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For this Question: Ram is studying in Class V and has four subjects, each subject carry 100 marks. He passed with flying colors in his exam, but when his neighbour asked how much percentage did he got in exam, he got stuck in calculation. Ram is a good student but he forgot how to calculate percentage. Help Ram to get him out of this problem.
Given four numbers a , b , c and d denoting the marks in four subjects . Calculate the overall percentage (floor value ) Ram got in exam .First line contains four variables a, b, c and d.
<b>Constraints</b>
1<= a, b, c, d <= 100
Print single line containing the percentage.Sample Input 1:
25 25 25 25
Sample Output 1:
25
Sample Input 2:
75 25 75 25
Sample Output 2:
50, 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));
String str[]=br.readLine().split(" ");
int a[]=new int[str.length];
int sum=0;
for(int i=0;i<str.length;i++)
{
a[i]=Integer.parseInt(str[i]);
sum=sum+a[i];
}
System.out.println(sum/4);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Ram is studying in Class V and has four subjects, each subject carry 100 marks. He passed with flying colors in his exam, but when his neighbour asked how much percentage did he got in exam, he got stuck in calculation. Ram is a good student but he forgot how to calculate percentage. Help Ram to get him out of this problem.
Given four numbers a , b , c and d denoting the marks in four subjects . Calculate the overall percentage (floor value ) Ram got in exam .First line contains four variables a, b, c and d.
<b>Constraints</b>
1<= a, b, c, d <= 100
Print single line containing the percentage.Sample Input 1:
25 25 25 25
Sample Output 1:
25
Sample Input 2:
75 25 75 25
Sample Output 2:
50, I have written this Solution Code: a,b,c,d = map(int,input().split())
print((a+b+c+d)*100//400), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an N*N matrix. Print the elements of the matrix in anticlockwise order (see the sample for better understanding).First line contains N.
N lines follow each containing N space seperated integers.
Constraints:-
2 <= N <= 500
1 <= Mat[i][j] <= 1000Output N*N integers in a single line separated by spaces, which are the elements of the matrix in anti-clockwise order.Sample Input
4
1 2 3 4
5 6 7 8
9 10 11 12
13 14 15 16
Sample output
1 5 9 13 14 15 16 12 8 4 3 2 6 10 11 7
Explanation:
We start from 1 , go down till 13 and then go right till 16 then go up till 4 , then we go left till 2 then down and so on in anti-clockwise fashion .
Sample Input
3
1 2 3
4 5 6
7 8 9
Sample output
1 4 7 8 9 6 3 2 5
, I have written this Solution Code: import java.util.*;
import java.lang.*;
import java.io.*;
class Main{
public static void main(String args[])throws IOException{
BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
int n = Integer.parseInt(in.readLine());
int x, y, start, end, num=n, firstcond, secondcond, thirdcond, fourthcond, flag;
if(n%2!=0)
num++;
num = num/2;
int[][] a = new int[n][n];
for(x=0; x<n; x++){
String nextLine[] = in.readLine().split(" ");
for(y=0; y<n; y++){
a[x][y] = Integer.parseInt(nextLine[y]);
}
}
start = 0;
end = n-1;
while(num>=1){
flag=0;
firstcond = secondcond = thirdcond = fourthcond = 1;
for(x=start, y=start;
(firstcond==1 || secondcond==1 || thirdcond==1 || fourthcond==1) && (x>=start && y>=start && x<=end && y<=end);
){
System.out.print(a[x][y] + " ");
if(firstcond==1){
x++;
if(x==end+1){
firstcond=0;
x--;
y++;
}
}else if(secondcond==1){
y++;
if(y==end+1){
secondcond=0;
y--;
x--;
}
}else if(thirdcond==1){
x--;
if(x==start-1){
if(flag==0)
break;
thirdcond=0;
x++;
y--;
}
flag=1;
}else{
y--;
if(y==start){
fourthcond=0;
x++;
y++;
}
}
}
start++;
end--;
num--;
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an N*N matrix. Print the elements of the matrix in anticlockwise order (see the sample for better understanding).First line contains N.
N lines follow each containing N space seperated integers.
Constraints:-
2 <= N <= 500
1 <= Mat[i][j] <= 1000Output N*N integers in a single line separated by spaces, which are the elements of the matrix in anti-clockwise order.Sample Input
4
1 2 3 4
5 6 7 8
9 10 11 12
13 14 15 16
Sample output
1 5 9 13 14 15 16 12 8 4 3 2 6 10 11 7
Explanation:
We start from 1 , go down till 13 and then go right till 16 then go up till 4 , then we go left till 2 then down and so on in anti-clockwise fashion .
Sample Input
3
1 2 3
4 5 6
7 8 9
Sample output
1 4 7 8 9 6 3 2 5
, I have written this Solution Code: n = int(input())
arr = []
for i in range(n):
j = input().split()
arr.append([int(xx) for xx in j])
def counterClockspiralPrint(m, n, arr) :
k = 0; l = 0
cnt = 0
total = m * n
while (k < m and l < n) :
if (cnt == total) :
break
for i in range(k, m) :
print(arr[i][l], end = " ")
cnt += 1
l += 1
if (cnt == total) :
break
for i in range (l, n) :
print( arr[m - 1][i], end = " ")
cnt += 1
m -= 1
if (cnt == total) :
break
if (k < m) :
for i in range(m - 1, k - 1, -1) :
print(arr[i][n - 1], end = " ")
cnt += 1
n -= 1
if (cnt == total) :
break
if (l < n) :
for i in range(n - 1, l - 1, -1) :
print( arr[k][i], end = " ")
cnt += 1
k += 1
counterClockspiralPrint(n, n, arr), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an N*N matrix. Print the elements of the matrix in anticlockwise order (see the sample for better understanding).First line contains N.
N lines follow each containing N space seperated integers.
Constraints:-
2 <= N <= 500
1 <= Mat[i][j] <= 1000Output N*N integers in a single line separated by spaces, which are the elements of the matrix in anti-clockwise order.Sample Input
4
1 2 3 4
5 6 7 8
9 10 11 12
13 14 15 16
Sample output
1 5 9 13 14 15 16 12 8 4 3 2 6 10 11 7
Explanation:
We start from 1 , go down till 13 and then go right till 16 then go up till 4 , then we go left till 2 then down and so on in anti-clockwise fashion .
Sample Input
3
1 2 3
4 5 6
7 8 9
Sample output
1 4 7 8 9 6 3 2 5
, I have written this Solution Code:
#include <bits/stdc++.h>
using namespace std;
#define N 1000
void counterClockspiralPrint( int m,
int n,
int arr[][N])
{
int i, k = 0, l = 0;
// k - starting row index
// m - ending row index
// l - starting column index
// n - ending column index
// i - iterator
// initialize the count
int cnt = 0;
// total number of
// elements in matrix
int total = m * n;
while (k < m && l < n)
{
if (cnt == total)
break;
// Print the first column
// from the remaining columns
for (i = k; i < m; ++i)
{
cout << arr[i][l] << " ";
cnt++;
}
l++;
if (cnt == total)
break;
// Print the last row from
// the remaining rows
for (i = l; i < n; ++i)
{
cout << arr[m - 1][i] << " ";
cnt++;
}
m--;
if (cnt == total)
break;
// Print the last column
// from the remaining columns
if (k < m)
{
for (i = m - 1; i >= k; --i)
{
cout << arr[i][n - 1] << " ";
cnt++;
}
n--;
}
if (cnt == total)
break;
// Print the first row
// from the remaining rows
if (l < n)
{
for (i = n - 1; i >= l; --i)
{
cout << arr[k][i] << " ";
cnt++;
}
k++;
}
}
}
int main()
{
int n;
cin>>n;
int arr[n][N];
for(int i=0;i<n;i++){
for(int j=0;j<n;j++){
cin>>arr[i][j];
}}
counterClockspiralPrint(n,n, arr);
return 0;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an N*N matrix. Print the elements of the matrix in anticlockwise order (see the sample for better understanding).First line contains N.
N lines follow each containing N space seperated integers.
Constraints:-
2 <= N <= 500
1 <= Mat[i][j] <= 1000Output N*N integers in a single line separated by spaces, which are the elements of the matrix in anti-clockwise order.Sample Input
4
1 2 3 4
5 6 7 8
9 10 11 12
13 14 15 16
Sample output
1 5 9 13 14 15 16 12 8 4 3 2 6 10 11 7
Explanation:
We start from 1 , go down till 13 and then go right till 16 then go up till 4 , then we go left till 2 then down and so on in anti-clockwise fashion .
Sample Input
3
1 2 3
4 5 6
7 8 9
Sample output
1 4 7 8 9 6 3 2 5
, I have written this Solution Code: function printAntiClockWise(mat)
{
let i, k = 0, l = 0;
let m = N;
let n = N;
let cnt = 0, total = m*n;
while(k < m && l < n)
{
if (cnt == total)
break;
// Print the first column
// from the remaining columns
for (i = k; i < m; ++i)
{
process.stdout.write(mat[i][l] + " ");
cnt++;
}
l++;
if (cnt == total)
break;
// Print the last row from
// the remaining rows
for (i = l; i < n; ++i)
{
process.stdout.write(mat[m - 1][i] + " ");
cnt++;
}
m--;
if (cnt == total)
break;
// Print the last column
// from the remaining columns
if (k < m)
{
for (i = m - 1; i >= k; --i)
{
process.stdout.write(mat[i][n - 1] + " ");
cnt++;
}
n--;
}
if (cnt == total)
break;
// Print the first row
// from the remaining rows
if (l < n)
{
for (i = n - 1; i >= l; --i)
{
process.stdout.write(mat[k][i] + " ");
cnt++;
}
k++;
}
}
}, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Ram is studying in Class V and has four subjects, each subject carry 100 marks. He passed with flying colors in his exam, but when his neighbour asked how much percentage did he got in exam, he got stuck in calculation. Ram is a good student but he forgot how to calculate percentage. Help Ram to get him out of this problem.
Given four numbers a , b , c and d denoting the marks in four subjects . Calculate the overall percentage (floor value ) Ram got in exam .First line contains four variables a, b, c and d.
<b>Constraints</b>
1<= a, b, c, d <= 100
Print single line containing the percentage.Sample Input 1:
25 25 25 25
Sample Output 1:
25
Sample Input 2:
75 25 75 25
Sample Output 2:
50, 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));
String str[]=br.readLine().split(" ");
int a[]=new int[str.length];
int sum=0;
for(int i=0;i<str.length;i++)
{
a[i]=Integer.parseInt(str[i]);
sum=sum+a[i];
}
System.out.println(sum/4);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Ram is studying in Class V and has four subjects, each subject carry 100 marks. He passed with flying colors in his exam, but when his neighbour asked how much percentage did he got in exam, he got stuck in calculation. Ram is a good student but he forgot how to calculate percentage. Help Ram to get him out of this problem.
Given four numbers a , b , c and d denoting the marks in four subjects . Calculate the overall percentage (floor value ) Ram got in exam .First line contains four variables a, b, c and d.
<b>Constraints</b>
1<= a, b, c, d <= 100
Print single line containing the percentage.Sample Input 1:
25 25 25 25
Sample Output 1:
25
Sample Input 2:
75 25 75 25
Sample Output 2:
50, I have written this Solution Code: a,b,c,d = map(int,input().split())
print((a+b+c+d)*100//400), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Newton is highly ambitious, and he currently has 10 rupees. He goes to the casino and plays against an infinitely rich man (never ending money) in order to win huge money.
The rules of the game are super simple. There are N turns and the following procedure is followed in each turn.
<ul>
<li>If one of the player has 0 money, nothing happens.
<li> Else, the losing player loses 1 rupee and the winning player wins 1 rupee.
</ul>
The probability of Newton winning in any turn is P/Q. Find the expected money he would have, if N is 5000<sup>5000<sup>5000</sup></sup>.The first and the only line of input contains two integers P and Q.
Constraints
1 <= P <= Q <= 1000000If the expected value is greater than 10^18, output 10^18, else output floor(expected value).Sample Input
1 1000000
Sample Output
0
Explanation
Given the probability of Newton winning is 0.000001.
It can be proved that the floor of expected money he has after infinite number of turns is 0., I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define sd(x) scanf("%d", &x)
#define sz(v) (int) v.size()
#define pr(v) For(i, 0, sz(v)) {cout<<v[i]<<" ";} cout<<endl;
#define slld(x) scanf("%lld", &x)
#define all(x) x.begin(), x.end()
#define For(i, st, en) for(ll 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 pb push_back
#define ll long long
#define ld long double
#define int long long
#define double long double
#define mp make_pair
#define F first
#define S second
typedef pair<int, int> pii;
typedef vector<int> vi;
#define pi 3.141592653589793238
const int MOD = 1e9+7;
const int INF = 1LL<<60;
const int N = 2e5+5;
// it's swapnil07 ;)
#ifdef SWAPNIL07
#define trace(...) __f(#__VA_ARGS__, __VA_ARGS__)
template <typename Arg1>
void __f(const char* name, Arg1&& arg1){
cout << name << " : " << arg1 << endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args){
const char* comma = strchr(names + 1, ',');cout.write(names, comma - names) << " : " << arg1<<" | ";__f(comma+1, args...);
}
int begtime = clock();
#define end_routine() cout << "\n\nTime elapsed: " << (clock() - begtime)*1000/CLOCKS_PER_SEC << " ms\n\n";
#else
#define endl '\n'
#define trace(...)
#define end_routine()
#endif
void solve(){
int p, q; cin>>p>>q;
q = q-p;
if(p < q){
cout<<0;
}
else if(p==q)
cout<<10;
else{
cout<<1000000000000000000LL;
}
}
signed main()
{
fast
#ifdef SWAPNIL07
freopen("input.txt","r",stdin);
freopen("output.txt","w",stdout);
#endif
int t=1;
// cin>>t;
while(t--){
solve();
cout<<"\n";
}
return 0;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given head which is a reference node to a doubly- linked list of integers and an integer K. Complete the function <b>swapNodes</b> which swaps the values of nodes at distance K from beginning and end. The first node is at distance 1 from beginning.User task:
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>swapNodes</b> that takes the head of the linked list and node distance K as parameters.
Constraints:
1 <= Number of nodes <= 10000.
1 <= K <= Number of nodesThe function need not return anything.Sample Input 1:-
5 2
1 2 3 4 5
Sample Output 1:-
1 4 3 2 5
Explanation:
The node at distance 2 from beginning is 2.
The node at distance 2 from end is 4.
We will swap 2 and 4.
Sample Input 2:-
3 2
1 2 3
Sample Output 2:-
1 2 3
Explanation:
The node at distance 2 from beginning is 2.
The node at distance 2 from end is 2.
We will swap 2 and 2. So, the linked list remains as it is., I have written this Solution Code: static void swapNodes(Node head,int k)
{
Node temp=head;
while(temp.next!=null)temp=temp.next;
k--;
while(k>0){
head=head.next;
temp=temp.prev;
k--;
}
int f=temp.data;
temp.data=head.data;
head.data=f;
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You went to shopping. You bought an item worth N rupees. What is the minimum change that you can get from the shopkeeper if you possess only 200 and 500 rupees notes.
Eg: If N = 678, the minimum change you can receive is 22, if you pay the shopkeeper a 500 and a 200 rupee note. You can show that no other combination can lead to a change lesser than this like (200, 200, 200, 200) or (500, 500).
Note: You have infinite number of 200 and 500 rupees notes. Enjoy, XD.The first and the only line of input contains an integer N.
Constraints
1 <= N <= 1000000000Output a single integer, the minimum amount of change that you will receive.Sample Input
678
Sample Output
22
Sample Input
900
Sample Output
0, I have written this Solution Code: def sample(n):
if n<200:
print(200-n)
elif n<400:
print(400-n)
elif n<500:
print(500-n)
else:
div=n//100
if div*100==n:
print(0)
else:
print((div+1)*100-n)
n=int(input())
sample(n), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You went to shopping. You bought an item worth N rupees. What is the minimum change that you can get from the shopkeeper if you possess only 200 and 500 rupees notes.
Eg: If N = 678, the minimum change you can receive is 22, if you pay the shopkeeper a 500 and a 200 rupee note. You can show that no other combination can lead to a change lesser than this like (200, 200, 200, 200) or (500, 500).
Note: You have infinite number of 200 and 500 rupees notes. Enjoy, XD.The first and the only line of input contains an integer N.
Constraints
1 <= N <= 1000000000Output a single integer, the minimum amount of change that you will receive.Sample Input
678
Sample Output
22
Sample Input
900
Sample Output
0, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define sd(x) scanf("%d", &x)
#define sz(v) (int) v.size()
#define pr(v) For(i, 0, sz(v)) {cout<<v[i]<<" ";} cout<<endl;
#define slld(x) scanf("%lld", &x)
#define all(x) x.begin(), x.end()
#define For(i, st, en) for(ll 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 pb push_back
#define ll long long
#define ld long double
#define int long long
#define double long double
#define mp make_pair
#define F first
#define S second
typedef pair<int, int> pii;
typedef vector<int> vi;
#define pi 3.141592653589793238
const int MOD = 1e9+7;
const int INF = 1LL<<60;
const int N = 2e5+5;
// it's swapnil07 ;)
#ifdef SWAPNIL07
#define trace(...) __f(#__VA_ARGS__, __VA_ARGS__)
template <typename Arg1>
void __f(const char* name, Arg1&& arg1){
cout << name << " : " << arg1 << endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args){
const char* comma = strchr(names + 1, ',');cout.write(names, comma - names) << " : " << arg1<<" | ";__f(comma+1, args...);
}
int begtime = clock();
#define end_routine() cout << "\n\nTime elapsed: " << (clock() - begtime)*1000/CLOCKS_PER_SEC << " ms\n\n";
#else
#define endl '\n'
#define trace(...)
#define end_routine()
#endif
void solve(){
int n; cin>>n;
if(n <= 200){
cout<<200-n;
return;
}
if(n <= 400){
cout<<400-n;
return;
}
int ans = (100-n%100)%100;
cout<<ans;
}
signed main()
{
fast
#ifdef SWAPNIL07
freopen("input.txt","r",stdin);
freopen("output.txt","w",stdout);
#endif
int t=1;
// cin>>t;
while(t--){
solve();
cout<<"\n";
}
return 0;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You went to shopping. You bought an item worth N rupees. What is the minimum change that you can get from the shopkeeper if you possess only 200 and 500 rupees notes.
Eg: If N = 678, the minimum change you can receive is 22, if you pay the shopkeeper a 500 and a 200 rupee note. You can show that no other combination can lead to a change lesser than this like (200, 200, 200, 200) or (500, 500).
Note: You have infinite number of 200 and 500 rupees notes. Enjoy, XD.The first and the only line of input contains an integer N.
Constraints
1 <= N <= 1000000000Output a single integer, the minimum amount of change that you will receive.Sample Input
678
Sample Output
22
Sample Input
900
Sample Output
0, I have written this Solution Code:
import java.util.*;
import java.lang.*;
import java.io.*;
class Main
{
public static void main (String[] args) throws java.lang.Exception
{
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
int ans=0;
if(n <= 200){
ans = 200-n;
}
else if(n <= 400){
ans=400-n;
}
else{
ans = (100-n%100)%100;
}
System.out.print(ans);
}
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array <b>arr[]</b> of <b>N</b> distinct integers, check if this array is Sorted and Rotated clockwise. A sorted array is not considered sorted and rotated, i.e., there should be at least one rotation.
<b>Note:-</b>
The array can be sorted both increasingly and decreasinglyThe first line of input contains the number of test cases T. Each test case contains 2 lines, the first line contains N, the number of elements in the array, and the second line contains N space-separated elements of the array.
<b>Constraints:</b>
1 <= T <= 50
3 <= N <= 10^3
1 <= A[i] <= 10^4
Print "<b>Yes</b>" if the given array is sorted and rotated, else Print "<b>No</b>", without Inverted commas.Sample Input:
2
4
3 4 1 2
3
1 3 2
Sample Output:
Yes
Yes
<b>Explanation:</b>
Testcase 1: The array is sorted (1, 2, 3, 4) and rotated twice (3, 4, 1, 2).
Testcase 2: The array is sorted (3, 2, 1) and rotated once (1, 3, 2)., 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];
bool check_increasing(int n){
int id = 0;
for(int i = 2; i <= n; i++){
if(a[i] < a[i-1]){
id = i;
break;
}
}
if(id == 0) return 0;
int cur = id+1;
while(cur != id+n){
if(a[cur] <= a[cur-1])
return 0;
cur++;
}
return 1;
}
bool check_decreasing(int n){
int id = 0;
for(int i = 2; i <= n; i++){
if(a[i] > a[i-1]){
id = i;
break;
}
}
if(id == 0) return 0;
int cur = id+1;
while(cur != id+n){
if(a[cur] >= a[cur-1])
return 0;
cur++;
}
return 1;
}
signed main() {
IOS;
int t; cin >> t;
while(t--){
int n; cin >> n;
for(int i = 1; i <= n; i++)
cin >> a[i], a[i+n] = a[i];
if(check_increasing(n))
cout << "Yes" << endl;
else if(check_decreasing(n))
cout << "Yes" << endl;
else
cout << "No" << endl;
}
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array <b>arr[]</b> of <b>N</b> distinct integers, check if this array is Sorted and Rotated clockwise. A sorted array is not considered sorted and rotated, i.e., there should be at least one rotation.
<b>Note:-</b>
The array can be sorted both increasingly and decreasinglyThe first line of input contains the number of test cases T. Each test case contains 2 lines, the first line contains N, the number of elements in the array, and the second line contains N space-separated elements of the array.
<b>Constraints:</b>
1 <= T <= 50
3 <= N <= 10^3
1 <= A[i] <= 10^4
Print "<b>Yes</b>" if the given array is sorted and rotated, else Print "<b>No</b>", without Inverted commas.Sample Input:
2
4
3 4 1 2
3
1 3 2
Sample Output:
Yes
Yes
<b>Explanation:</b>
Testcase 1: The array is sorted (1, 2, 3, 4) and rotated twice (3, 4, 1, 2).
Testcase 2: The array is sorted (3, 2, 1) and rotated once (1, 3, 2)., I have written this Solution Code:
cases = int(input())
for _ in range(cases):
size = int(input())
orig = list(map(int,input().split()))
givenList = orig.copy()
givenList.sort()
flag = False
for _ in range(size-1):
ele = givenList.pop()
givenList.insert(0,ele)
if givenList == orig:
print("Yes")
flag = True
break
if not flag:
givenList.sort()
givenList.reverse()
for _ in range(size-1):
ele = givenList.pop()
givenList.insert(0,ele)
if givenList == orig:
print("Yes")
flag = True
break
if not flag:
print("No"), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array <b>arr[]</b> of <b>N</b> distinct integers, check if this array is Sorted and Rotated clockwise. A sorted array is not considered sorted and rotated, i.e., there should be at least one rotation.
<b>Note:-</b>
The array can be sorted both increasingly and decreasinglyThe first line of input contains the number of test cases T. Each test case contains 2 lines, the first line contains N, the number of elements in the array, and the second line contains N space-separated elements of the array.
<b>Constraints:</b>
1 <= T <= 50
3 <= N <= 10^3
1 <= A[i] <= 10^4
Print "<b>Yes</b>" if the given array is sorted and rotated, else Print "<b>No</b>", without Inverted commas.Sample Input:
2
4
3 4 1 2
3
1 3 2
Sample Output:
Yes
Yes
<b>Explanation:</b>
Testcase 1: The array is sorted (1, 2, 3, 4) and rotated twice (3, 4, 1, 2).
Testcase 2: The array is sorted (3, 2, 1) and rotated once (1, 3, 2)., I have written this Solution Code: import java.util.*;
import java.io.*;
import java.lang.*;
class Main {
public static void main (String[] args) throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int t = Integer.parseInt(br.readLine().trim()); //Inputting the testcases
while(t-->0){
long n = Long.parseLong(br.readLine());
int arr[] = new int[(int)n];
String inputLine[] = br.readLine().trim().split("\\s+");
for(long i=0; i<n; i++){
arr[(int)i] = Integer.parseInt(inputLine[(int)i]);
}
long mini = Integer.MAX_VALUE, maxi = Integer.MIN_VALUE;
long max_index = 0, min_index = 0;
for(long i=0; i<n; i++){
if(maxi < arr[(int)i]){
maxi = arr[(int)i];
max_index = i;
}
if(mini > arr[(int)i]){
mini = arr[(int)i];
min_index = i;
}
}
int flag = 0;
if(max_index == min_index -1)
flag = 1;
else if(min_index == max_index - 1)
flag = -1;
if(flag == 1){
for(long i = 1; flag==1 && i<=max_index; ++i){
if(arr[(int)i-1] >= arr[(int)i])
flag = 0;
}
for(long i = min_index+1; flag==1 && i<n; ++i){
if(arr[(int)i-1] >= arr[(int)i])
flag = 0;
}
if(arr[0]<=arr[(int)n-1])
flag = 0;
} else if(flag == -1){
for(long i = 1; flag ==-1 && i<=min_index; ++i){
if(arr[(int)i-1] <= arr[(int)i])
flag = 0;
}
for(long i = max_index+1; flag==-1 && i<n; ++i){
if(arr[(int)i-1] <= arr[(int)i])
flag = 0;
}
if(arr[0]>=arr[(int)n-1])
flag = 0;
}
if(flag == 0)
System.out.println("No");
else
System.out.println("Yes");
}
}
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given two integers a and b, your task is to check following conditions:-
1. If a <= 10 and b >= 10 (Logical AND).
2. Atleast one from a or b will be even (Logical OR).
3. if a is not equal to b (Logical NOT).The first line of the input contains 2 integers a and b.
<b>Constraints:</b>
1 <= a, b <= 100Print the string <b>"true"</b> if the condition holds in each function else <b>"false"</b> .
Sample Input:-
3 12
Sample Output:-
true true true
Explanation
So a = 3 and b = 12, so a<=10 and b>=10 hence first condition true, a is not even but b is even so atleast one of them is even hence true, third a != b which is also true hence the final output comes true true true.
Sample Input:-
10 10
Sample Output:-
true true false
, I have written this Solution Code: a, b = list(map(int, input().split(" ")))
print(str(a <= 10 and b >= 10).lower(), end=' ')
print(str(a % 2 == 0 or b % 2 == 0).lower(), end=' ')
print(str(not a == b).lower()), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given two integers a and b, your task is to check following conditions:-
1. If a <= 10 and b >= 10 (Logical AND).
2. Atleast one from a or b will be even (Logical OR).
3. if a is not equal to b (Logical NOT).The first line of the input contains 2 integers a and b.
<b>Constraints:</b>
1 <= a, b <= 100Print the string <b>"true"</b> if the condition holds in each function else <b>"false"</b> .
Sample Input:-
3 12
Sample Output:-
true true true
Explanation
So a = 3 and b = 12, so a<=10 and b>=10 hence first condition true, a is not even but b is even so atleast one of them is even hence true, third a != b which is also true hence the final output comes true true true.
Sample Input:-
10 10
Sample Output:-
true true false
, I have written this Solution Code: import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;
class Main {
static boolean Logical_AND(int a, int b){
if(a<=10 && b>=10){
return true;}
return false;}
static boolean Logical_OR(int a, int b){
if(a%2==0 || b%2==0){
return true;}
return false;}
static boolean Logical_NOT(int a, int b){
if(a!=b){
return true;}
return false;}
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int a=in.nextInt();
int b=in.nextInt();
System.out.print(Logical_AND(a, b)+" ");
System.out.print(Logical_OR(a,b)+" ");
System.out.print(Logical_NOT(a,b)+" ");
}
}, In this Programming Language: Java, 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: 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: Given an array A of N integers. You have to find whether A has some subsequence of length at least 3 that is a palindrome.First line of the input contains an integer N.
The second line of the input contains N space seperated integers.
Constraints:
1 <= N <= 5000
1 <= A<sub>i</sub> <= 10<sup>9</sup>Print "YES" if A has some subseqeuence of length at least 3 that is a palindrome, else print "NO", without the quotes.Sample Input:
5
1 5 2 2 1
Sample Output:
YES
Explanation:
We pick the elements at indices 1, 3, 4, 5 (1- indexed) making the subsequence: [1, 2, 2, 1], which is a palindrome., I have written this Solution Code: def sunpal(n,a):
ok=False
for i in range(n):
for j in range(i+2,n):
if a[i]==a[j]:
ok=True
return("YES" if ok else "NO")
if __name__=="__main__":
n=int(input())
a=input().split()
res=sunpal(n,a)
print(res), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A of N integers. You have to find whether A has some subsequence of length at least 3 that is a palindrome.First line of the input contains an integer N.
The second line of the input contains N space seperated integers.
Constraints:
1 <= N <= 5000
1 <= A<sub>i</sub> <= 10<sup>9</sup>Print "YES" if A has some subseqeuence of length at least 3 that is a palindrome, else print "NO", without the quotes.Sample Input:
5
1 5 2 2 1
Sample Output:
YES
Explanation:
We pick the elements at indices 1, 3, 4, 5 (1- indexed) making the subsequence: [1, 2, 2, 1], which is a palindrome., I have written this Solution Code: #include<bits/stdc++.h>
using namespace std;
#define fast ios_base::sync_with_stdio(false); cin.tie(NULL);
#define int long long
#define pb push_back
#define ff first
#define ss second
#define endl '\n'
#define all(a) a.begin(), a.end()
#define rall(a) a.rbegin(), a.rend()
using T = pair<int, int>;
typedef long double ld;
const int mod = 1e9 + 7;
const int INF = 1e18;
void solve(){
int N;
cin >> N;
map<int, int> mp;
string ans = "NO";
for(int i = 1; i <= N; i++) {
int a;
cin >> a;
if(mp[a] != 0 and mp[a] <= i - 2) {
ans = "YES";
}
if(mp[a] == 0) mp[a] = i;
}
cout << ans;
}
signed main(){
fast
int t = 1;
// cin >> t;
for(int i = 1; i <= t; i++){
solve();
if(i != t) cout << endl;
}
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: There are n cities in the universe and our beloved Spider-Man is in city 1. He doesn't like to travel by vehicles, so he shot webs forming edges between some pairs of cities. Eventually, there were m edges and each had some cost associated with it.
Spider-Man now defines the cost of a path p from cities p<sub>1</sub> to p<sub>k</sub> as w<sub>1</sub> + 2w<sub>2</sub> + 3w<sub>3</sub> . . . + (k-1)*w<sub>k-1</sub>, where w<sub>i</sub> is the cost of an edge from p<sub>i</sub> to p<sub>i+1</sub>.
Thus, the minimum distance between cities i and j is the smallest cost of a path starting from i and ending at j.
Find the minimum distance from city 1 to all the cities i (1 ≤ i ≤ n). If there exists no way to go from city 1 to city i, print -1.
<b>Note: </b>
All the edges are bidirectional. There may be multiple edges and self-loops in the input.The first line contains two space separated integers n and m - the number of nodes and edges respectively.
The next m lines contain three-space separated integers x, y, w - representing an edge between x and y with cost w.
<b>Constraints:</b>
1 ≤ n ≤ 3000
0 ≤ m ≤ 10000
1 ≤ x, y ≤ n
1 ≤ w ≤ 10<sup>9</sup>Output n lines. In the i<sup>th</sup> line, output the minimum distance from city 1 to the i<sup>th</sup> city. If there exists no such path, output -1.Sample Input
6 5
2 4 3
2 3 4
2 1 2
2 5 6
1 5 2
Sample Output
0
2
10
8
2
-1
Explanation:
Shortest path from 1 to 3 is (1->2->3) with total weight= 1*2+2*4=10
Shortest path from 1 to 5 is (1->5) with total weight= 1*2=2
There doesn't exist any path from 1 to 6 so print -1
, I have written this Solution Code:
import sys
from collections import defaultdict
from heapq import heappush, heappop
n, m = map(int, input().split())
d = defaultdict(list)
dist = [sys.maxsize]*n
dist[0] = 0
for _ in range(m):
start, dest, wt = map(int, input().split())
d[start-1].append((dest-1, wt))
d[dest-1].append((start-1, wt))
heap = [(0, 0, 0)]
while heap:
count, cost, u= heappop(heap)
for vertex, weight in d[u]:
if dist[vertex] > cost + weight*(count+1):
dist[vertex] = cost + weight*(count+1)
heappush(heap, (count+1, dist[vertex], vertex))
for d in dist:
if d == sys.maxsize:
print(-1)
else:
print(d), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: There are n cities in the universe and our beloved Spider-Man is in city 1. He doesn't like to travel by vehicles, so he shot webs forming edges between some pairs of cities. Eventually, there were m edges and each had some cost associated with it.
Spider-Man now defines the cost of a path p from cities p<sub>1</sub> to p<sub>k</sub> as w<sub>1</sub> + 2w<sub>2</sub> + 3w<sub>3</sub> . . . + (k-1)*w<sub>k-1</sub>, where w<sub>i</sub> is the cost of an edge from p<sub>i</sub> to p<sub>i+1</sub>.
Thus, the minimum distance between cities i and j is the smallest cost of a path starting from i and ending at j.
Find the minimum distance from city 1 to all the cities i (1 ≤ i ≤ n). If there exists no way to go from city 1 to city i, print -1.
<b>Note: </b>
All the edges are bidirectional. There may be multiple edges and self-loops in the input.The first line contains two space separated integers n and m - the number of nodes and edges respectively.
The next m lines contain three-space separated integers x, y, w - representing an edge between x and y with cost w.
<b>Constraints:</b>
1 ≤ n ≤ 3000
0 ≤ m ≤ 10000
1 ≤ x, y ≤ n
1 ≤ w ≤ 10<sup>9</sup>Output n lines. In the i<sup>th</sup> line, output the minimum distance from city 1 to the i<sup>th</sup> city. If there exists no such path, output -1.Sample Input
6 5
2 4 3
2 3 4
2 1 2
2 5 6
1 5 2
Sample Output
0
2
10
8
2
-1
Explanation:
Shortest path from 1 to 3 is (1->2->3) with total weight= 1*2+2*4=10
Shortest path from 1 to 5 is (1->5) with total weight= 1*2=2
There doesn't exist any path from 1 to 6 so print -1
, I have written this Solution Code: #include<bits/stdc++.h>
using namespace std;
#define int long long
const int INF =1e18;
vector<tuple<int, int, int>> adj;
void solve()
{
int n, m;
cin>>n>>m;
// assert(1<=n && n<=3000);
// assert(0<=m && m<=10000);
adj.resize(n);
for(int i = 0;i<m;i++)
{
int x, y, w;
cin>>x>>y>>w;
x--;
y--;
// assert(0<=x && x<n);
// assert(0<=y && y<n);
// assert(1<=w && w<=1e9);
adj.push_back({x, y, w});
adj.push_back({y, x, w});
}
vector<int> dp_old(n, INF);
vector<int> dp_new(n, INF);
dp_old[0] = 0;
for(int i = 1;i<=n;i++)
{
fill(dp_new.begin(), dp_new.end(), INF);
for(auto [x, y,w]:adj)
{
dp_new[y]= min({dp_new[y], dp_old[x] + i * w});
}
for(int j = 0;j<n;j++)
dp_new[j] = min(dp_new[j], dp_old[j]);
swap(dp_new, dp_old);
}
for(int i = 0;i<n;i++)
{
if(dp_old[i] == INF)
dp_old[i] = -1;
cout<<dp_old[i]<<"\n";
}
}
signed main()
{
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cerr.tie(NULL);
#ifndef ONLINE_JUDGE
if (fopen("INPUT.txt", "r"))
{
freopen("INPUT.txt", "r", stdin);
freopen("OUTPUT.txt", "w", stdout);
}
#endif
solve();
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: There are n cities in the universe and our beloved Spider-Man is in city 1. He doesn't like to travel by vehicles, so he shot webs forming edges between some pairs of cities. Eventually, there were m edges and each had some cost associated with it.
Spider-Man now defines the cost of a path p from cities p<sub>1</sub> to p<sub>k</sub> as w<sub>1</sub> + 2w<sub>2</sub> + 3w<sub>3</sub> . . . + (k-1)*w<sub>k-1</sub>, where w<sub>i</sub> is the cost of an edge from p<sub>i</sub> to p<sub>i+1</sub>.
Thus, the minimum distance between cities i and j is the smallest cost of a path starting from i and ending at j.
Find the minimum distance from city 1 to all the cities i (1 ≤ i ≤ n). If there exists no way to go from city 1 to city i, print -1.
<b>Note: </b>
All the edges are bidirectional. There may be multiple edges and self-loops in the input.The first line contains two space separated integers n and m - the number of nodes and edges respectively.
The next m lines contain three-space separated integers x, y, w - representing an edge between x and y with cost w.
<b>Constraints:</b>
1 ≤ n ≤ 3000
0 ≤ m ≤ 10000
1 ≤ x, y ≤ n
1 ≤ w ≤ 10<sup>9</sup>Output n lines. In the i<sup>th</sup> line, output the minimum distance from city 1 to the i<sup>th</sup> city. If there exists no such path, output -1.Sample Input
6 5
2 4 3
2 3 4
2 1 2
2 5 6
1 5 2
Sample Output
0
2
10
8
2
-1
Explanation:
Shortest path from 1 to 3 is (1->2->3) with total weight= 1*2+2*4=10
Shortest path from 1 to 5 is (1->5) with total weight= 1*2=2
There doesn't exist any path from 1 to 6 so print -1
, I have written this Solution Code: import java.io.*;import java.util.*;import java.math.*;import static java.lang.Math.*;import static java.
util.Map.*;import static java.util.Arrays.*;import static java.util.Collections.*;
import static java.lang.System.*;
public class Main
{
public void tq()throws Exception
{
st=new StringTokenizer(bq.readLine());
int tq=1;
sb=new StringBuilder(2000000);
o:
while(tq-->0)
{
int n=i();
int m=i();
LinkedList<int[]> l[]=new LinkedList[n];
for(int x=0;x<n;x++)l[x]=new LinkedList<>();
for(int x=0;x<m;x++)
{
int a=i()-1;
int b=i()-1;
int c=i();
l[a].add(new int[]{b,c});
l[b].add(new int[]{a,c});
}
long d[]=new long[n];
for(int x=0;x<n;x++)d[x]=maxl;
d[0]=0l;
PriorityQueue<long[]> p=new PriorityQueue<>(5000,(a,b)->a[2]-b[2]<1l?-1:1);
p.add(new long[]{0l,0,0});
while(p.size()>0)
{
long r[]=p.poll();
long di=r[0];
int no=(int)r[1];
long mu=r[2];
for(int e[]:l[no])
{
int node=e[0];
int w=e[1];
long de=di+w*(mu+1);
if(d[node]>de)
{
d[node]=de;
p.add(new long[]{de,node,mu+1});
}
}
}
for(long x:d)
{
sl(x==maxl?-1:x);
}
}
p(sb);
}
int di[][]={{-1,0},{1,0},{0,-1},{0,1}};
int de[][]={{-1,0},{1,0},{0,-1},{0,1},{-1,-1},{1,1},{-1,1},{1,-1}};
long mod=1000000007l;int max=Integer.MAX_VALUE,min=Integer.MIN_VALUE;long maxl=Long.MAX_VALUE, minl=Long.
MIN_VALUE;BufferedReader bq=new BufferedReader(new InputStreamReader(in));StringTokenizer st;
StringBuilder sb;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];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];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];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 l(int v){return 31-Integer.numberOfLeadingZeros(v);}
long l(long v){return 63-Long.numberOfLeadingZeros(v);}int sq(int a){return (int)sqrt(a);}long sq(long a)
{return (long)sqrt(a);}long gcd(long a,long b){while(b>0l){long c=a%b;a=b;b=c;}return a;}int gcd(int a,int b)
{while(b>0){int c=a%b;a=b;b=c;}return a;}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(bq.readLine());return Integer.parseInt(st.nextToken());}long l()throws IOException
{if(!st.hasMoreTokens())st=new StringTokenizer(bq.readLine());return Long.parseLong(st.nextToken());}String
s()throws IOException {if (!st.hasMoreTokens())st=new StringTokenizer(bq.readLine());return st.nextToken();}
double d()throws IOException{if(!st.hasMoreTokens())st=new StringTokenizer(bq.readLine());return Double.
parseDouble(st.nextToken());}void p(Object p){out.print(p);}void p(String p){out.print(p);}void p(int p)
{out.print(p);}void p(double p){out.print(p);}void p(long p){out.print(p);}void p(char p){out.print(p);}void
p(boolean p){out.print(p);}void pl(Object p){out.println(p);}void pl(String p){out.println(p);}void pl(int p)
{out.println(p);}void pl(char p){out.println(p);}void pl(double p){out.println(p);}void pl(long p){out.
println(p);}void pl(boolean p)
{out.println(p);}void pl(){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(bq.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
(bq.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(bq.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(bq.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(bq.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
(bq.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(bq.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(bq.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=bq.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(' ');}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(' ');}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(' ');}out.println(sb);}
void p(double ar[])
{StringBuilder sb=new StringBuilder(2*ar.length);for(double a:ar){sb.append(a);
sb.append(' ');}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(' ');}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 an index K, return the K-th row of <a href = "https://en.wikipedia.org/wiki/Pascal%27s_triangle">Pascal’s triangle</a>.
You must print the K-th row modulo 10<sup>9</sup> + 7.
<b>The rows of Pascal's triangle are conventionally enumerated starting with row K=0 at the top (the 0th row) </b>The only line of input contains the input K.
<b>Constraints:-</b>
0 ≤ K ≤ 3000
Print k-th row of Pascal's Triangle containing k+1 integers modulo 10^9+7.Sample Input :
3
Sample Output:
1 3 3 1, 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 k = sc.nextInt();
int digit = 1;
int m = 1000000007;;
int arr[] = new int[k+1];
arr[0] = 1;
arr[1] = 1;
for(int i=2; i<=k; i++){
for(int j=i;j>=1;j--){
arr[j] += arr[j-1];
arr[j] = arr[j]%m;
}
}
for(int i=0;i<=k;i++){
System.out.print(arr[i] + " ");
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an index K, return the K-th row of <a href = "https://en.wikipedia.org/wiki/Pascal%27s_triangle">Pascal’s triangle</a>.
You must print the K-th row modulo 10<sup>9</sup> + 7.
<b>The rows of Pascal's triangle are conventionally enumerated starting with row K=0 at the top (the 0th row) </b>The only line of input contains the input K.
<b>Constraints:-</b>
0 ≤ K ≤ 3000
Print k-th row of Pascal's Triangle containing k+1 integers modulo 10^9+7.Sample Input :
3
Sample Output:
1 3 3 1, I have written this Solution Code: def generateNthRow (N):
prev = 1
print(prev, end = '')
for i in range(1, N + 1):
curr = (prev * (N - i + 1)) // i
print("", curr%1000000007, end = '')
prev = curr
N = int(input())
generateNthRow(N), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an index K, return the K-th row of <a href = "https://en.wikipedia.org/wiki/Pascal%27s_triangle">Pascal’s triangle</a>.
You must print the K-th row modulo 10<sup>9</sup> + 7.
<b>The rows of Pascal's triangle are conventionally enumerated starting with row K=0 at the top (the 0th row) </b>The only line of input contains the input K.
<b>Constraints:-</b>
0 ≤ K ≤ 3000
Print k-th row of Pascal's Triangle containing k+1 integers modulo 10^9+7.Sample Input :
3
Sample Output:
1 3 3 1, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define mod 1000000007
signed main()
{
int t,n;
t=1;
vector<vector<long long> > v(3002);
for(int i=0; i<=3001; i++)
{
for (int j=0; j<=i; j++)
{
if (j==0 || j==i) v[i].push_back(1);
else v[i].push_back((v[i-1][j]%mod + v[i-1][j-1]%mod)%mod);
}
}
while (t--)
{
cin>>n;
n=n+1;
for (int i=0; i<v[n-1].size(); i++) cout<<v[n-1][i]<<" ";
cout<<"\n";
}
return 0;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Consider the integer sequence A = {1, 2, 3, ...., N} i.e. the first N natural numbers in order.
You are now given two integers, L and S. Determine whether there exists a subarray with length L and sum S after removing <b>at most one</b> element from A.
A <b>subarray</b> of an array is a non-empty sequence obtained by removing zero or more elements from the front of the array, and zero or more elements from the back of the array.The first line contains a single integer T, the number of test cases.
T lines follow. Each line describes a single test case and contains three integers: N, L, and S.
<b>Constraints:</b>
1 <= T <= 100
2 <= N <= 10<sup>9</sup>
1 <= L <= N-1
1 <= S <= 10<sup>18</sup> <b> (Note that S will not fit in a 32-bit integer) </b>For each testcase, print "YES" (without quotes) if a required subarray can exist, and "NO" (without quotes) otherwise.Sample Input:
3
5 3 11
5 3 5
5 3 6
Sample Output:
YES
NO
YES
Sample Explanation:
For the first test case, we can remove 3 from A to obtain A = {1, 2, 4, 5} where {2, 4, 5} is a required subarray of size 3 and sum 11., I have written this Solution Code: //HEADER FILES AND NAMESPACES
#include<bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
// #include <sys/resource.h>
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>;
template <typename T>
using ordered_multiset = tree<T, null_type, less_equal<T>, rb_tree_tag, tree_order_statistics_node_update>;
// DEFINE STATEMENTS
const long long INF = 1e18;
const int INFINT = INT_MAX/2;
#define num1 1000000007
#define num2 998244353
#define REP(i,a,n) for(ll i=a;i<n;i++)
#define REPd(i,a,n) for(ll i=a; i>=n; i--)
#define pb push_back
#define f first
#define s second
#define fix(f,n) std::fixed<<std::setprecision(n)<<f
#define all(x) x.begin(), x.end()
#define M_PI 3.14159265358979323846
#define epsilon (double)(0.000000001)
#define popcount __builtin_popcountll
#define fileio(x, y) freopen(x, "r", stdin); freopen(y, "w", stdout);
#define out(x) cout << ((x) ? "YES\n" : "NO\n")
#define CASE(x, y) cout << "Case #" << x << ":" << " \n"[y]
#define start_clock() auto start_time = std::chrono::high_resolution_clock::now(); auto end_time = start_time;
#define measure() end_time = std::chrono::high_resolution_clock::now(); cerr << (end_time - start_time)/std::chrono::milliseconds(1) << "ms" << endl;
#define reset_clock() start_time = std::chrono::high_resolution_clock::now();
typedef long long ll;
typedef long double ld;
typedef vector<long long> vll;
typedef pair<long long, long long> pll;
typedef vector<pair<long long, long long>> vpll;
typedef vector<int> vii;
// DEBUG FUNCTIONS
#ifdef LOCALY
template<typename T>
void __p(T a) {
cout<<a;
}
template<typename T, typename F>
void __p(pair<T, F> a) {
cout<<"{";
__p(a.first);
cout<<",";
__p(a.second);
cout<<"}";
}
template<typename T>
void __p(std::vector<T> a) {
cout<<"{";
for(auto it=a.begin(); it<a.end(); it++)
__p(*it),cout<<",}"[it+1==a.end()];
}
template<typename T>
void __p(std::set<T> a) {
cout<<"{";
for(auto it=a.begin(); it!=a.end();){
__p(*it);
cout<<",}"[++it==a.end()];
}
}
template<typename T>
void __p(std::multiset<T> a) {
cout<<"{";
for(auto it=a.begin(); it!=a.end();){
__p(*it);
cout<<",}"[++it==a.end()];
}
}
template<typename T, typename F>
void __p(std::map<T,F> a) {
cout<<"{\n";
for(auto it=a.begin(); it!=a.end();++it)
{
__p(it->first);
cout << ": ";
__p(it->second);
cout<<"\n";
}
cout << "}\n";
}
template<typename T, typename ...Arg>
void __p(T a1, Arg ...a) {
__p(a1);
__p(a...);
}
template<typename Arg1>
void __f(const char *name, Arg1 &&arg1) {
cout<<name<<" : ";
__p(arg1);
cout<<endl;
}
template<typename Arg1, typename ... Args>
void __f(const char *names, Arg1 &&arg1, Args &&... args) {
int bracket=0,i=0;
for(;; i++)
if(names[i]==','&&bracket==0)
break;
else if(names[i]=='(')
bracket++;
else if(names[i]==')')
bracket--;
const char *comma=names+i;
cout.write(names,comma-names)<<" : ";
__p(arg1);
cout<<" | ";
__f(comma+1,args...);
}
#define trace(...) cout<<"Line:"<<__LINE__<<" ", __f(#__VA_ARGS__, __VA_ARGS__)
#else
#define trace(...)
#define error(...)
#endif
// DEBUG FUNCTIONS END
// CUSTOM HASH TO SPEED UP UNORDERED MAP AND TO AVOID FORCED CLASHES
struct custom_hash {
static uint64_t splitmix64(uint64_t x) {
x += 0x9e3779b97f4a7c15;
x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;
x = (x ^ (x >> 27)) * 0x94d049bb133111eb;
return x ^ (x >> 31);
}
size_t operator()(uint64_t x) const {
static const uint64_t FIXED_RANDOM = chrono::steady_clock::now().time_since_epoch().count();
return splitmix64(x + FIXED_RANDOM);
}
};
mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count()); // FOR RANDOM NUMBER GENERATION
ll mod_exp(ll a, ll b, ll c)
{
ll res=1; a=a%c;
while(b>0)
{
if(b%2==1)
res=(res*a)%c;
b/=2;
a=(a*a)%c;
}
return res;
}
ll norm(ll a,ll b)
{
return (((a = a%b) < 0) ? a + b : a);
}
ll gcdExtended(ll,ll,ll *,ll *);
ll modInverse(ll a, ll m)
{
ll x, y;
ll g = gcdExtended(a, m, &x, &y);
g++; g--; //this line was added just to remove compiler warning
ll res = (x%m + m) % m;
return res;
}
ll gcdExtended(ll a, ll b, ll *x, ll *y)
{
if (a == 0)
{
*x = 0, *y = 1;
return b;
}
ll x1, y1;
ll gcd = gcdExtended(b%a, a, &x1, &y1);
*x = y1 - (b/a) * x1;
*y = x1;
return gcd;
}
struct Graph
{
vector<vector<int>> adj;
Graph(int n): adj(n+1) {}
void add_edge(int a, int b, bool directed = false)
{
adj[a].pb(b);
if(!directed) adj[b].pb(a);
}
};
struct LCA
{
vll tin, tout, level;
vector<vll> up;
ll timer;
void _dfs(vector<vector<int>> &adj, ll x, ll par=-1)
{
up[x][0] = par;
REP(i, 1, 20)
{
if(up[x][i-1] == -1) break;
up[x][i] = up[up[x][i-1]][i-1];
}
tin[x] = ++timer;
for(auto &p: adj[x])
{
if(p != par)
{
level[p] = level[x]+1; _dfs(adj, p, x);
}
}
tout[x] = ++timer;
}
LCA(Graph &G, ll root)
{
int n = G.adj.size();
tin.resize(n); tout.resize(n); up.resize(n, vll(20, -1)); level.resize(n, 0);
timer = 0;
//does not handle forests, easy to modify to handle
_dfs(G.adj, root);
}
ll find_kth(ll x, ll k)
{
ll cur = x;
REP(i, 0, 20)
{
if((k>>i)&1) cur = up[cur][i];
if(cur == -1) break;
}
return cur;
}
bool is_ancestor(ll x, ll y)
{
return tin[x]<=tin[y]&&tout[x]>=tout[y];
}
ll find_lca(ll x, ll y)
{
if(is_ancestor(x, y)) return x;
if(is_ancestor(y, x)) return y;
ll best = x;
REPd(i, 19, 0)
{
if(up[x][i] == -1) continue;
else if(is_ancestor(up[x][i], y)) best = up[x][i];
else x = up[x][i];
}
return best;
}
ll dist(ll a, ll b)
{
return level[a] + level[b] - 2*level[find_lca(a, b)];
}
};
long long ap_sum(long long st, long long len) {
//diff is always 1
long long en = st + len - 1;
return (len * (st + en))/2;
}
signed main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
int t = 1;
cin >> t;
for(int i=0; i<t; i++) {
long long N, L, S;
cin >> N >> L >> S;
if(S < ap_sum(1, L) || S > ap_sum(N - L + 1, L)) {
cout << "NO\n";
} else {
cout << "YES\n";
}
}
return 0;
}
/*
*/, In this Programming Language: C++, 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: Implement pow(X, N), which calculates x raised to the power N i.e. (X^N).
Try using a recursive approachThe first line contains T, denoting the number of test cases. Each test case contains a single line containing X, and N.
<b>Constraints:</b>
1 ≤ T ≤ 100
-10.00 ≤ X ≤10.00
-10 ≤ N ≤ 10
For each test case, you need to print the value of X^N. Print up to two places of decimal.
<b>Note:</b>
Please take care that the output can be very large but it will not exceed double the data type value.Input:
1
2.00 -2
Output:
0.25
<b>Explanation:</b>
2^(-2) = 1/2^2 = 1/4 = 0.25, I have written this Solution Code:
// X and Y are numbers
// ignore number of testcases variable
function pow(X, Y) {
// write code here
// console.log the output in a single line,like example
console.log(Math.pow(X, Y).toFixed(2))
}
, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Implement pow(X, N), which calculates x raised to the power N i.e. (X^N).
Try using a recursive approachThe first line contains T, denoting the number of test cases. Each test case contains a single line containing X, and N.
<b>Constraints:</b>
1 ≤ T ≤ 100
-10.00 ≤ X ≤10.00
-10 ≤ N ≤ 10
For each test case, you need to print the value of X^N. Print up to two places of decimal.
<b>Note:</b>
Please take care that the output can be very large but it will not exceed double the data type value.Input:
1
2.00 -2
Output:
0.25
<b>Explanation:</b>
2^(-2) = 1/2^2 = 1/4 = 0.25, I have written this Solution Code: def power(N,K):
return ("{0:.2f}".format(N**K))
T=int(input())
for i in range(T):
X,N = map(float,input().strip().split())
print(power(X,N)), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Implement pow(X, N), which calculates x raised to the power N i.e. (X^N).
Try using a recursive approachThe first line contains T, denoting the number of test cases. Each test case contains a single line containing X, and N.
<b>Constraints:</b>
1 ≤ T ≤ 100
-10.00 ≤ X ≤10.00
-10 ≤ N ≤ 10
For each test case, you need to print the value of X^N. Print up to two places of decimal.
<b>Note:</b>
Please take care that the output can be very large but it will not exceed double the data type value.Input:
1
2.00 -2
Output:
0.25
<b>Explanation:</b>
2^(-2) = 1/2^2 = 1/4 = 0.25, I have written this Solution Code: import java.util.*;
import java.io.*;
import java.lang.*;
class Main
{
public static void main (String[] args)throws Exception {
BufferedReader read = new BufferedReader(new InputStreamReader(System.in));
int t = Integer.parseInt(read.readLine());
while(t-- > 0)
{
String str[] = read.readLine().trim().split(" ");
double X = Double.parseDouble(str[0]);
int N = Integer.parseInt(str[1]);
System.out.println(String.format("%.2f", myPow(X, N)));
}
}
public static double myPow(double x, int n) {
if (n == Integer.MIN_VALUE)
n = - (Integer.MAX_VALUE - 1);
if (n == 0)
return 1.0;
else if (n < 0)
return 1 / myPow(x, -n);
else if (n % 2 == 1)
return x * myPow(x, n - 1);
else {
double sqrt = myPow(x, n / 2);
return sqrt * sqrt;
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Implement pow(X, N), which calculates x raised to the power N i.e. (X^N).
Try using a recursive approachThe first line contains T, denoting the number of test cases. Each test case contains a single line containing X, and N.
<b>Constraints:</b>
1 ≤ T ≤ 100
-10.00 ≤ X ≤10.00
-10 ≤ N ≤ 10
For each test case, you need to print the value of X^N. Print up to two places of decimal.
<b>Note:</b>
Please take care that the output can be very large but it will not exceed double the data type value.Input:
1
2.00 -2
Output:
0.25
<b>Explanation:</b>
2^(-2) = 1/2^2 = 1/4 = 0.25, I have written this Solution Code: #include "bits/stdc++.h"
using namespace std;
#define ld long double
#define int long long int
#define speed ios::sync_with_stdio(false); cin.tie(0); cout.tie(0)
#define endl '\n'
const int N = 2e5 + 5;
const int mod = 1e9 + 7;
const int inf = 1e9 + 9;
ld power(ld x, ld n){
if(n == 0)
return 1;
else
return x*power(x, n-1);
}
signed main() {
speed;
int t; cin >> t;
while(t--){
double x; int n;
cin >> x >> n;
if(n < 0)
x = 1.0/x, n *= -1;
cout << setprecision(2) << fixed << power(x, n) << endl;
}
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array <b>A[]</b> having <b>N</b> positive integers. You need to arrange these elements in increasing order using <b>Quick Sort</b> algorithm.<b>User Task:</b>
Since this is a functional problem you don't have to worry about the input. You just have to complete the function <b>quickSort()</b> which contains following arguments.
<b>A[]:</b> input array
<b>start:</b> starting index of array
<b>end</b>: ending index of array
Constraints
1 <= T <= 1000
1 <= N <= 10^4
1 <= A[i] <= 10^5
<b>Sum of "N" over all testcases does not exceed 10^5</b>For each testcase you need to return the sorted array. The driver code will do the rest.Sample Input:
2
3
3 1 2
3
4 5 6
Sample Output:
1 2 3
4 5 6, I have written this Solution Code: def partition(array, low, high):
pivot = array[high]
i = low - 1
for j in range(low, high):
if array[j] <= pivot:
i = i + 1
(array[i], array[j]) = (array[j], array[i])
(array[i + 1], array[high]) = (array[high], array[i + 1])
return i + 1
def quick_sort(array, low, high):
if low < high:
pi = partition(array, low, high)
quick_sort(array, low, pi - 1)
quick_sort(array, pi + 1, high)
t=int(input())
for i in range(t):
n=int(input())
a=input().strip().split()
a=[int(i) for i in a]
quick_sort(a, 0, n - 1)
for i in a:
print(i,end=" ")
print(), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array <b>A[]</b> having <b>N</b> positive integers. You need to arrange these elements in increasing order using <b>Quick Sort</b> algorithm.<b>User Task:</b>
Since this is a functional problem you don't have to worry about the input. You just have to complete the function <b>quickSort()</b> which contains following arguments.
<b>A[]:</b> input array
<b>start:</b> starting index of array
<b>end</b>: ending index of array
Constraints
1 <= T <= 1000
1 <= N <= 10^4
1 <= A[i] <= 10^5
<b>Sum of "N" over all testcases does not exceed 10^5</b>For each testcase you need to return the sorted array. The driver code will do the rest.Sample Input:
2
3
3 1 2
3
4 5 6
Sample Output:
1 2 3
4 5 6, I have written this Solution Code:
public static int[] quickSort(int arr[], int low, int high)
{
if (low < high)
{
/* pi is partitioning index, arr[pi] is
now at right place */
int pi = partition(arr, low, high);
// Recursively sort elements before
// partition and after partition
quickSort(arr, low, pi-1);
quickSort(arr, pi+1, high);
}
return arr;
}
public static int partition(int arr[], int low, int high)
{
int pivot = arr[high];
int i = (low-1); // index of smaller element
for (int j=low; j<high; j++)
{
// If current element is smaller than the pivot
if (arr[j] < pivot)
{
i++;
// swap arr[i] and arr[j]
int temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
}
}
// swap arr[i+1] and arr[high] (or pivot)
int temp = arr[i+1];
arr[i+1] = arr[high];
arr[high] = temp;
return i+1;
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array <b>A[]</b> having <b>N</b> positive integers. You need to arrange these elements in increasing order using <b>Quick Sort</b> algorithm.<b>User Task:</b>
Since this is a functional problem you don't have to worry about the input. You just have to complete the function <b>quickSort()</b> which contains following arguments.
<b>A[]:</b> input array
<b>start:</b> starting index of array
<b>end</b>: ending index of array
Constraints
1 <= T <= 1000
1 <= N <= 10^4
1 <= A[i] <= 10^5
<b>Sum of "N" over all testcases does not exceed 10^5</b>For each testcase you need to return the sorted array. The driver code will do the rest.Sample Input:
2
3
3 1 2
3
4 5 6
Sample Output:
1 2 3
4 5 6, I have written this Solution Code: function quickSort(arr, low, high)
{
if(low < high)
{
let pi = partition(arr, low, high);
quickSort(arr, low, pi-1);
quickSort(arr, pi+1, high);
}
return arr;
}
function partition(arr, low, high)
{
let pivot = arr[high];
let i = (low-1); // index of smaller element
for (let j=low; j<high; j++)
{
// If current element is smaller than the pivot
if (arr[j] < pivot)
{
i++;
// swap arr[i] and arr[j]
let temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
}
}
// swap arr[i+1] and arr[high] (or pivot)
let temp = arr[i+1];
arr[i+1] = arr[high];
arr[high] = temp;
return i+1;
}, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara is solving a math problem in which she has been given an integer N and her task is to find the number of operations required to convert N into 1.
Where in one operation you replace the number with its second-highest divisor.<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>DivisorProblem()</b> that takes integer N as argument.
Constraints:-
1 <= N <= 100000Return the number of operations required.Sample Input:-
100
Sample Output:-
4
Explanation:-
100 - > 50
50 - > 25
25 - > 5
5 - > 1
Sample Input:-
10
Sample Output:-
2, I have written this Solution Code: int DivisorProblem(int N){
int ans=0;
while(N>1){
int cnt=2;
while(N%cnt!=0){
cnt++;
}
N/=cnt;
ans++;
}
return ans;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara is solving a math problem in which she has been given an integer N and her task is to find the number of operations required to convert N into 1.
Where in one operation you replace the number with its second-highest divisor.<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>DivisorProblem()</b> that takes integer N as argument.
Constraints:-
1 <= N <= 100000Return the number of operations required.Sample Input:-
100
Sample Output:-
4
Explanation:-
100 - > 50
50 - > 25
25 - > 5
5 - > 1
Sample Input:-
10
Sample Output:-
2, I have written this Solution Code: def DivisorProblem(N):
ans=0
while N>1:
cnt=2
while N%cnt!=0:
cnt=cnt+1
N = N//cnt
ans=ans+1
return ans
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara is solving a math problem in which she has been given an integer N and her task is to find the number of operations required to convert N into 1.
Where in one operation you replace the number with its second-highest divisor.<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>DivisorProblem()</b> that takes integer N as argument.
Constraints:-
1 <= N <= 100000Return the number of operations required.Sample Input:-
100
Sample Output:-
4
Explanation:-
100 - > 50
50 - > 25
25 - > 5
5 - > 1
Sample Input:-
10
Sample Output:-
2, I have written this Solution Code: static int DivisorProblem(int N){
int ans=0;
while(N>1){
int cnt=2;
while(N%cnt!=0){
cnt++;
}
N/=cnt;
ans++;
}
return ans;
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara is solving a math problem in which she has been given an integer N and her task is to find the number of operations required to convert N into 1.
Where in one operation you replace the number with its second-highest divisor.<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>DivisorProblem()</b> that takes integer N as argument.
Constraints:-
1 <= N <= 100000Return the number of operations required.Sample Input:-
100
Sample Output:-
4
Explanation:-
100 - > 50
50 - > 25
25 - > 5
5 - > 1
Sample Input:-
10
Sample Output:-
2, I have written this Solution Code: int DivisorProblem(int N){
int ans=0;
while(N>1){
int cnt=2;
while(N%cnt!=0){
cnt++;
}
N/=cnt;
ans++;
}
return ans;
}, In this Programming Language: C, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Your task is to implement a stack using a linked list and perform given queries
Note:-if stack is already empty than pop operation will do nothing and 0 will be printed as a top element of stack if it is empty.User task:
Since this will be a functional problem, you don't have to take input. You just have to complete the functions:
<b>push()</b>:- that takes the integer to be added as a parameter.
<b>pop()</b>:- that takes no parameter.
<b>top()</b> :- that takes no parameter.
Constraints:
1 <= N(number of queries) <= 10<sup>3</sup>You don't need to print anything else other than in top function in which you require to print the top most element of your stack in a new line, if the stack is empty you just need to print 0.Input:
7
push 1
push 2
top
pop
top
pop
top
Output:
2
1
0
, I have written this Solution Code:
Node top = null;
public void push(int x)
{
Node temp = new Node(x);
temp.next = top;
top = temp;
}
public void pop()
{
if (top == null) {
}
else {
top = (top).next;}
}
public void top()
{
// check for stack underflow
if (top == null) {
System.out.println("0");
}
else {
Node temp = top;
System.out.println(temp.val);
}
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given two integers a and b, your task is to check following conditions:-
1. If a <= 10 and b >= 10 (Logical AND).
2. Atleast one from a or b will be even (Logical OR).
3. if a is not equal to b (Logical NOT).The first line of the input contains 2 integers a and b.
<b>Constraints:</b>
1 <= a, b <= 100Print the string <b>"true"</b> if the condition holds in each function else <b>"false"</b> .
Sample Input:-
3 12
Sample Output:-
true true true
Explanation
So a = 3 and b = 12, so a<=10 and b>=10 hence first condition true, a is not even but b is even so atleast one of them is even hence true, third a != b which is also true hence the final output comes true true true.
Sample Input:-
10 10
Sample Output:-
true true false
, I have written this Solution Code: a, b = list(map(int, input().split(" ")))
print(str(a <= 10 and b >= 10).lower(), end=' ')
print(str(a % 2 == 0 or b % 2 == 0).lower(), end=' ')
print(str(not a == b).lower()), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given two integers a and b, your task is to check following conditions:-
1. If a <= 10 and b >= 10 (Logical AND).
2. Atleast one from a or b will be even (Logical OR).
3. if a is not equal to b (Logical NOT).The first line of the input contains 2 integers a and b.
<b>Constraints:</b>
1 <= a, b <= 100Print the string <b>"true"</b> if the condition holds in each function else <b>"false"</b> .
Sample Input:-
3 12
Sample Output:-
true true true
Explanation
So a = 3 and b = 12, so a<=10 and b>=10 hence first condition true, a is not even but b is even so atleast one of them is even hence true, third a != b which is also true hence the final output comes true true true.
Sample Input:-
10 10
Sample Output:-
true true false
, I have written this Solution Code: import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;
class Main {
static boolean Logical_AND(int a, int b){
if(a<=10 && b>=10){
return true;}
return false;}
static boolean Logical_OR(int a, int b){
if(a%2==0 || b%2==0){
return true;}
return false;}
static boolean Logical_NOT(int a, int b){
if(a!=b){
return true;}
return false;}
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int a=in.nextInt();
int b=in.nextInt();
System.out.print(Logical_AND(a, b)+" ");
System.out.print(Logical_OR(a,b)+" ");
System.out.print(Logical_NOT(a,b)+" ");
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Loki is one mischievous guy. He would always find different ways to make things difficult for everyone. After spending hours sorting a coded array of size N (in increasing order), you realise it’s been tampered with by none other than Loki. Like a clock, he has moved the array thus tampering the data. The task is to find the <b>minimum</b> element in it.The first line of input contains a single integer T denoting the number of test cases. Then T test cases follow. Each test case consist of two lines. The first line of each test case consists of an integer N, where N is the size of array.
The second line of each test case contains N space separated integers denoting array elements.
Constraints:
1 <= T <= 100
1 <= N <= 10^5
1 <= A[i] <= 10^6
<b>Sum of "N" over all testcases does not exceed 10^5</b>Corresponding to each test case, in a new line, print the minimum element in the array.Input:
3
10
2 3 4 5 6 7 8 9 10 1
5
3 4 5 1 2
8
10 20 30 45 50 60 4 6
Output:
1
1
4
Explanation:
Testcase 1: The array is rotated once anti-clockwise. So minium element is at last index (n-1) which is 1.
Testcase 2: The array is rotated and the minimum element present is at index (n-2) which is 1.
Testcase 3: The array is rotated and the minimum element present is 4., I have written this Solution Code: def findMin(arr, low, high):
if high < low:
return arr[0]
if high == low:
return arr[low]
mid = int((low + high)/2)
if mid < high and arr[mid+1] < arr[mid]:
return arr[mid+1]
if mid > low and arr[mid] < arr[mid - 1]:
return arr[mid]
if arr[high] > arr[mid]:
return findMin(arr, low, mid-1)
return findMin(arr, mid+1, high)
T =int(input())
for i in range(T):
N = int(input())
arr = list(input().split())
for k in range(len(arr)):
arr[k] = int(arr[k])
print(findMin(arr,0,N-1)), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Loki is one mischievous guy. He would always find different ways to make things difficult for everyone. After spending hours sorting a coded array of size N (in increasing order), you realise it’s been tampered with by none other than Loki. Like a clock, he has moved the array thus tampering the data. The task is to find the <b>minimum</b> element in it.The first line of input contains a single integer T denoting the number of test cases. Then T test cases follow. Each test case consist of two lines. The first line of each test case consists of an integer N, where N is the size of array.
The second line of each test case contains N space separated integers denoting array elements.
Constraints:
1 <= T <= 100
1 <= N <= 10^5
1 <= A[i] <= 10^6
<b>Sum of "N" over all testcases does not exceed 10^5</b>Corresponding to each test case, in a new line, print the minimum element in the array.Input:
3
10
2 3 4 5 6 7 8 9 10 1
5
3 4 5 1 2
8
10 20 30 45 50 60 4 6
Output:
1
1
4
Explanation:
Testcase 1: The array is rotated once anti-clockwise. So minium element is at last index (n-1) which is 1.
Testcase 2: The array is rotated and the minimum element present is at index (n-2) which is 1.
Testcase 3: The array is rotated and the minimum element present is 4., 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 = 1e6 + 5;
const int mod = 1e9 + 7;
const int inf = 1e9 + 9;
int a[N];
signed main() {
IOS;
int t; cin >> t;
while(t--){
int n; cin >> n;
for(int i = 1; i <= n; i++)
cin >> a[i];
int l = 0, h = n+1;
if(a[1] < a[n]){
cout << a[1] << endl;
continue;
}
while(l+1 < h){
int m = (l + h) >> 1;
if(a[m] >= a[1])
l = m;
else
h = m;
}
cout << a[h] << endl;
}
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Loki is one mischievous guy. He would always find different ways to make things difficult for everyone. After spending hours sorting a coded array of size N (in increasing order), you realise it’s been tampered with by none other than Loki. Like a clock, he has moved the array thus tampering the data. The task is to find the <b>minimum</b> element in it.The first line of input contains a single integer T denoting the number of test cases. Then T test cases follow. Each test case consist of two lines. The first line of each test case consists of an integer N, where N is the size of array.
The second line of each test case contains N space separated integers denoting array elements.
Constraints:
1 <= T <= 100
1 <= N <= 10^5
1 <= A[i] <= 10^6
<b>Sum of "N" over all testcases does not exceed 10^5</b>Corresponding to each test case, in a new line, print the minimum element in the array.Input:
3
10
2 3 4 5 6 7 8 9 10 1
5
3 4 5 1 2
8
10 20 30 45 50 60 4 6
Output:
1
1
4
Explanation:
Testcase 1: The array is rotated once anti-clockwise. So minium element is at last index (n-1) which is 1.
Testcase 2: The array is rotated and the minimum element present is at index (n-2) which is 1.
Testcase 3: The array is rotated and the minimum element present is 4., I have written this Solution Code: import java.util.*;
import java.io.*;
import java.lang.*;
class Main{
public static void main (String[] args) {
//code
Scanner s = new Scanner(System.in);
int t = s.nextInt();
for(int j=0;j<t;j++){
int al = s.nextInt();
int a[] = new int[al];
for(int i=0;i<al;i++){
a[i] = s.nextInt();
}
binSearchSmallest(a);
}
}
public static void binSearchSmallest(int a[]) {
int s=0;
int e = a.length - 1;
int mid = 0;
while(s<=e){
mid = (s+e)/2;
if(a[s]<a[e]){
System.out.println(a[s]);
return;
}
if(a[mid]>=a[s]){
s=mid+1;
}
else{
e=mid;
}
if(s == e){
System.out.println(a[s]);
return;
}
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Loki is one mischievous guy. He would always find different ways to make things difficult for everyone. After spending hours sorting a coded array of size N (in increasing order), you realise it’s been tampered with by none other than Loki. Like a clock, he has moved the array thus tampering the data. The task is to find the <b>minimum</b> element in it.The first line of input contains a single integer T denoting the number of test cases. Then T test cases follow. Each test case consist of two lines. The first line of each test case consists of an integer N, where N is the size of array.
The second line of each test case contains N space separated integers denoting array elements.
Constraints:
1 <= T <= 100
1 <= N <= 10^5
1 <= A[i] <= 10^6
<b>Sum of "N" over all testcases does not exceed 10^5</b>Corresponding to each test case, in a new line, print the minimum element in the array.Input:
3
10
2 3 4 5 6 7 8 9 10 1
5
3 4 5 1 2
8
10 20 30 45 50 60 4 6
Output:
1
1
4
Explanation:
Testcase 1: The array is rotated once anti-clockwise. So minium element is at last index (n-1) which is 1.
Testcase 2: The array is rotated and the minimum element present is at index (n-2) which is 1.
Testcase 3: The array is rotated and the minimum element present is 4., I have written this Solution Code: // arr is the input array
function findMin(arr,n) {
// write code here
// do not console.log
// return the number
let min = arr[0]
for(let i=1;i<arr.length;i++){
if(min > arr[i]){
min = arr[i]
}
}
return min
}, In this Programming Language: JavaScript, 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: English Team has now adopted a rotation policy for two of their players: Dom and Leach.
On the first day, both of them played but, from the second day onwards, Dom plays every second day, while Leach plays every third day.
For example, on:
Day 1 - Both players play,
Day 2 - Neither of them plays,
Day 3 - Only Dom plays,
Day 4 - Only Leach plays,
Day 5 - Only Dom plays,
Day 6 - Neither of them plays,
Day 7 - Both the players play.. and so on.
Find the number of days in the interval [A, B] (A and B, both inclusive) when neither Dom nor Leach plays.<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>RotationPolicy()</b> that takes integers A, and B as arguments.
Constraints:-
1 <= A, B <=100000Return the number of days when neither of the two players played the game.Sample Input:-
3 8
Sample Output:-
2
Sample Input:-
1 4
Sample Output:-
1, I have written this Solution Code:
int RotationPolicy(int A, int B){
int cnt=0;
for(int i=A;i<=B;i++){
if((i-1)%2!=0 && (i-1)%3!=0){cnt++;}
}
return cnt;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: English Team has now adopted a rotation policy for two of their players: Dom and Leach.
On the first day, both of them played but, from the second day onwards, Dom plays every second day, while Leach plays every third day.
For example, on:
Day 1 - Both players play,
Day 2 - Neither of them plays,
Day 3 - Only Dom plays,
Day 4 - Only Leach plays,
Day 5 - Only Dom plays,
Day 6 - Neither of them plays,
Day 7 - Both the players play.. and so on.
Find the number of days in the interval [A, B] (A and B, both inclusive) when neither Dom nor Leach plays.<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>RotationPolicy()</b> that takes integers A, and B as arguments.
Constraints:-
1 <= A, B <=100000Return the number of days when neither of the two players played the game.Sample Input:-
3 8
Sample Output:-
2
Sample Input:-
1 4
Sample Output:-
1, I have written this Solution Code:
int RotationPolicy(int A, int B){
int cnt=0;
for(int i=A;i<=B;i++){
if((i-1)%2!=0 && (i-1)%3!=0){cnt++;}
}
return cnt;
}
, In this Programming Language: C, Now tell me if this Code is compilable or not? | Compilable |
For this Question: English Team has now adopted a rotation policy for two of their players: Dom and Leach.
On the first day, both of them played but, from the second day onwards, Dom plays every second day, while Leach plays every third day.
For example, on:
Day 1 - Both players play,
Day 2 - Neither of them plays,
Day 3 - Only Dom plays,
Day 4 - Only Leach plays,
Day 5 - Only Dom plays,
Day 6 - Neither of them plays,
Day 7 - Both the players play.. and so on.
Find the number of days in the interval [A, B] (A and B, both inclusive) when neither Dom nor Leach plays.<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>RotationPolicy()</b> that takes integers A, and B as arguments.
Constraints:-
1 <= A, B <=100000Return the number of days when neither of the two players played the game.Sample Input:-
3 8
Sample Output:-
2
Sample Input:-
1 4
Sample Output:-
1, I have written this Solution Code: static int RotationPolicy(int A, int B){
int cnt=0;
for(int i=A;i<=B;i++){
if((i-1)%2!=0 && (i-1)%3!=0){cnt++;}
}
return cnt;
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: English Team has now adopted a rotation policy for two of their players: Dom and Leach.
On the first day, both of them played but, from the second day onwards, Dom plays every second day, while Leach plays every third day.
For example, on:
Day 1 - Both players play,
Day 2 - Neither of them plays,
Day 3 - Only Dom plays,
Day 4 - Only Leach plays,
Day 5 - Only Dom plays,
Day 6 - Neither of them plays,
Day 7 - Both the players play.. and so on.
Find the number of days in the interval [A, B] (A and B, both inclusive) when neither Dom nor Leach plays.<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>RotationPolicy()</b> that takes integers A, and B as arguments.
Constraints:-
1 <= A, B <=100000Return the number of days when neither of the two players played the game.Sample Input:-
3 8
Sample Output:-
2
Sample Input:-
1 4
Sample Output:-
1, I have written this Solution Code:
def RotationPolicy(A, B):
cnt=0
for i in range (A,B+1):
if(i-1)%2!=0 and (i-1)%3!=0:
cnt=cnt+1
return cnt
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: English Team has now adopted a rotation policy for two of their players: Dom and Leach.
On the first day, both of them played but, from the second day onwards, Dom plays every second day, while Leach plays every third day.
For example, on:
Day 1 - Both players play,
Day 2 - Neither of them plays,
Day 3 - Only Dom plays,
Day 4 - Only Leach plays,
Day 5 - Only Dom plays,
Day 6 - Neither of them plays,
Day 7 - Both the players play.. and so on.
Find the number of days in the interval [A, B] (A and B, both inclusive) when neither Dom nor Leach plays.<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>RotationPolicy()</b> that takes integers A, and B as arguments.
Constraints:-
1 <= A, B <=100000Return the number of days when neither of the two players played the game.Sample Input:-
3 8
Sample Output:-
2
Sample Input:-
1 4
Sample Output:-
1, I have written this Solution Code: function RotationPolicy(a, b) {
// write code here
// do no console.log the answer
// return the output using return keyword
let count = 0
for (let i = a; i <= b; i++) {
if((i-1)%2 !== 0 && (i-1)%3 !==0){
count++
}
}
return count
}
, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara is giving a True False exam consisting of 10 questions. In which she knows that exactly X of the given questions are True and the rest are false (thanks to her friend) but she does not know the order. This time she decided to fill it optimally so that in any case she can get the maximum of minimum marks possible.
Given the value of X, your task is to tell Sara what can be the maximum of minimum marks she can get. i. e out of every minimum mark for a different number of True and False what will be the maximum of that minimum.<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>maximumMarks()</b> that takes integer X as argument.
Constraints:-
0 <= X <= 10Return the maximum marks of all the minimum possible.Sample Input:-
X = 5
Sample Output:-
5
Explanation:-
One of the possible solutions is she can mark all the answers true.
Sample Input:-
1
Sample Output:-
9, I have written this Solution Code: static int maximumMarks(int X){
if(X>5){
return X;
}
return 10-X;
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara is giving a True False exam consisting of 10 questions. In which she knows that exactly X of the given questions are True and the rest are false (thanks to her friend) but she does not know the order. This time she decided to fill it optimally so that in any case she can get the maximum of minimum marks possible.
Given the value of X, your task is to tell Sara what can be the maximum of minimum marks she can get. i. e out of every minimum mark for a different number of True and False what will be the maximum of that minimum.<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>maximumMarks()</b> that takes integer X as argument.
Constraints:-
0 <= X <= 10Return the maximum marks of all the minimum possible.Sample Input:-
X = 5
Sample Output:-
5
Explanation:-
One of the possible solutions is she can mark all the answers true.
Sample Input:-
1
Sample Output:-
9, I have written this Solution Code:
int maximumMarks(int X){
if(X>5){
return X;
}
return 10-X;
}
, In this Programming Language: C, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara is giving a True False exam consisting of 10 questions. In which she knows that exactly X of the given questions are True and the rest are false (thanks to her friend) but she does not know the order. This time she decided to fill it optimally so that in any case she can get the maximum of minimum marks possible.
Given the value of X, your task is to tell Sara what can be the maximum of minimum marks she can get. i. e out of every minimum mark for a different number of True and False what will be the maximum of that minimum.<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>maximumMarks()</b> that takes integer X as argument.
Constraints:-
0 <= X <= 10Return the maximum marks of all the minimum possible.Sample Input:-
X = 5
Sample Output:-
5
Explanation:-
One of the possible solutions is she can mark all the answers true.
Sample Input:-
1
Sample Output:-
9, I have written this Solution Code: def maximumMarks(X):
if X>5:
return (X)
return (10-X)
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara is giving a True False exam consisting of 10 questions. In which she knows that exactly X of the given questions are True and the rest are false (thanks to her friend) but she does not know the order. This time she decided to fill it optimally so that in any case she can get the maximum of minimum marks possible.
Given the value of X, your task is to tell Sara what can be the maximum of minimum marks she can get. i. e out of every minimum mark for a different number of True and False what will be the maximum of that minimum.<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>maximumMarks()</b> that takes integer X as argument.
Constraints:-
0 <= X <= 10Return the maximum marks of all the minimum possible.Sample Input:-
X = 5
Sample Output:-
5
Explanation:-
One of the possible solutions is she can mark all the answers true.
Sample Input:-
1
Sample Output:-
9, I have written this Solution Code:
int maximumMarks(int X){
if(X>5){
return X;
}
return 10-X;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number N for each i (1 < = i < = N), you have to print the number except :-
For each multiple of 3, print "Newton" instead of the number.
For each multiple of 5, print "School" instead of the number.
For numbers that are multiples of both 3 and 5, print "NewtonSchool" instead of the number.The first line of the input contains N.
<b>Constraints</b>
1 < = N < = 1000
Print N space separated number or Newton School according to the condition.Sample Input:-
3
Sample Output:-
1 2 Newton
Sample Input:-
5
Sample Output:-
1 2 Newton 4 School, I have written this Solution Code: n=int(input())
for i in range(1,n+1):
if i%3==0 and i%5==0:
print("NewtonSchool",end=" ")
elif i%3==0:
print("Newton",end=" ")
elif i%5==0:
print("School",end=" ")
else:
print(i,end=" "), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number N for each i (1 < = i < = N), you have to print the number except :-
For each multiple of 3, print "Newton" instead of the number.
For each multiple of 5, print "School" instead of the number.
For numbers that are multiples of both 3 and 5, print "NewtonSchool" instead of the number.The first line of the input contains N.
<b>Constraints</b>
1 < = N < = 1000
Print N space separated number or Newton School according to the condition.Sample Input:-
3
Sample Output:-
1 2 Newton
Sample Input:-
5
Sample Output:-
1 2 Newton 4 School, I have written this Solution Code: import java.util.*;
import java.lang.*;
import java.io.*;
class Main
{
static void NewtonSchool(int n){
for(int i=1;i<=n;i++){
if(i%3==0 && i%5==0){System.out.print("NewtonSchool ");}
else if(i%5==0){System.out.print("School ");}
else if(i%3==0){System.out.print("Newton ");}
else{System.out.print(i+" ");}
}
}
public static void main (String[] args) throws java.lang.Exception
{
Scanner sc = new Scanner(System.in);
int x= sc.nextInt();
NewtonSchool(x);
}
}, In this Programming Language: Java, 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: Sara loves patterns, today she encounters an interesting pattern and wants to write a code that can print the pattern of a given height N. As Sara is weak in programming help her to code it.
The pattern for height 6:-
0 4 8 12 16 20
6 10 14 18 22 26
12 16 20 24 28 32
18 22 26 30 34 38
24 28 32 36 40 44
30 34 38 42 46 50<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 integer N as an argument.
Constraints:-
1 <= N <= 100Print the given pattern.Sample Input:-
3
Sample Output:-
0 4 8
6 10 14
12 16 20
Sample Input:-
5
Sample Output:-
0 4 8 12 16
6 10 14 18 22
12 16 20 24 28
18 22 26 30 34
24 28 32 36 40, I have written this Solution Code: static void Pattern(int N){
int x=0;
for(int i=0;i<N;i++){
for(int j=0;j<N;j++){
System.out.print(x+4*j+" ");
}
System.out.println();
x+=6;
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara loves patterns, today she encounters an interesting pattern and wants to write a code that can print the pattern of a given height N. As Sara is weak in programming help her to code it.
The pattern for height 6:-
0 4 8 12 16 20
6 10 14 18 22 26
12 16 20 24 28 32
18 22 26 30 34 38
24 28 32 36 40 44
30 34 38 42 46 50<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 integer N as an argument.
Constraints:-
1 <= N <= 100Print the given pattern.Sample Input:-
3
Sample Output:-
0 4 8
6 10 14
12 16 20
Sample Input:-
5
Sample Output:-
0 4 8 12 16
6 10 14 18 22
12 16 20 24 28
18 22 26 30 34
24 28 32 36 40, I have written this Solution Code: def Pattern(N):
x=0
for i in range (0,N):
for j in range (0,N):
print(x+4*j,end=' ')
print()
x = x+6
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara loves patterns, today she encounters an interesting pattern and wants to write a code that can print the pattern of a given height N. As Sara is weak in programming help her to code it.
The pattern for height 6:-
0 4 8 12 16 20
6 10 14 18 22 26
12 16 20 24 28 32
18 22 26 30 34 38
24 28 32 36 40 44
30 34 38 42 46 50<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 integer N as an argument.
Constraints:-
1 <= N <= 100Print the given pattern.Sample Input:-
3
Sample Output:-
0 4 8
6 10 14
12 16 20
Sample Input:-
5
Sample Output:-
0 4 8 12 16
6 10 14 18 22
12 16 20 24 28
18 22 26 30 34
24 28 32 36 40, I have written this Solution Code: void Pattern(int N){
int x=0;
for(int i=0;i<N;i++){
for(int j=0;j<N;j++){
printf("%d ",x+4*j);
}
printf("\n");
x+=6;
}
}, In this Programming Language: C, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara loves patterns, today she encounters an interesting pattern and wants to write a code that can print the pattern of a given height N. As Sara is weak in programming help her to code it.
The pattern for height 6:-
0 4 8 12 16 20
6 10 14 18 22 26
12 16 20 24 28 32
18 22 26 30 34 38
24 28 32 36 40 44
30 34 38 42 46 50<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 integer N as an argument.
Constraints:-
1 <= N <= 100Print the given pattern.Sample Input:-
3
Sample Output:-
0 4 8
6 10 14
12 16 20
Sample Input:-
5
Sample Output:-
0 4 8 12 16
6 10 14 18 22
12 16 20 24 28
18 22 26 30 34
24 28 32 36 40, I have written this Solution Code: void Pattern(int N){
int x=0;
for(int i=0;i<N;i++){
for(int j=0;j<N;j++){
printf("%d ",x+4*j);
}
printf("\n");
x+=6;
}
}, 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 out the integer N.You don't have to worry about the input, you just have to complete the function <b>printIntger()</b>
<b>Constraints:-</b>
1 ≤ N ≤ 10<sup>9</sup>Print the integer N.Sample Input:-
3
Sample Output:
3
Sample Input:
56
Sample Output:
56, I have written this Solution Code: static void printInteger(int N){
System.out.println(N);
}, 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 out the integer N.You don't have to worry about the input, you just have to complete the function <b>printIntger()</b>
<b>Constraints:-</b>
1 ≤ N ≤ 10<sup>9</sup>Print the integer N.Sample Input:-
3
Sample Output:
3
Sample Input:
56
Sample Output:
56, I have written this Solution Code: void printInteger(int x){
printf("%d", x);
}, 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 out the integer N.You don't have to worry about the input, you just have to complete the function <b>printIntger()</b>
<b>Constraints:-</b>
1 ≤ N ≤ 10<sup>9</sup>Print the integer N.Sample Input:-
3
Sample Output:
3
Sample Input:
56
Sample Output:
56, I have written this Solution Code: void printIntger(int n)
{
cout<<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 out the integer N.You don't have to worry about the input, you just have to complete the function <b>printIntger()</b>
<b>Constraints:-</b>
1 ≤ N ≤ 10<sup>9</sup>Print the integer N.Sample Input:-
3
Sample Output:
3
Sample Input:
56
Sample Output:
56, I have written this Solution Code: n=int(input())
print (n), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Nutan and Tusla are both students at Newton School. They are both among the best students in the class. In order to know who is better among them, a game was organised. The game consisted of L rounds, where L is an odd integer. The student winning more rounds than the other was declared the winner.
You would be given a string of odd length L in which each character is 'N' or 'T'. If the i<sup>th</sup> character is 'N', then the i<sup>th</sup> round was won by Nutan, else if the character is 'T' it was won by Tusla. Print "Nutan'' if Nutan has won more rounds than Tusla, else print "Tusla'' if Tusla has won more rounds than Nutan.
Note: You have to print everything without quotes.The first line of the input contains a single integer L — the number of rounds (1 ≤ L ≤ 100 and L is odd).
The second line contains a string S of length L. Each character of S is either 'N' or 'T'.Print "Nutan" or "Tusla" according to the input.Sample Input:
3
NNT
Sample Output:
Nutan
Explanation:
Nutan has won two games while Tusla has only won a single game, so the overall winner is Nutan., I have written this Solution Code: import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;
import static java.lang.System.out;
public class Main {
static class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader() {
br = new BufferedReader(new InputStreamReader(System.in));
}
String next() {
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
long nextLong() {
return Long.parseLong(next());
}
double nextDouble() {
return Double.parseDouble(next());
}
String nextLine() {
String str = "";
try {
str = br.readLine();
} catch (IOException e) {
e.printStackTrace();
}
return str;
}
}
public static void main(String[] args) {
FastReader reader = new FastReader();
int n = reader.nextInt();
String S = reader.next();
int ncount = 0;
int tcount = 0;
for (char c : S.toCharArray()) {
if (c == 'N') ncount++;
else tcount++;
}
if (ncount > tcount) {
out.print("Nutan\n");
} else {
out.print("Tusla\n");
}
out.flush();
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Nutan and Tusla are both students at Newton School. They are both among the best students in the class. In order to know who is better among them, a game was organised. The game consisted of L rounds, where L is an odd integer. The student winning more rounds than the other was declared the winner.
You would be given a string of odd length L in which each character is 'N' or 'T'. If the i<sup>th</sup> character is 'N', then the i<sup>th</sup> round was won by Nutan, else if the character is 'T' it was won by Tusla. Print "Nutan'' if Nutan has won more rounds than Tusla, else print "Tusla'' if Tusla has won more rounds than Nutan.
Note: You have to print everything without quotes.The first line of the input contains a single integer L — the number of rounds (1 ≤ L ≤ 100 and L is odd).
The second line contains a string S of length L. Each character of S is either 'N' or 'T'.Print "Nutan" or "Tusla" according to the input.Sample Input:
3
NNT
Sample Output:
Nutan
Explanation:
Nutan has won two games while Tusla has only won a single game, so the overall winner is Nutan., I have written this Solution Code: n = int(input())
s = input()
a1 = s.count('N')
a2 = s.count('T')
if(a1 > a2):
print("Nutan")
else:
print('Tusla'), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Nutan and Tusla are both students at Newton School. They are both among the best students in the class. In order to know who is better among them, a game was organised. The game consisted of L rounds, where L is an odd integer. The student winning more rounds than the other was declared the winner.
You would be given a string of odd length L in which each character is 'N' or 'T'. If the i<sup>th</sup> character is 'N', then the i<sup>th</sup> round was won by Nutan, else if the character is 'T' it was won by Tusla. Print "Nutan'' if Nutan has won more rounds than Tusla, else print "Tusla'' if Tusla has won more rounds than Nutan.
Note: You have to print everything without quotes.The first line of the input contains a single integer L — the number of rounds (1 ≤ L ≤ 100 and L is odd).
The second line contains a string S of length L. Each character of S is either 'N' or 'T'.Print "Nutan" or "Tusla" according to the input.Sample Input:
3
NNT
Sample Output:
Nutan
Explanation:
Nutan has won two games while Tusla has only won a single game, so the overall winner is Nutan., I have written this Solution Code: //Author: Xzirium
//Time and Date: 02:18:28 24 March 2022
//Optional FAST
//#pragma GCC optimize("Ofast")
//#pragma GCC optimize("unroll-loops")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,fma,abm,mmx,avx,avx2,tune=native")
//Required Libraries
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/detail/standard_policies.hpp>
//Required namespaces
using namespace std;
using namespace __gnu_pbds;
//Required defines
#define endl '\n'
#define READ(X) cin>>X;
#define READV(X) long long X; cin>>X;
#define READAR(A,N) long long A[N]; for(long long i=0;i<N;i++) {cin>>A[i];}
#define rz(A,N) A.resize(N);
#define sz(X) (long long)(X.size())
#define pb push_back
#define pf push_front
#define fi first
#define se second
#define FORI(a,b,c) for(long long a=b;a<c;a++)
#define FORD(a,b,c) for(long long a=b;a>c;a--)
//Required typedefs
template <typename T> using ordered_set = tree<T,null_type,less<T>,rb_tree_tag,tree_order_statistics_node_update>;
template <typename T> using ordered_set1 = tree<T,null_type,greater<T>,rb_tree_tag,tree_order_statistics_node_update>;
typedef long long ll;
typedef long double ld;
typedef pair<int,int> pii;
typedef pair<long long,long long> pll;
//Required Constants
const long long inf=(long long)1e18;
const long long MOD=(long long)(1e9+7);
const long long INIT=(long long)(1e6+1);
const long double PI=3.14159265358979;
// Required random number generators
// mt19937 gen_rand_int(chrono::steady_clock::now().time_since_epoch().count());
// mt19937_64 gen_rand_ll(chrono::steady_clock::now().time_since_epoch().count());
//Required Functions
ll power(ll b, ll e)
{
ll r = 1ll;
for(; e > 0; e /= 2, (b *= b) %= MOD)
if(e % 2) (r *= b) %= MOD;
return r;
}
ll modInverse(ll a)
{
return power(a,MOD-2);
}
//Work
int main()
{
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
//-----------------------------------------------------------------------------------------------------------//
READV(N);
string S;
cin>>S;
ll n=0,t=0;
FORI(i,0,N)
{
if(S[i]=='N')
{
n++;
}
else if(S[i]=='T')
{
t++;
}
}
if(n>t)
{
cout<<"Nutan"<<endl;
}
else
{
cout<<"Tusla"<<endl;
}
//-----------------------------------------------------------------------------------------------------------//
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given an AI platform and asked to analyse the piece of code through it. As u studied about that platform you knew that it can analyse the code upto 1000 characters length. You are given the length of characters and asked to print YES or NO on the basis of whether the code can be analysed or not respectively.The first line of the input contains N (Length of the characters)
<b>Constraints</b>
1 ≤ N ≤ 100000Print YES or NO on the basis of whether the code can be analysed or not.Sample Input
10001
Sample Output
NO, I have written this Solution Code: #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 <class T>
using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
template <class key, class value, class cmp = std::less<key>>
using ordered_map = tree<key, value, cmp, rb_tree_tag, tree_order_statistics_node_update>;
// find_by_order(k) returns iterator to kth element starting from 0;
// order_of_key(k) returns count of elements strictly smaller than k;
#ifdef LOCAL
#define debug(...) cerr << "[" << #__VA_ARGS__ << "]:", debug_out(__VA_ARGS__)
#else
#define debug(...) 2351
#endif
#define int long long
mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
inline int64_t random_long(long long l = LLONG_MIN, long long r = LLONG_MAX) {
uniform_int_distribution<int64_t> generator(l, r);
return generator(rng);
}
int32_t main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
auto start = std::chrono::high_resolution_clock::now();
int n;
cin >> n;
if (n <= 1000) {
cout << "YES\n";
} else {
cout << "NO\n";
}
auto stop = std::chrono::high_resolution_clock::now();
auto duration = std::chrono::duration_cast<std::chrono::nanoseconds>(stop - start);
cerr << "Time taken : " << ((long double)duration.count()) / ((long double)1e9) << "s " << endl;
return 0;
};
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given an AI platform and asked to analyse the piece of code through it. As u studied about that platform you knew that it can analyse the code upto 1000 characters length. You are given the length of characters and asked to print YES or NO on the basis of whether the code can be analysed or not respectively.The first line of the input contains N (Length of the characters)
<b>Constraints</b>
1 ≤ N ≤ 100000Print YES or NO on the basis of whether the code can be analysed or not.Sample Input
10001
Sample Output
NO, 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();
if(n<=1000){
System.out.print("YES");
}
else
System.out.print("NO");
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: John has N candies. He want to crush all of them. He feels that it would be boring to crush the candies randomly, so he devices a method to crush them. He divides these candies in minimum number of groups such than no group contains more than 3 candy. He crushes one candy from each group. If there are G groups in a single step, then the cost incurred in crushing a single candy for that step is G dollars. After candy from each group is crushed, he takes all the remaining candies and repeats the process until he has no candies left. He hasn't started crushing yet, but he wants to know how much total cost would be incurred. Can you help him?1 <= N <= 10^9return the cost from the functionSample Input 1:
4
Explanation:
Query 1: First step John divides the candies in two groups of 3 and 1 candy respectively. Crushing one- one candy from both group would cost him 2x2 = 4 dollars. He is now left with 2 candies. He divides it into one group. He crushes one candy for 1 dollar. Now, he is left with 1 candy. He crushes the last candy for 1 dollar. So, the total cost incurred is 4+1+1 = 6 dollars., I have written this Solution Code: // ignore number of testcases/queries
// n is the number as provided in input
function findCost(n) {
// write code here
// do not console.log the answer
// return answer as a number
if (n == 0) return 0;
let g = Math.floor((n - 1) / 3 )+ 1;
return g*g + findCost(n - g);
}
, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: John has N candies. He want to crush all of them. He feels that it would be boring to crush the candies randomly, so he devices a method to crush them. He divides these candies in minimum number of groups such than no group contains more than 3 candy. He crushes one candy from each group. If there are G groups in a single step, then the cost incurred in crushing a single candy for that step is G dollars. After candy from each group is crushed, he takes all the remaining candies and repeats the process until he has no candies left. He hasn't started crushing yet, but he wants to know how much total cost would be incurred. Can you help him?1 <= N <= 10^9return the cost from the functionSample Input 1:
4
Explanation:
Query 1: First step John divides the candies in two groups of 3 and 1 candy respectively. Crushing one- one candy from both group would cost him 2x2 = 4 dollars. He is now left with 2 candies. He divides it into one group. He crushes one candy for 1 dollar. Now, he is left with 1 candy. He crushes the last candy for 1 dollar. So, the total cost incurred is 4+1+1 = 6 dollars., 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 testCase = sc.nextInt();
while(testCase != 0){
int g= 0;
int n= sc.nextInt();
int temp = n;
while(temp > 0){
int t= temp/3;
int rem= temp - (t*3);
if(rem ==0){
temp=temp - t;
g= g + (t*t);
}
else{
temp = temp - (t+1);
g=g+(t+1)*(t+1);
}
}
System.out.println(g);
--testCase;
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: The factor of a number is been divided into two types:-
1. Prime factors
2. Factors that can be further divided into more factors.
A number N is called a super number if all the factors other than the primes one's are divisible by each and every prime factor of N.
Given a Number N check if it is a Super number or not.
Note:- 1 will not be counted as a factor<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>superNumber()</b> that takes integer N argument.
Constraints:-
1 <= N <= 100Return 1 if the given number is super else return 0.Sample Input:-
15
Sample Output:-
1
Explanation:-
Prime factors:- 3 and 5
15 is only factor, which is divisible by both 3 and 5
Sample Input:-
12
Sample Output:-
0
Explanation:-
Prime factors:-2 and 3
4 is a factor and it is not divisible by 3, I have written this Solution Code:
int superNumber(int N){
int x=1;
int a[101];
for(int i=0;i<=100;i++){
a[i]=0;
}
int i=2;
while(N>1){i=2;
while(N%i!=0){
i++;
}
a[i]++;
N/=i;
}
int cnt1=0,cnt2=0;
for(int i=2;i<=100;i++){
if(a[i]>0){cnt1++;}
if(a[i]>1){cnt2++;}
}
// cout<<cnt1<<" "<<cnt2;
if((cnt1==2 && cnt2==0) || (cnt2>=0 && cnt1==1)){return 1;}
return 0;
}
, In this Programming Language: C, Now tell me if this Code is compilable or not? | Compilable |
For this Question: The factor of a number is been divided into two types:-
1. Prime factors
2. Factors that can be further divided into more factors.
A number N is called a super number if all the factors other than the primes one's are divisible by each and every prime factor of N.
Given a Number N check if it is a Super number or not.
Note:- 1 will not be counted as a factor<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>superNumber()</b> that takes integer N argument.
Constraints:-
1 <= N <= 100Return 1 if the given number is super else return 0.Sample Input:-
15
Sample Output:-
1
Explanation:-
Prime factors:- 3 and 5
15 is only factor, which is divisible by both 3 and 5
Sample Input:-
12
Sample Output:-
0
Explanation:-
Prime factors:-2 and 3
4 is a factor and it is not divisible by 3, I have written this Solution Code: def superNumber(N):
x=2
p=0
cnt1=0
cnt2=0
while N>1:
x=2
p=0
while N%x!=0:
x=x+1
cnt1=cnt1+1
N=N/x
while(N%x==0):
N=N/x
p=1;
cnt2=cnt2+p
if((cnt1==1) or (cnt2==0 and cnt1==2)):
return 1
return 0
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: The factor of a number is been divided into two types:-
1. Prime factors
2. Factors that can be further divided into more factors.
A number N is called a super number if all the factors other than the primes one's are divisible by each and every prime factor of N.
Given a Number N check if it is a Super number or not.
Note:- 1 will not be counted as a factor<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>superNumber()</b> that takes integer N argument.
Constraints:-
1 <= N <= 100Return 1 if the given number is super else return 0.Sample Input:-
15
Sample Output:-
1
Explanation:-
Prime factors:- 3 and 5
15 is only factor, which is divisible by both 3 and 5
Sample Input:-
12
Sample Output:-
0
Explanation:-
Prime factors:-2 and 3
4 is a factor and it is not divisible by 3, I have written this Solution Code: static int superNumber(int N){
int x=1;
int a[] = new int[101];
for(int i=0;i<=100;i++){
a[i]=0;
}
int i=2;
while(N>1){i=2;
while(N%i!=0){
i++;
}
a[i]++;
N/=i;
}
int cnt1=0,cnt2=0;
for(i=2;i<=100;i++){
if(a[i]>0){cnt1++;}
if(a[i]>1){cnt2++;}
}
// cout<<cnt1<<" "<<cnt2;
if((cnt1==2 && cnt2==0) || (cnt2>=0 && cnt1==1)){return 1;}
return 0;}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: The factor of a number is been divided into two types:-
1. Prime factors
2. Factors that can be further divided into more factors.
A number N is called a super number if all the factors other than the primes one's are divisible by each and every prime factor of N.
Given a Number N check if it is a Super number or not.
Note:- 1 will not be counted as a factor<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>superNumber()</b> that takes integer N argument.
Constraints:-
1 <= N <= 100Return 1 if the given number is super else return 0.Sample Input:-
15
Sample Output:-
1
Explanation:-
Prime factors:- 3 and 5
15 is only factor, which is divisible by both 3 and 5
Sample Input:-
12
Sample Output:-
0
Explanation:-
Prime factors:-2 and 3
4 is a factor and it is not divisible by 3, I have written this Solution Code:
int superNumber(int N){
int x=1;
int a[101];
for(int i=0;i<=100;i++){
a[i]=0;
}
int i=2;
while(N>1){i=2;
while(N%i!=0){
i++;
}
a[i]++;
N/=i;
}
int cnt1=0,cnt2=0;
for(int i=2;i<=100;i++){
if(a[i]>0){cnt1++;}
if(a[i]>1){cnt2++;}
}
// cout<<cnt1<<" "<<cnt2;
if((cnt1==2 && cnt2==0) || (cnt2>=0 && cnt1==1)){return 1;}
return 0;
}
, In this Programming Language: C++, 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: 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: Given an array, sort the array in reverse order by simply swapping its adjacent elements.First line of the input contains an integer, N, which denotes the length of the array. Next N inputs are elements of the array that is to be sorted in descending order.
Constraints
1<=N<=1000
-10000<=Arr[i]<=100000Output sorted array in descending order where each element is space separated.Sample Input:
6
3 1 2 7 9 87
Sample Output:
87 9 7 3 2 1, I have written this Solution Code: function bubbleSort(arr, n) {
// write code here
// do not console.log the answer
// return sorted array
return arr.sort((a, b) => b - a)
}
, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array, sort the array in reverse order by simply swapping its adjacent elements.First line of the input contains an integer, N, which denotes the length of the array. Next N inputs are elements of the array that is to be sorted in descending order.
Constraints
1<=N<=1000
-10000<=Arr[i]<=100000Output sorted array in descending order where each element is space separated.Sample Input:
6
3 1 2 7 9 87
Sample Output:
87 9 7 3 2 1, I have written this Solution Code: def bubbleSort(arr):
arr.sort(reverse = True)
return arr
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array, sort the array in reverse order by simply swapping its adjacent elements.First line of the input contains an integer, N, which denotes the length of the array. Next N inputs are elements of the array that is to be sorted in descending order.
Constraints
1<=N<=1000
-10000<=Arr[i]<=100000Output sorted array in descending order where each element is space separated.Sample Input:
6
3 1 2 7 9 87
Sample Output:
87 9 7 3 2 1, I have written this Solution Code:
import java.util.*;
import java.lang.*;
import java.io.*;
class Main
{
public static void main (String[] args) throws java.lang.Exception
{
Scanner sc = new Scanner(System.in);
int n=sc.nextInt();
int a[] = new int[n];
for(int i=0;i<n;i++){
a[i]=sc.nextInt();
}
int t;
for(int i=1;i<n;i++){
if(a[i]>a[i-1]){
for(int j=i;j>0;j--){
if(a[j]>a[j-1]){
t=a[j];
a[j]=a[j-1];
a[j-1]=t;
}
}
}
}
for(int i=0;i<n;i++){
System.out.print(a[i]+" ");
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array, sort the array in reverse order by simply swapping its adjacent elements.First line of the input contains an integer, N, which denotes the length of the array. Next N inputs are elements of the array that is to be sorted in descending order.
Constraints
1<=N<=1000
-10000<=Arr[i]<=100000Output sorted array in descending order where each element is space separated.Sample Input:
6
3 1 2 7 9 87
Sample Output:
87 9 7 3 2 1, 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
const int N = 3e5+5;
#define read(type) readInt<type>()
#define max1 100001
#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;
const ll inf = 0x3f3f3f3f3f3f3f3f;
const ll mod = 998244353;
using vl = vector<ll>;
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;
}
long long phi[max1], result[max1],F[max1];
// Precomputation of phi[] numbers. Refer below link
// for details : https://goo.gl/LUqdtY
void computeTotient()
{
// Refer https://goo.gl/LUqdtY
phi[1] = 1;
for (int i=2; i<max1; i++)
{
if (!phi[i])
{
phi[i] = i-1;
for (int j = (i<<1); j<max1; j+=i)
{
if (!phi[j])
phi[j] = j;
phi[j] = (phi[j]/i)*(i-1);
}
}
}
for(int i=1;i<=100000;i++)
{
for(int j=i;j<=100000;j+=i)
{ int p=j/i;
F[j]+=(i*phi[p])%mod;
F[j]%=mod;
}
}
}
int gcd(int a, int b, int& x, int& y) {
if (b == 0) {
x = 1;
y = 0;
return a;
}
int x1, y1;
int d = gcd(b, a % b, x1, y1);
x = y1;
y = x1 - y1 * (a / b);
return d;
}
bool find_any_solution(int a, int b, int c, int &x0, int &y0, int &g) {
g = gcd(abs(a), abs(b), x0, y0);
if (c % g) {
return false;
}
x0 *= c / g;
y0 *= c / g;
if (a < 0) x0 = -x0;
if (b < 0) y0 = -y0;
return true;
}
int main() {
int n;
cin>>n;
int a[n];
for(int i=0;i<n;i++){
cin>>a[i];
}
sort(a,a+n,greater<int>());
FOR(i,n){
out1(a[i]);}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Write a program to find the compound interest for given principal amount P, time Tm(in years), and interest rate R.
<b>Note:</b> Compound interest is the interest you earn on interest.
This can be illustrated by using basic math: if you have $100 and it earns 5% interest each year, you'll have $105 at the end of the first year. At the end of the second year, you'll have $110.25The input contains three integers P, R, and Tm.
<b>Constraints:- </b>
1 < = P < = 10^3
1 < = R < = 100
1 < = Tm < = 20Print the compound interest by <b> 2 decimal places </b>.Sample Input:
100 1 2
Sample Output:-
2.01
Sample Input:
1 99 2
Sample Output:-
2.96, I have written this Solution Code: def compound_interest(principle, rate, time):
Amount = principle * (pow((1 + rate / 100), time))
CI = Amount - principle
print( '%.2f'%CI)
principle,rate,time=map(int, input().split())
compound_interest(principle,rate,time), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
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