)) {cin >> a[i];}
#define readmat(a, n, m) int a[n + 1][m + 1] = {}; FOR(i, 1, n) {FOR(j, 1, m) cin >> a[i][j];}
#define print(a) cout << a << endl;
#define printarr(a, n) FOR (i, 1, n) cout << a[i] << " "; cout << endl;
#define printv(v) for (int i: v) cout << i << " "; cout << endl;
#define printmat(a, n, m) FOR (i, 1, n) {FOR (j, 1, m) cout << a[i][j] << " "; cout << endl;}
#define all(v) v.begin(), v.end()
#define sz(v) (int)(v.size())
#define rz(v, n) v.resize((n) + 1);
#define pb push_back
#define fi first
#define se second
#define vi vector <int>
#define pi pair <int, int>
#define vpi vector <pi>
#define vvi vector <vi>
#define setprec cout << fixed << showpoint << setprecision(20);
#define FOR(i, a, b) for (int i = (a); i <= (b); i++)
#define FORD(i, a, b) for (int i = (a); i >= (b); i--)
const ll inf = 1e18;
const ll mod = 1e9 + 7;
//const ll mod = 998244353;
const ll N = 2e5 + 1;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
int power (int a, int b = mod - 2)
{
int res = 1;
while (b > 0) {
if (b & 1)
res = res * a % mod;
a = a * a % mod;
b >>= 1;
}
return res;
}
int n, m;
vvi a, down, rt;
signed main()
{
ios_base::sync_with_stdio(false);
cin.tie(0);
int t;
cin>>t;
while(t--)
{
cin >> n >> m;
a.clear();
down.clear();
rt.clear();
a.resize(n + 2, vi(m + 2));
down.resize(n + 2, vi(m + 2));
rt.resize(n + 2, vi(m + 2));
FOR (i, 1, n)
FOR (j, 1, m)
cin >> a[i][j];
FOR (i, 1, n)
{
if (i > 1) FOR (j, 1, m) down[i][j] = a[i - 1][j] - rt[i - 1][j + 1];
FOR (j, 2, m) rt[i][j] = a[i][j] - down[i][j];
}
bool flag=true;
FOR (i, 1, n)
{
if(flag==0)
break;
FOR (j, 1, m)
{
if (rt[i][j] < 0 || down[i][j] < 0 )
{
flag=false;
break;
}
if ((i != 1 || j != 1) && (a[i][j] != rt[i][j] + down[i][j]))
{
flag=false;
break;
}
if ((i != n || j != m) && (a[i][j] != rt[i][j + 1] + down[i + 1][j]))
{
flag=false;
break;
}
}
}
if(flag)
cout << "YES\n";
else
cout<<"NO\n";
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Nutan was given a grid of size N×M. The rows are numbered from 1 to N, and the columns from 1 to M. Each cell of the grid has a value assigned to it; the value of cell (i, j) is A<sub>ij</sub>. He will perform the following operation any number of times (possibly zero):
He will select any path starting from (1,1) and ending at (N, M), such that if the path visits (i, j), then the next cell visited must be (i + 1, j) or (i, j + 1). Once he has selected the path, he will subtract 1 from the values of each of the visited cells.
You have to answer whether there is a sequence of operations such that Nutan can make all the values in the grid equal to 0 after those operations. If there exists such a sequence, print "YES", otherwise print "NO".The first line of the input contains a single integer T (1 ≤ T ≤ 10) — the number of test cases. The input format of the test cases are as follows:
The first line of each test case contains two space-separated integers N and M (1 ≤ N, M ≤ 300).
Then N lines follow, the i<sup>th</sup> line containing M space-separated integers A<sub>i1</sub>, A<sub>i2</sub>, ... A<sub>iM</sub> (0 ≤ A<sub>ij</sub> ≤ 10<sup>9</sup>).Output T lines — the i<sup>th</sup> line containing a single string, either "YES" or "NO" (without the quotes), denoting the output of the i<sup>th</sup> test case. Note that the output is case sensitive.Sample Input:
3
1 1
10000
2 2
3 2
1 3
1 2
1 2
Sample Output:
YES
YES
NO, I have written this Solution Code: import static java.lang.Math.max;
import static java.lang.Math.min;
import static java.lang.Math.abs;
import java.util.*;
import java.io.*;
import java.math.*;
public class Main {
public static void process() throws IOException {
int n = sc.nextInt(), m = sc.nextInt();
int arr[][] = new int[n][m];
int mat[][] = new int[n][m];
for(int i = 0; i<n; i++)arr[i] = sc.readArray(m);
mat[0][0] = arr[0][0];
int i = 0, j = 0;
while(i<n && j<n) {
if(arr[i][j] != mat[i][j]) {
System.out.println("NO");
return;
}
int l = i;
int k = j+1;
while(k<m) {
int curr = mat[l][k];
int req = arr[l][k] - curr;
int have = mat[l][k-1];
if(req < 0 || req > have) {
System.out.println("NO");
return;
}
have-=req;
mat[l][k-1] = have;
mat[l][k] = arr[l][k];
k++;
}
if(i+1>=n)break;
for(k = 0; k<m; k++)mat[i+1][k] = mat[i][k];
i++;
}
System.out.println("YES");
}
private static long INF = 2000000000000000000L, M = 1000000007, MM = 998244353;
private static int N = 0;
private static void google(int tt) {
System.out.print("Case #" + (tt) + ": ");
}
static FastScanner sc;
static FastWriter out;
public static void main(String[] args) throws IOException {
boolean oj = true;
if (oj) {
sc = new FastScanner();
out = new FastWriter(System.out);
} else {
sc = new FastScanner("input.txt");
out = new FastWriter("output.txt");
}
long s = System.currentTimeMillis();
int t = 1;
t = sc.nextInt();
int TTT = 1;
while (t-- > 0) {
process();
}
out.flush();
}
private static boolean oj = System.getProperty("ONLINE_JUDGE") != null;
private static void tr(Object... o) {
if (!oj)
System.err.println(Arrays.deepToString(o));
}
static class Pair implements Comparable<Pair> {
int x, y;
Pair(int x, int y) {
this.x = x;
this.y = y;
}
@Override
public int compareTo(Pair o) {
return Integer.compare(this.x, o.x);
}
}
static int ceil(int x, int y) {
return (x % y == 0 ? x / y : (x / y + 1));
}
static long ceil(long x, long y) {
return (x % y == 0 ? x / y : (x / y + 1));
}
static long sqrt(long z) {
long sqz = (long) Math.sqrt(z);
while (sqz * 1L * sqz < z) {
sqz++;
}
while (sqz * 1L * sqz > z) {
sqz--;
}
return sqz;
}
static int log2(int N) {
int result = (int) (Math.log(N) / Math.log(2));
return result;
}
public static long gcd(long a, long b) {
if (a > b)
a = (a + b) - (b = a);
if (a == 0L)
return b;
return gcd(b % a, a);
}
public static long lcm(long a, long b) {
return (a * b) / gcd(a, b);
}
public static int lower_bound(int[] arr, int x) {
int low = 0, high = arr.length - 1, mid = -1;
int ans = -1;
while (low <= high) {
mid = (low + high) / 2;
if (arr[mid] > x) {
high = mid - 1;
} else {
ans = mid;
low = mid + 1;
}
}
return ans;
}
public static int upper_bound(int[] arr, int x) {
int low = 0, high = arr.length - 1, mid = -1;
int ans = arr.length;
while (low < high) {
mid = (low + high) / 2;
if (arr[mid] >= x) {
ans = mid;
high = mid - 1;
} else {
low = mid + 1;
}
}
return ans;
}
static void ruffleSort(int[] a) {
Random get = new Random();
for (int i = 0; i < a.length; i++) {
int r = get.nextInt(a.length);
int temp = a[i];
a[i] = a[r];
a[r] = temp;
}
Arrays.sort(a);
}
static void ruffleSort(long[] a) {
Random get = new Random();
for (int i = 0; i < a.length; i++) {
int r = get.nextInt(a.length);
long temp = a[i];
a[i] = a[r];
a[r] = temp;
}
Arrays.sort(a);
}
static void reverseArray(int[] a) {
int n = a.length;
int arr[] = new int[n];
for (int i = 0; i < n; i++)
arr[i] = a[n - i - 1];
for (int i = 0; i < n; i++)
a[i] = arr[i];
}
static void reverseArray(long[] a) {
int n = a.length;
long arr[] = new long[n];
for (int i = 0; i < n; i++)
arr[i] = a[n - i - 1];
for (int i = 0; i < n; i++)
a[i] = arr[i];
}
public static void push(TreeMap<Integer, Integer> map, int k, int v) {
if (!map.containsKey(k))
map.put(k, v);
else
map.put(k, map.get(k) + v);
}
public static void pull(TreeMap<Integer, Integer> map, int k, int v) {
int lol = map.get(k);
if (lol == v)
map.remove(k);
else
map.put(k, lol - v);
}
public static int[] compress(int[] arr) {
ArrayList<Integer> ls = new ArrayList<Integer>();
for (int x : arr)
ls.add(x);
Collections.sort(ls);
HashMap<Integer, Integer> map = new HashMap<Integer, Integer>();
int boof = 1;
for (int x : ls)
if (!map.containsKey(x))
map.put(x, boof++);
int[] brr = new int[arr.length];
for (int i = 0; i < arr.length; i++)
brr[i] = map.get(arr[i]);
return brr;
}
public static class FastWriter {
private static final int BUF_SIZE = 1 << 13;
private final byte[] buf = new byte[BUF_SIZE];
private final OutputStream out;
private int ptr = 0;
private FastWriter() {
out = null;
}
public FastWriter(OutputStream os) {
this.out = os;
}
public FastWriter(String path) {
try {
this.out = new FileOutputStream(path);
} catch (FileNotFoundException e) {
throw new RuntimeException("FastWriter");
}
}
public FastWriter write(byte b) {
buf[ptr++] = b;
if (ptr == BUF_SIZE)
innerflush();
return this;
}
public FastWriter write(char c) {
return write((byte) c);
}
public FastWriter write(char[] s) {
for (char c : s) {
buf[ptr++] = (byte) c;
if (ptr == BUF_SIZE)
innerflush();
}
return this;
}
public FastWriter write(String s) {
s.chars().forEach(c -> {
buf[ptr++] = (byte) c;
if (ptr == BUF_SIZE)
innerflush();
});
return this;
}
private static int countDigits(int l) {
if (l >= 1000000000)
return 10;
if (l >= 100000000)
return 9;
if (l >= 10000000)
return 8;
if (l >= 1000000)
return 7;
if (l >= 100000)
return 6;
if (l >= 10000)
return 5;
if (l >= 1000)
return 4;
if (l >= 100)
return 3;
if (l >= 10)
return 2;
return 1;
}
public FastWriter write(int x) {
if (x == Integer.MIN_VALUE) {
return write((long) x);
}
if (ptr + 12 >= BUF_SIZE)
innerflush();
if (x < 0) {
write((byte) '-');
x = -x;
}
int d = countDigits(x);
for (int i = ptr + d - 1; i >= ptr; i--) {
buf[i] = (byte) ('0' + x % 10);
x /= 10;
}
ptr += d;
return this;
}
private static int countDigits(long l) {
if (l >= 1000000000000000000L)
return 19;
if (l >= 100000000000000000L)
return 18;
if (l >= 10000000000000000L)
return 17;
if (l >= 1000000000000000L)
return 16;
if (l >= 100000000000000L)
return 15;
if (l >= 10000000000000L)
return 14;
if (l >= 1000000000000L)
return 13;
if (l >= 100000000000L)
return 12;
if (l >= 10000000000L)
return 11;
if (l >= 1000000000L)
return 10;
if (l >= 100000000L)
return 9;
if (l >= 10000000L)
return 8;
if (l >= 1000000L)
return 7;
if (l >= 100000L)
return 6;
if (l >= 10000L)
return 5;
if (l >= 1000L)
return 4;
if (l >= 100L)
return 3;
if (l >= 10L)
return 2;
return 1;
}
public FastWriter write(long x) {
if (x == Long.MIN_VALUE) {
return write("" + x);
}
if (ptr + 21 >= BUF_SIZE)
innerflush();
if (x < 0) {
write((byte) '-');
x = -x;
}
int d = countDigits(x);
for (int i = ptr + d - 1; i >= ptr; i--) {
buf[i] = (byte) ('0' + x % 10);
x /= 10;
}
ptr += d;
return this;
}
public FastWriter write(double x, int precision) {
if (x < 0) {
write('-');
x = -x;
}
x += Math.pow(10, -precision) / 2;
write((long) x).write(".");
x -= (long) x;
for (int i = 0; i < precision; i++) {
x *= 10;
write((char) ('0' + (int) x));
x -= (int) x;
}
return this;
}
public FastWriter writeln(char c) {
return write(c).writeln();
}
public FastWriter writeln(int x) {
return write(x).writeln();
}
public FastWriter writeln(long x) {
return write(x).writeln();
}
public FastWriter writeln(double x, int precision) {
return write(x, precision).writeln();
}
public FastWriter write(int... xs) {
boolean first = true;
for (int x : xs) {
if (!first)
write(' ');
first = false;
write(x);
}
return this;
}
public FastWriter write(long... xs) {
boolean first = true;
for (long x : xs) {
if (!first)
write(' ');
first = false;
write(x);
}
return this;
}
public FastWriter writeln() {
return write((byte) '\n');
}
public FastWriter writeln(int... xs) {
return write(xs).writeln();
}
public FastWriter writeln(long... xs) {
return write(xs).writeln();
}
public FastWriter writeln(char[] line) {
return write(line).writeln();
}
public FastWriter writeln(char[]... map) {
for (char[] line : map)
write(line).writeln();
return this;
}
public FastWriter writeln(String s) {
return write(s).writeln();
}
private void innerflush() {
try {
out.write(buf, 0, ptr);
ptr = 0;
} catch (IOException e) {
throw new RuntimeException("innerflush");
}
}
public void flush() {
innerflush();
try {
out.flush();
} catch (IOException e) {
throw new RuntimeException("flush");
}
}
public FastWriter print(byte b) {
return write(b);
}
public FastWriter print(char c) {
return write(c);
}
public FastWriter print(char[] s) {
return write(s);
}
public FastWriter print(String s) {
return write(s);
}
public FastWriter print(int x) {
return write(x);
}
public FastWriter print(long x) {
return write(x);
}
public FastWriter print(double x, int precision) {
return write(x, precision);
}
public FastWriter println(char c) {
return writeln(c);
}
public FastWriter println(int x) {
return writeln(x);
}
public FastWriter println(long x) {
return writeln(x);
}
public FastWriter println(double x, int precision) {
return writeln(x, precision);
}
public FastWriter print(int... xs) {
return write(xs);
}
public FastWriter print(long... xs) {
return write(xs);
}
public FastWriter println(int... xs) {
return writeln(xs);
}
public FastWriter println(long... xs) {
return writeln(xs);
}
public FastWriter println(char[] line) {
return writeln(line);
}
public FastWriter println(char[]... map) {
return writeln(map);
}
public FastWriter println(String s) {
return writeln(s);
}
public FastWriter println() {
return writeln();
}
}
static class FastScanner {
private int BS = 1 << 16;
private char NC = (char) 0;
private byte[] buf = new byte[BS];
private int bId = 0, size = 0;
private char c = NC;
private double cnt = 1;
private BufferedInputStream in;
public FastScanner() {
in = new BufferedInputStream(System.in, BS);
}
public FastScanner(String s) {
try {
in = new BufferedInputStream(new FileInputStream(new File(s)), BS);
} catch (Exception e) {
in = new BufferedInputStream(System.in, BS);
}
}
private char getChar() {
while (bId == size) {
try {
size = in.read(buf);
} catch (Exception e) {
return NC;
}
if (size == -1)
return NC;
bId = 0;
}
return (char) buf[bId++];
}
public int nextInt() {
return (int) nextLong();
}
public int[] readArray(int N) {
int[] res = new int[N];
for (int i = 0; i < N; i++) {
res[i] = (int) nextLong();
}
return res;
}
public long[] readArrayLong(int N) {
long[] res = new long[N];
for (int i = 0; i < N; i++) {
res[i] = nextLong();
}
return res;
}
public int[][] readArrayMatrix(int N, int M, int Index) {
if (Index == 0) {
int[][] res = new int[N][M];
for (int i = 0; i < N; i++) {
for (int j = 0; j < M; j++)
res[i][j] = (int) nextLong();
}
return res;
}
int[][] res = new int[N][M];
for (int i = 1; i <= N; i++) {
for (int j = 1; j <= M; j++)
res[i][j] = (int) nextLong();
}
return res;
}
public long[][] readArrayMatrixLong(int N, int M, int Index) {
if (Index == 0) {
long[][] res = new long[N][M];
for (int i = 0; i < N; i++) {
for (int j = 0; j < M; j++)
res[i][j] = nextLong();
}
return res;
}
long[][] res = new long[N][M];
for (int i = 1; i <= N; i++) {
for (int j = 1; j <= M; j++)
res[i][j] = nextLong();
}
return res;
}
public long nextLong() {
cnt = 1;
boolean neg = false;
if (c == NC)
c = getChar();
for (; (c < '0' || c > '9'); c = getChar()) {
if (c == '-')
neg = true;
}
long res = 0;
for (; c >= '0' && c <= '9'; c = getChar()) {
res = (res << 3) + (res << 1) + c - '0';
cnt *= 10;
}
return neg ? -res : res;
}
public double nextDouble() {
double cur = nextLong();
return c != '.' ? cur : cur + nextLong() / cnt;
}
public double[] readArrayDouble(int N) {
double[] res = new double[N];
for (int i = 0; i < N; i++) {
res[i] = nextDouble();
}
return res;
}
public String next() {
StringBuilder res = new StringBuilder();
while (c <= 32)
c = getChar();
while (c > 32) {
res.append(c);
c = getChar();
}
return res.toString();
}
public String nextLine() {
StringBuilder res = new StringBuilder();
while (c <= 32)
c = getChar();
while (c != '\n') {
res.append(c);
c = getChar();
}
return res.toString();
}
public boolean hasNext() {
if (c > 32)
return true;
while (true) {
c = getChar();
if (c == NC)
return false;
else if (c > 32)
return true;
}
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: A number s called rare if all of its digits are divisible by K. Given a number N your task is to check if the given number is rare or not.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>Rare()</b> that takes integer N and K as arguments.
Constraints:-
1 <= N <= 100000
1 <= K <= 9Return 1 if the given number is rare else return 0.Sample Input:-
2468 2
Sample Output:-
1
Sample Input:-
234 2
Sample Output:-
0
Explanation :
3 is not divisible by 2., I have written this Solution Code: class Solution {
public static int Rare(int n, int k){
while(n>0){
if((n%10)%k!=0){
return 0;
}
n/=10;
}
return 1;
}
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: A number s called rare if all of its digits are divisible by K. Given a number N your task is to check if the given number is rare or not.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>Rare()</b> that takes integer N and K as arguments.
Constraints:-
1 <= N <= 100000
1 <= K <= 9Return 1 if the given number is rare else return 0.Sample Input:-
2468 2
Sample Output:-
1
Sample Input:-
234 2
Sample Output:-
0
Explanation :
3 is not divisible by 2., I have written this Solution Code: def Rare(N,K):
while N>0:
if(N%10)%K!=0:
return 0
N=N//10
return 1
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: A number s called rare if all of its digits are divisible by K. Given a number N your task is to check if the given number is rare or not.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>Rare()</b> that takes integer N and K as arguments.
Constraints:-
1 <= N <= 100000
1 <= K <= 9Return 1 if the given number is rare else return 0.Sample Input:-
2468 2
Sample Output:-
1
Sample Input:-
234 2
Sample Output:-
0
Explanation :
3 is not divisible by 2., I have written this Solution Code:
int Rare(int n, int k){
while(n){
if((n%10)%k!=0){
return 0;
}
n/=10;
}
return 1;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: A number s called rare if all of its digits are divisible by K. Given a number N your task is to check if the given number is rare or not.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>Rare()</b> that takes integer N and K as arguments.
Constraints:-
1 <= N <= 100000
1 <= K <= 9Return 1 if the given number is rare else return 0.Sample Input:-
2468 2
Sample Output:-
1
Sample Input:-
234 2
Sample Output:-
0
Explanation :
3 is not divisible by 2., I have written this Solution Code:
int Rare(int n, int k){
while(n){
if((n%10)%k!=0){
return 0;
}
n/=10;
}
return 1;
}, In this Programming Language: C, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a natural number N, your task is to print all the digits of the number in English words. The words have to separate by space and in lowercase English letters.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>Print_Digit()</b> that takes integer N as a parameter.
<b>Constraints:-</b>
1 ≤ N ≤ 10<sup>7</sup>Print the digits of the number as shown in the example.
<b>Note:-</b>
Print all digits in lowercase English lettersSample Input:-
1024
Sample Output:-
one zero two four
Sample Input:-
2
Sample Output:-
two, I have written this Solution Code: def Print_Digit(n):
dc = {1: "one", 2: "two", 3: "three", 4: "four",
5: "five", 6: "six", 7: "seven", 8: "eight", 9: "nine", 0: "zero"}
final_list = []
while (n > 0):
final_list.append(dc[int(n%10)])
n = int(n / 10)
for val in final_list[::-1]:
print(val, end=' '), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a natural number N, your task is to print all the digits of the number in English words. The words have to separate by space and in lowercase English letters.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>Print_Digit()</b> that takes integer N as a parameter.
<b>Constraints:-</b>
1 ≤ N ≤ 10<sup>7</sup>Print the digits of the number as shown in the example.
<b>Note:-</b>
Print all digits in lowercase English lettersSample Input:-
1024
Sample Output:-
one zero two four
Sample Input:-
2
Sample Output:-
two, I have written this Solution Code: class Solution {
public static void Print_Digits(int N){
if(N==0){return;}
Print_Digits(N/10);
int x=N%10;
if(x==1){System.out.print("one ");}
else if(x==2){System.out.print("two ");}
else if(x==3){System.out.print("three ");}
else if(x==4){System.out.print("four ");}
else if(x==5){System.out.print("five ");}
else if(x==6){System.out.print("six ");}
else if(x==7){System.out.print("seven ");}
else if(x==8){System.out.print("eight ");}
else if(x==9){System.out.print("nine ");}
else if(x==0){System.out.print("zero ");}
}
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of integers of size N. Print the third largest integer in the array in linear time.
Third largest element of the array is the third number in the array when sorted in non- increasing order.A single integer n.
Next line contains n integers separated by space.
<b>Constraints </b>
3 <= n <= 10<sup>5</sup>
-10<sup>9</sup> <= arr[i] <= 10<sup>9</sup>A single integer denoting required answer.Input:
6
-2 7 3 1 6 5
Output:
5
Explanation :
sort in non- increasing order : 7 6 5 3 1 -2, I have written this Solution Code: import java.io.*;
import java.util.*;
public class Main {
public static void main(String[] args) {
InputStream inputStream = System.in;
OutputStream outputStream = System.out;
InputReader in = new InputReader(inputStream);
OutputWriter out = new OutputWriter(outputStream);
int n=Integer.parseInt(in.next());
int a[] = new int[n];
for(int i=0;i<n;i++){
a[i]=Integer.parseInt(in.next());
}
int x=-1,y=-1,z=-1;
for(int i=0;i<n;i++){
if(x==-1 || a[i] >= a[x]){
z=y;
y=x;
x=i;
}
else if(y==-1 || a[i] >= a[y]){
z=y;
y=i;
}
else if(z == -1 || a[i] >= a[z]){
z=i;
}
}
out.print(a[z]);
out.close();
}
static class InputReader {
BufferedReader reader;
StringTokenizer tokenizer;
public InputReader(InputStream stream) {
reader = new BufferedReader(new InputStreamReader(stream), 32768);
tokenizer = null;
}
public String next() {
while (tokenizer == null || !tokenizer.hasMoreTokens()) {
try {
tokenizer = new StringTokenizer(reader.readLine());
} catch (IOException e) {
throw new RuntimeException(e);
}
}
return tokenizer.nextToken();
}
public int nextInt() {
return Integer.parseInt(next());
}
}
static class OutputWriter {
private final PrintWriter writer;
public OutputWriter(OutputStream outputStream) {
writer = new PrintWriter(new BufferedWriter(new OutputStreamWriter(outputStream)));
}
public OutputWriter(Writer writer) {
this.writer = new PrintWriter(writer);
}
public void print(Object... objects) {
for (int i = 0; i < objects.length; i++) {
if (i != 0) {
writer.print(' ');
}
writer.print(objects[i]);
}
}
public void println(Object... objects) {
print(objects);
writer.println();
}
public void close() {
writer.close();
}
public void println(int i) {
writer.println(i);
}
}
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a year(an integer variable) as input, find if it is a leap year or not.
Note: Leap year is the year that is multiple of 4. But, multiples of 100 which are not multiples of 400 are not leap years.The input contains a single integer N
<b>Constraint:</b>
1 <= n <= 10<sup>4</sup>Print "YES" if the year is a leap year else print "NO".Sample Input:
2000
Sample Output:
YES
Sample Input:
2003
Sample Output:
NO
Sample Input:
1900
Sample Output:
NO, I have written this Solution Code:
function LeapYear(year){
// write code here
// return the output using return keyword
// do not use console.log here
if ((0 != year % 4) || ((0 == year % 100) && (0 != year % 400))) {
return 0;
} else {
return 1
}
}, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a year(an integer variable) as input, find if it is a leap year or not.
Note: Leap year is the year that is multiple of 4. But, multiples of 100 which are not multiples of 400 are not leap years.The input contains a single integer N
<b>Constraint:</b>
1 <= n <= 10<sup>4</sup>Print "YES" if the year is a leap year else print "NO".Sample Input:
2000
Sample Output:
YES
Sample Input:
2003
Sample Output:
NO
Sample Input:
1900
Sample Output:
NO, I have written this Solution Code: year = int(input())
if year % 4 == 0 and not year % 100 == 0 or year % 400 == 0:
print("YES")
else:
print("NO"), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a year(an integer variable) as input, find if it is a leap year or not.
Note: Leap year is the year that is multiple of 4. But, multiples of 100 which are not multiples of 400 are not leap years.The input contains a single integer N
<b>Constraint:</b>
1 <= n <= 10<sup>4</sup>Print "YES" if the year is a leap year else print "NO".Sample Input:
2000
Sample Output:
YES
Sample Input:
2003
Sample Output:
NO
Sample Input:
1900
Sample Output:
NO, I have written this Solution Code: import java.util.Scanner;
class Main {
public static void main (String[] args)
{
//Capture the user's input
Scanner scanner = new Scanner(System.in);
//Storing the captured value in a variable
int side = scanner.nextInt();
int area = LeapYear(side);
if(area==1){
System.out.println("YES");}
else{
System.out.println("NO");}
}
static int LeapYear(int year){
if(year%400==0){return 1;}
if(year%100 != 0 && year%4==0){return 1;}
else {
return 0;}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of N integers, your task is to find the maximum value of the sum of its subarray modulo M, i. e., find the sum of each subarray mod M and print the maximum value of this after modulo operation.The first line of input contains two space-separated integers N and M, the next line of input contains N space-separated integers depicting value of the array.
<b>Constraints:-</b>
1 < = N < = 100000
1 < = M < = 10000000000
1 < = Arr[i] < = 10000000000Print the maximum value of sum modulo m.Sample Input:-
5 13
6 6 11 15 2
Sample Output:-
12
Explanation:
[6, 6] is subarray is maximum sum modulo 13
Sample Input:-
3 15
1 2 3
Sample Output:-
6
Explanation:
Max sum occurs when we take the whole array, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args)throws IOException {
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
String[] s = br.readLine().split(" ");
int N = Integer.parseInt(s[0]);
int M = Integer.parseInt(s[1]);
s = br.readLine().split(" ");
int[] prefix = new int[N];
int preSum = 0;
for(int i=0; i<N; i++){
int curr = Integer.parseInt(s[i]);
preSum += curr;
prefix[i] = preSum;
}
if(M>=prefix[N-1]){
System.out.println(prefix[N-1]);
return;
}
int maxSum = 0;
for(int i=0; i<N; i++){
for(int j=0; j<i; j++){
int curr = prefix[i]%M;
maxSum = Math.max(maxSum, curr);
curr = (prefix[i]-prefix[j])%M;
maxSum = Math.max(maxSum, curr);
if(maxSum==M-1){
break;
}
}
}
System.out.println(maxSum);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
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