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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: Alexa Loves to play with 0's and 2's. Among the positive integers that consist of 0's and 2's when written in base 10, he wanted to find the Kth smallest integer. Help him find that.Input is given from Standard Input in the following format:
KPrint the answer as an integer.
Here, the exact value must be printed as an integer, even if it is big. Unnecessary leading zeros such as 0523 are not allowed.<b>Sample Input 1</b>
3
<b>Sample Output 1</b>
22
<b>Sample Input 2</b>
11
<b>Sample Output 2</b>
2022
<b>Sample Input 3</b>
923423423420220108
<b>Sample Output 3</b>
220022020000202020002022022000002020002222002200002022002200
Explanation:
Positive numbers consisting of 0's and 2's are 2,20,22,200,202,220 and so on. Therefore,the 3rd smallest such number is 22., I have written this Solution Code: #include<bits/stdc++.h>
using namespace std;
string convert(long long x){
string res;
while(x>0){
res.push_back('0'+(x%2));
x/=2;
}
reverse(res.begin(),res.end());
return res;
}
void output(string s){
for(auto &nx : s){
if(nx=='1'){cout << '2';}
else{cout << '0';}
}
cout << '\n';
}
int main(){
long long k;
cin >> k;
output(convert(k));
return 0;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A of size n, and an integer k. Find the maximum force by involving only k elements. The Force of an element is the square of its value.
<b>Note:</b>
Elements are not needed to be continuous.The first line of the input contains two integers denoting n and k.
The next line contains n integers denoting elements of the array.
<b>Constraints:</b>
1 < = k < = n < = 1000
-10^7 <= A[i] <= 10^7Output the maximum force.Sample Input 1:
4 4
1 2 3 4
Sample Output 1:
30
Sample Input 2:
2 1
1 10
Sample Output 2:
100
<b>Explanation:</b>
Force = 1*1 + 2*2 + 3*3 + 4*4 = 30, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int m = 100001;
int main(){
int n,k;
cin>>n>>k;
long long a[n],sum=0;
for(int i=0;i<n;i++){
cin>>a[i];
if(a[i]<0){
a[i]=-a[i];
}
}
sort(a,a+n);
for(int i=0;i<k;i++){
sum+=a[n-i-1]*a[n-i-1];
}
cout<<sum;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A of size n, and an integer k. Find the maximum force by involving only k elements. The Force of an element is the square of its value.
<b>Note:</b>
Elements are not needed to be continuous.The first line of the input contains two integers denoting n and k.
The next line contains n integers denoting elements of the array.
<b>Constraints:</b>
1 < = k < = n < = 1000
-10^7 <= A[i] <= 10^7Output the maximum force.Sample Input 1:
4 4
1 2 3 4
Sample Output 1:
30
Sample Input 2:
2 1
1 10
Sample Output 2:
100
<b>Explanation:</b>
Force = 1*1 + 2*2 + 3*3 + 4*4 = 30, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws IOException {
InputStreamReader ir = new InputStreamReader(System.in);
BufferedReader br = new BufferedReader(ir);
String[] NK = br.readLine().split(" ");
String[] inputs = br.readLine().split(" ");
int N = Integer.parseInt(NK[0]);
int K = Integer.parseInt(NK[1]);
long[] arr = new long[N];
long answer = 0;
for(int i = 0; i < N; i++){
arr[i] = Math.abs(Long.parseLong(inputs[i]));
}
quicksort(arr, 0, N-1);
for(int i = (N-K); i < N;i++){
answer += (arr[i]*arr[i]);
}
System.out.println(answer);
}
static void quicksort(long[] arr, int start, int end){
if(start < end){
int pivot = partition(arr, start, end);
quicksort(arr, start, pivot-1);
quicksort(arr, pivot+1, end);
}
}
static int partition(long[] arr, int start, int end){
long pivot = arr[end];
int i = start - 1;
for(int j = start; j < end; j++){
if(arr[j] < pivot){
i++;
swap(arr, i, j);
}
}
swap(arr, i+1, end);
return (i+1);
}
static void swap(long[] arr, int i, int j){
long temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A of size n, and an integer k. Find the maximum force by involving only k elements. The Force of an element is the square of its value.
<b>Note:</b>
Elements are not needed to be continuous.The first line of the input contains two integers denoting n and k.
The next line contains n integers denoting elements of the array.
<b>Constraints:</b>
1 < = k < = n < = 1000
-10^7 <= A[i] <= 10^7Output the maximum force.Sample Input 1:
4 4
1 2 3 4
Sample Output 1:
30
Sample Input 2:
2 1
1 10
Sample Output 2:
100
<b>Explanation:</b>
Force = 1*1 + 2*2 + 3*3 + 4*4 = 30, I have written this Solution Code: x,y = map(int,input().split())
arr = list(map(int,input().split()))
s=0
for i in range(x):
if arr[i]<0:
arr[i]=abs(arr[i])
arr=sorted(arr,reverse=True)
for i in range(0,y):
s = s+arr[i]*arr[i]
print(s)
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You will be given 2 parameters: a low and high number. Your goal is to print all numbers between low and high,
and for each of these numbers print whether or not the number is divisible by 3. If the number is divisible by 3,
print the word "div3" directly after the number.2 numbers, one will be low and other high.
0<=low<=high<=10000If the number is divisible by 3, print the word "div3" directly after the number.Sample input:-
1 6
Sample output:-
1 2 3 div3 4 5 6 div3, 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 low = sc.nextInt();
int high = sc.nextInt();
for(int i = low; i <= high; i++){
if(i%3 == 0){
System.out.print(i);
System.out.print(" ");
System.out.print("div"+3);
System.out.print(" ");
}
else{
System.out.print(i);
System.out.print(" ");
}
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You will be given 2 parameters: a low and high number. Your goal is to print all numbers between low and high,
and for each of these numbers print whether or not the number is divisible by 3. If the number is divisible by 3,
print the word "div3" directly after the number.2 numbers, one will be low and other high.
0<=low<=high<=10000If the number is divisible by 3, print the word "div3" directly after the number.Sample input:-
1 6
Sample output:-
1 2 3 div3 4 5 6 div3, I have written this Solution Code: inp = input("").split(" ")
init = []
for i in range(int(inp[0]),int(inp[1])+1):
if(i%3 == 0):
init.append(str(i)+" div3")
else:
init.append(str(i))
print(" ".join(init)), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You will be given 2 parameters: a low and high number. Your goal is to print all numbers between low and high,
and for each of these numbers print whether or not the number is divisible by 3. If the number is divisible by 3,
print the word "div3" directly after the number.2 numbers, one will be low and other high.
0<=low<=high<=10000If the number is divisible by 3, print the word "div3" directly after the number.Sample input:-
1 6
Sample output:-
1 2 3 div3 4 5 6 div3, I have written this Solution Code: function test_divisors(low, high) {
// we'll store all numbers and strings within an array
// instead of printing directly to the console
const output = [];
for (let i = low; i <= high; i++) {
// simply store the current number in the output array
output.push(i);
// check if the current number is evenly divisible by 3
if (i % 3 === 0) { output.push('div3'); }
}
// return all numbers and strings
console.log(output.join(" "));
}
, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a string, convert it to a list containing all of its elements only once without any repetition, even if there is repetition in the string.
hint - you might use sets for this.The input would be a string containing 5 characters. Each character can be any alpha numeric value.The output would be a sorted list with no repeating elementsSample input
abbbc
Sample output
['a', 'b', 'c']
Explanation-
the string 'abbbc' contains elements 'a','b', and 'c' , therefore they are printed out in the list without any duplication., I have written this Solution Code: my_list = []
strin = input()
for i in range(5):
my_list.append(strin[i])
print(sorted(list(set(my_list)))), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Let's assume some functional definitions for this problem.
We take prime(x) as the set of all prime divisors of x. For example, prime(140)={2,5,7}, prime(169)={13}.
Let f(x,p) be the maximum possible integer p**k where k is an integer such that x is divisible by p**k.
(Here a**b means a raised to the power b or pow(a, b))
For example:
f(99,3)=9 (99 is divisible by 3**2=9 but not divisible by 3**3=27),
f(63,7)=7 (63 is divisible by 7**1=7 but not divisible by 7**2=49).
Let g(x,y) be the product of f(y,p) for all p in prime(x).
For example:
g(30,70)=f(70,2)*f(70,3)*f(70,5)=2*1*5=10,
g(525,63)=f(63,3)*f(63,5)*f(63,7)=9*1*7=63.
You are given two integers x and n. Calculate g(x,1)*g(x,2)*…*g(x,n) mod (1000000007).
(Read modulo exponentiation before attempting this problem)The only line contains integers x and n — the numbers used in formula.
Constraints
2 ≤ x ≤ 1000000000
1 ≤ n ≤ 1000000000000000000Print the answer corresponding to the input.Sample Input 1
10 2
Sample Output 1
2
Sample Input 2
20190929 1605
Sample Output 2
363165664
Explanation
In the first example, g(10,1)=f(1,2)⋅f(1,5)=1, g(10,2)=f(2,2)⋅f(2,5)=2., I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
static long mod =1000000007;
public static boolean isPrime(long m){
if (m <2)
return false;
for(int i =2 ;i*i<=m;i++)
{
if (m%i == 0)
return false;
}
return true;
}
public static void main (String[] args) throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
String s [] =br.readLine().trim().split(" ");
long x = Long.parseLong(s[0]);
long n = Long.parseLong(s[1]);
long ans = 1;
for(int i = 2; i*i <= x; i++){
if(x%i != 0) continue;
ans = (ans*f(n, i)) % mod;
while(x%i == 0)
x /= i;
}
if(x > 1)
ans = (ans*f(n, x)) % mod;
System.out.println(ans);
}
static long f(long n, long p){
long ans = 1;
long cur = 1;
while(cur <= n/p){
cur = cur*p;
long z = power(p, n/cur);
ans = (ans*z) % mod;
}
return ans;
}
public static long power(long no,long pow){
long p = 1000000007;
long result = 1;
while(pow > 0)
{
if ( (pow & 1) == 1)
result = ((result%p) * (no%p))%p;
no = ((no%p) * (no%p))%p;
pow >>= 1;
}
return result;
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Let's assume some functional definitions for this problem.
We take prime(x) as the set of all prime divisors of x. For example, prime(140)={2,5,7}, prime(169)={13}.
Let f(x,p) be the maximum possible integer p**k where k is an integer such that x is divisible by p**k.
(Here a**b means a raised to the power b or pow(a, b))
For example:
f(99,3)=9 (99 is divisible by 3**2=9 but not divisible by 3**3=27),
f(63,7)=7 (63 is divisible by 7**1=7 but not divisible by 7**2=49).
Let g(x,y) be the product of f(y,p) for all p in prime(x).
For example:
g(30,70)=f(70,2)*f(70,3)*f(70,5)=2*1*5=10,
g(525,63)=f(63,3)*f(63,5)*f(63,7)=9*1*7=63.
You are given two integers x and n. Calculate g(x,1)*g(x,2)*…*g(x,n) mod (1000000007).
(Read modulo exponentiation before attempting this problem)The only line contains integers x and n — the numbers used in formula.
Constraints
2 ≤ x ≤ 1000000000
1 ≤ n ≤ 1000000000000000000Print the answer corresponding to the input.Sample Input 1
10 2
Sample Output 1
2
Sample Input 2
20190929 1605
Sample Output 2
363165664
Explanation
In the first example, g(10,1)=f(1,2)⋅f(1,5)=1, g(10,2)=f(2,2)⋅f(2,5)=2., I have written this Solution Code: import math
mod = 1000000007
def modExpo(x, n):
if n <= 0:
return 1
if n % 2 == 0:
return modExpo((x * x) % mod, n // 2)
return (x * modExpo((x * x) % mod, (n - 1) // 2)) % mod
def calc(n, p):
prod = 1
pPow = 1
while pPow <= (n // p):
pPow *= p
add = 0
while pPow != 1:
quo = n // pPow
quo -= add
power = modExpo(pPow, quo)
prod = (prod * power) % mod
add += quo
pPow //= p
return prod
x, n = map(int, input().split())
res = 1
i = 2
while (i * i) <= x:
if x % i == 0:
res = (res * calc(n, i)) % mod
while x % i == 0:
x //= i
i += 1
if x > 1:
res = (res * calc(n, x)) % mod
print(res), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Let's assume some functional definitions for this problem.
We take prime(x) as the set of all prime divisors of x. For example, prime(140)={2,5,7}, prime(169)={13}.
Let f(x,p) be the maximum possible integer p**k where k is an integer such that x is divisible by p**k.
(Here a**b means a raised to the power b or pow(a, b))
For example:
f(99,3)=9 (99 is divisible by 3**2=9 but not divisible by 3**3=27),
f(63,7)=7 (63 is divisible by 7**1=7 but not divisible by 7**2=49).
Let g(x,y) be the product of f(y,p) for all p in prime(x).
For example:
g(30,70)=f(70,2)*f(70,3)*f(70,5)=2*1*5=10,
g(525,63)=f(63,3)*f(63,5)*f(63,7)=9*1*7=63.
You are given two integers x and n. Calculate g(x,1)*g(x,2)*…*g(x,n) mod (1000000007).
(Read modulo exponentiation before attempting this problem)The only line contains integers x and n — the numbers used in formula.
Constraints
2 ≤ x ≤ 1000000000
1 ≤ n ≤ 1000000000000000000Print the answer corresponding to the input.Sample Input 1
10 2
Sample Output 1
2
Sample Input 2
20190929 1605
Sample Output 2
363165664
Explanation
In the first example, g(10,1)=f(1,2)⋅f(1,5)=1, g(10,2)=f(2,2)⋅f(2,5)=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 power(int a, int b){
int ans = 1;
b %= (mod-1);
while(b){
if(b&1)
ans = (ans*a) % mod;
b >>= 1;
a = (a*a) % mod;
}
return ans;
}
int f(int n, int p){
int ans = 1;
int cur = 1;
while(cur <= n/p){
cur = cur*p;
int z = power(p, n/cur);
ans = (ans*z) % mod;
}
return ans;
}
signed main() {
IOS;
int x, n, ans = 1;
cin >> x >> n;
for(int i = 2; i*i <= x; i++){
if(x%i != 0) continue;
ans = (ans*f(n, i)) % mod;
while(x%i == 0)
x /= i;
}
if(x > 1)
ans = (ans*f(n, x)) % mod;
cout << ans;
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given two non-negative integers, A and B. You are required to print the smallest non-negative integer such that it is not equal to both A and B.The input consists of two space separated integers A and B.
Constraints:
0 ≤ A ≤ 10
0 ≤ B ≤ 10
A ≠ B
Print a single integer denoting the answer.Sample Input 1:
0 1
Sample Output 1:
2
Sample Input 2:
4 5
Sample Output 2:
0, 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();
Set<Integer> set=new HashSet<>();
String s[]=bu.readLine().split(" ");
for(String x:s) set.add(Integer.parseInt(x));
int mex=0;
while(set.contains(mex)) mex++;
System.out.println(mex);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given two non-negative integers, A and B. You are required to print the smallest non-negative integer such that it is not equal to both A and B.The input consists of two space separated integers A and B.
Constraints:
0 ≤ A ≤ 10
0 ≤ B ≤ 10
A ≠ B
Print a single integer denoting the answer.Sample Input 1:
0 1
Sample Output 1:
2
Sample Input 2:
4 5
Sample Output 2:
0, I have written this Solution Code: p=input()
q=list(p.split(" "))
for i in range(0,int(q[1])+2):
if(int(q[0])!=i and int(q[1])!=i):
print(i)
break, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given two non-negative integers, A and B. You are required to print the smallest non-negative integer such that it is not equal to both A and B.The input consists of two space separated integers A and B.
Constraints:
0 ≤ A ≤ 10
0 ≤ B ≤ 10
A ≠ B
Print a single integer denoting the answer.Sample Input 1:
0 1
Sample Output 1:
2
Sample Input 2:
4 5
Sample Output 2:
0, I have written this Solution Code: //Author: Xzirium
//Time and Date: 15:00:01 23 April 2022
#include <bits/stdc++.h>
using namespace std;
int main()
{
#ifndef ONLINE_JUDGE
if (fopen("INPUT.txt", "r"))
{
freopen ("INPUT.txt" , "r" , stdin);
//freopen ("OUTPUT.txt" , "w" , stdout);
}
#endif
int A,B;
cin>>A>>B;
for(int i=0 ; i<=10 ; i++)
{
if(i!=A && i!=B)
{
cout<<i<<endl;
break;
}
}
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: Given an array containing N integers and an integer K. Your task is to find the length of the longest Sub-Array with sum of the elements equal to the given value K.The first line of input contains an integer T denoting the number of test cases. Then T test cases follow. Each test case consists of two lines. The first line of each test case contains two integers N and K and the second line contains N space-separated elements of the array.
Constraints:-
1<=T<=500
1<=N,K<=10^5
-10^5<=A[i]<=10^5
Sum of N over all test cases does not exceed 10^5For each test case, print the required length of the longest Sub-Array in a new line. If no such sub-array can be formed print 0.Sample Input:
3
6 15
10 5 2 7 1 9
6 -5
-5 8 -14 2 4 12
3 6
-1 2 3
Sample Output:
4
5
0, I have written this Solution Code: def lenOfLongSubarr(arr, N, K):
mydict = dict()
sum = 0
maxLen = 0
for i in range(N):
sum += arr[i]
if (sum == K):
maxLen = i + 1
elif (sum - K) in mydict:
maxLen = max(maxLen, i - mydict[sum - K])
if sum not in mydict:
mydict[sum] = i
return maxLen
if __name__ == '__main__':
T = int(input())
#N,K=list(map(int,input().split()))
for i in range(T):
N,k= [int(N)for N in input("").split()]
arr=list(map(int,input().split()))
N = len(arr)
print(lenOfLongSubarr(arr, N, k)), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array containing N integers and an integer K. Your task is to find the length of the longest Sub-Array with sum of the elements equal to the given value K.The first line of input contains an integer T denoting the number of test cases. Then T test cases follow. Each test case consists of two lines. The first line of each test case contains two integers N and K and the second line contains N space-separated elements of the array.
Constraints:-
1<=T<=500
1<=N,K<=10^5
-10^5<=A[i]<=10^5
Sum of N over all test cases does not exceed 10^5For each test case, print the required length of the longest Sub-Array in a new line. If no such sub-array can be formed print 0.Sample Input:
3
6 15
10 5 2 7 1 9
6 -5
-5 8 -14 2 4 12
3 6
-1 2 3
Sample Output:
4
5
0, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main(){
int t;
cin>>t;
while(t--){
unordered_map<long long,int> um;
int n,k;
cin>>n>>k;
long arr[n];
int maxLen=0;
for(int i=0;i<n;i++){cin>>arr[i];}
long long sum=0;
for(int i=0;i<n;i++){
sum += arr[i];
// when subarray starts from index '0'
if (sum == k)
maxLen = i + 1;
// make an entry for 'sum' if it is
// not present in 'um'
if (um.find(sum) == um.end())
um[sum] = i;
// check if 'sum-k' is present in 'um'
// or not
if (um.find(sum - k) != um.end()) {
// update maxLength
if (maxLen < (i - um[sum - k]))
maxLen = i - um[sum - k];
}
}
cout<<maxLen<<endl;
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array containing N integers and an integer K. Your task is to find the length of the longest Sub-Array with sum of the elements equal to the given value K.The first line of input contains an integer T denoting the number of test cases. Then T test cases follow. Each test case consists of two lines. The first line of each test case contains two integers N and K and the second line contains N space-separated elements of the array.
Constraints:-
1<=T<=500
1<=N,K<=10^5
-10^5<=A[i]<=10^5
Sum of N over all test cases does not exceed 10^5For each test case, print the required length of the longest Sub-Array in a new line. If no such sub-array can be formed print 0.Sample Input:
3
6 15
10 5 2 7 1 9
6 -5
-5 8 -14 2 4 12
3 6
-1 2 3
Sample Output:
4
5
0, 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 t=sc.nextInt();
while(t-->0){
if(t%10==0){
System.gc();
}
int arrsize=sc.nextInt();
int k=sc.nextInt();
int[] arr=new int[arrsize];
for(int i=0;i<arrsize;i++){
arr[i]=sc.nextInt();
}
int subsize=0;
int sum=0;
HashMap<Integer, Integer> hash=new HashMap<>();
for(int i=0;i<arrsize;i++){
sum+=arr[i];
if(sum==k){
subsize=i+1;
}
if(!hash.containsKey(sum)){
hash.put(sum,i);
}
if(hash.containsKey(sum-k)){
if(subsize<(i-hash.get(sum-k))){
subsize=i-hash.get(sum-k);
}
}
}
System.out.println(subsize);
}
}
}, 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 a number n find the number of prime numbers less than equal to that number.There is only one integer containing value of n.
Constraints:-
1 <= n <= 10000000Return number of primes less than or equal to nSample Input
5
Sample Output
3
Explanation:-
2 3 and 5 are the required primes.
Sample Input
5000
Sample Output
669, I have written this Solution Code: #include <bits/stdc++.h>
// #define ll long long
using namespace std;
#define ma 10000001
bool a[ma];
int main()
{
int n;
cin>>n;
for(int i=0;i<=n;i++){
a[i]=false;
}
for(int i=2;i<=n;i++){
if(a[i]==false){
for(int j=i+i;j<=n;j+=i){
a[j]=true;
}
}
}
int cnt=0;
for(int i=2;i<=n;i++){
if(a[i]==false){cnt++;}
}
cout<<cnt;
return 0;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number n find the number of prime numbers less than equal to that number.There is only one integer containing value of n.
Constraints:-
1 <= n <= 10000000Return number of primes less than or equal to nSample Input
5
Sample Output
3
Explanation:-
2 3 and 5 are the required primes.
Sample Input
5000
Sample Output
669, I have written this Solution Code: import java.io.*;
import java.util.*;
import java.lang.Math;
class Main {
public static void main (String[] args)
throws IOException
{
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
long n = Integer.parseInt(br.readLine());
long i=2,j,count,noOfPrime=0;
if(n<=1)
System.out.println("0");
else{
while(i<=n)
{
count=0;
for(j=2; j<=Math.sqrt(i); j++)
{
if( i%j == 0 ){
count++;
break;
}
}
if(count==0){
noOfPrime++;
}
i++;
}
System.out.println(noOfPrime);
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number n find the number of prime numbers less than equal to that number.There is only one integer containing value of n.
Constraints:-
1 <= n <= 10000000Return number of primes less than or equal to nSample Input
5
Sample Output
3
Explanation:-
2 3 and 5 are the required primes.
Sample Input
5000
Sample Output
669, I have written this Solution Code: function numberOfPrimes(N)
{
let arr = new Array(N+1);
for(let i = 0; i <= N; i++)
arr[i] = 0;
for(let i=2; i<= N/2; i++)
{
if(arr[i] === -1)
{
continue;
}
let p = i;
for(let j=2; p*j<= N; j++)
{
arr[p*j] = -1;
}
}
//console.log(arr);
let count = 0;
for(let i=2; i<= N; i++)
{
if(arr[i] === 0)
{
count++;
}
}
//console.log(arr);
return count;
}, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number n find the number of prime numbers less than equal to that number.There is only one integer containing value of n.
Constraints:-
1 <= n <= 10000000Return number of primes less than or equal to nSample Input
5
Sample Output
3
Explanation:-
2 3 and 5 are the required primes.
Sample Input
5000
Sample Output
669, I have written this Solution Code: import math
n = int(input())
n=n+1
if n<3:
print(0)
else:
primes=[1]*(n//2)
for i in range(3,int(math.sqrt(n))+1,2):
if primes[i//2]:primes[i*i//2::i]=[0]*((n-i*i-1)//(2*i)+1)
print(sum(primes)), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a Binary Tree, find diameter of it. The <b>diameter</b> of a tree is the number of nodes on the longest path between two leaves in the tree. The diagram below shows two trees each with diameter nine, the leaves that form the ends of a longest path are shaded (note that there is more than one path in each tree of length nine, but no path longer than nine nodes).<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 getDiameter() that takes "root" node as parameter.
<b>Constraints:</b>
1 <= T <= 100
1 <= N <= 10^4
1 <= node values <= 10^3
<b>Sum of "N" over all testcases does not exceed 10^5</b>
For <b>Custom Input:</b>
First line of input should contains the number of test cases T. For each test case, there will be two lines of input.
First line contains number of nodes N. Second line will be a string representing the tree as described below:
The values in the string are in the order of level order traversal of the tree where, numbers denote node values, and a character “N” denotes NULL child.
<b>Note:</b> If a node has been declared Null using 'N', no information about its children will be given further in the array.Return the diameter of the tree.Sample Input:
2
3
1 2 3
5
10 20 30 40 60
Sample Output:
3
4
Explanation:
Test Case1: The tree is
1
/ \
2 3
The diameter is of 3 length.
Test Case2: The tree is
10
/ \
20 30
/ \
40 60
The diameter is of 4 length., I have written this Solution Code: static int dia = 0;
public static int util(Node root) {
if (root == null) return 0;
int l = util(root.left); // height of left subtree
int r = util(root.right); // height of right subtree
if (l + r + 1 > dia) dia = l + r + 1; // l+r+1 is a possible max dia
return 1 + Math.max(l, r); // returning height of subtree
}
public static int getDiameter(Node root) {
dia = 0;
util(root);
return dia;
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A of size N, find the maximum value of A<sub>i</sub> - A<sub>j</sub> over all i, j such that 1 <= i, j <= N.First line of the input contains a single integer N.
The second line of the input contains N space seperated integers.
Constraints:
1 <= N <= 10<sup>5</sup>
1 <= A<sub>i</sub> <= 10<sup>9</sup>Print the maximum value of A<sub>i</sub> - A<sub>j</sub> over all i, j such that 1 <= i, j <= N.Sample Input:
5
6 7 4 5 2
Sample Output:
5
Explaination:
We take i = 2, j = 5. A<sub>2</sub> - A<sub>5</sub> = 7 - 2 = 5., I have written this Solution Code: n = int(input())
arr = [int(x) for x in input().split()]
arr.sort()
print(arr[n-1] - arr[0]), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A of size N, find the maximum value of A<sub>i</sub> - A<sub>j</sub> over all i, j such that 1 <= i, j <= N.First line of the input contains a single integer N.
The second line of the input contains N space seperated integers.
Constraints:
1 <= N <= 10<sup>5</sup>
1 <= A<sub>i</sub> <= 10<sup>9</sup>Print the maximum value of A<sub>i</sub> - A<sub>j</sub> over all i, j such that 1 <= i, j <= N.Sample Input:
5
6 7 4 5 2
Sample Output:
5
Explaination:
We take i = 2, j = 5. A<sub>2</sub> - A<sub>5</sub> = 7 - 2 = 5., 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 = 2e9;
void solve(){
int n;
cin >> n;
vector<int> a(n);
for(auto &i : a) cin >> i;
sort(all(a));
cout << a.back() - a[0];
}
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: 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: Tono has two integer intervals [A, B] and [C, D] (the intervals are inclusive of both numbers). She wonders if Solo will be able to solve the following simple problem:
Choose an integer X from first interval and an integer Y from second interval. Now for all combinations of X and Y, what is the maximum value of X*Y.
As Solo is small, please help her solve the problem.The first and the only line of input contains 4 integers A, B, C, and D.
Constraints
-10<sup>9</sup> <= A <= B <= 10<sup>9</sup>
-10<sup>9</sup> <= C <= D <= 10<sup>9</sup>Output a single integer, the maximum value of the product.Sample Input
1 2 1 1
Sample Output
2
Explanation: Choose 2 from [1, 2] and 1 from [1, 1].
Sample Input
3 5 -4 -2
Sample Output
-6
Sample Input
-1000000000 0 -1000000000 0
Sample Output
1000000000000000000, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws IOException{
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
String[] arr=br.readLine().split(" ");
int a = Integer.parseInt(arr[0]);
int b = Integer.parseInt(arr[1]);
int c = Integer.parseInt(arr[2]);
int d= Integer.parseInt(arr[3]);
int secondMax = Math.max(a,b) < 0 ? Math.min(c,d) : Math.max(c,d);
int firstMax = Math.max(c,d) < 0 ? Math.min(a,b) : Math.max(a,b);
if(Math.abs(Math.min(a,b)) > Math.max(a,b) && Math.abs(Math.min(c,d)) > Math.max(c,d)){
long num= (long)Math.min(a,b) * Math.min(c,d);
System.out.println(num);
return;
}
long prod = (long)firstMax*secondMax;
System.out.println(prod);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Tono has two integer intervals [A, B] and [C, D] (the intervals are inclusive of both numbers). She wonders if Solo will be able to solve the following simple problem:
Choose an integer X from first interval and an integer Y from second interval. Now for all combinations of X and Y, what is the maximum value of X*Y.
As Solo is small, please help her solve the problem.The first and the only line of input contains 4 integers A, B, C, and D.
Constraints
-10<sup>9</sup> <= A <= B <= 10<sup>9</sup>
-10<sup>9</sup> <= C <= D <= 10<sup>9</sup>Output a single integer, the maximum value of the product.Sample Input
1 2 1 1
Sample Output
2
Explanation: Choose 2 from [1, 2] and 1 from [1, 1].
Sample Input
3 5 -4 -2
Sample Output
-6
Sample Input
-1000000000 0 -1000000000 0
Sample Output
1000000000000000000, I have written this Solution Code: A,B,C,D=list(map(int,input().split()))
p=[A*C,A*D,B*C,B*D]
print(max(p)), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Tono has two integer intervals [A, B] and [C, D] (the intervals are inclusive of both numbers). She wonders if Solo will be able to solve the following simple problem:
Choose an integer X from first interval and an integer Y from second interval. Now for all combinations of X and Y, what is the maximum value of X*Y.
As Solo is small, please help her solve the problem.The first and the only line of input contains 4 integers A, B, C, and D.
Constraints
-10<sup>9</sup> <= A <= B <= 10<sup>9</sup>
-10<sup>9</sup> <= C <= D <= 10<sup>9</sup>Output a single integer, the maximum value of the product.Sample Input
1 2 1 1
Sample Output
2
Explanation: Choose 2 from [1, 2] and 1 from [1, 1].
Sample Input
3 5 -4 -2
Sample Output
-6
Sample Input
-1000000000 0 -1000000000 0
Sample Output
1000000000000000000, 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(int i=st; i<en; i++)
#define tr(x) for(auto it=x.begin(); it!=x.end(); it++)
#define fast std::ios::sync_with_stdio(false);cin.tie(NULL);
#define 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 a, b, c, d; cin>>a>>b>>c>>d;
int a1 = max(a*c, b*d);
int a2 = max(a*d, b*c);
cout<<max(a1, a2);
}
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: Given a number N for each I (1 < = I < = N), you have to print the number except:-
For each multiple of 3, print "Fizz" instead of the number.
For each multiple of 5, print "Buzz" instead of the number.
For numbers that are multiples of both 3 and 5, print "FizzBuzz" instead of the number.The input contains a single integer N.
<b>Constraints:-</b>
1 ≤ N ≤ 1000Print N space separated number or Fizz buzz according to the condition.Sample Input:-
3
Sample Output:-
1 2 Fizz
Sample Input:-
5
Sample Output:-
1 2 Fizz 4 Buzz, I have written this Solution Code: void fizzbuzz(int n){
for(int i=1;i<=n;i++){
if(i%3==0 && i%5==0){cout<<"FizzBuzz"<<" ";}
else if(i%5==0){cout<<"Buzz ";}
else if(i%3==0){cout<<"Fizz ";}
else{cout<<i<<" ";}
}
}, 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 "Fizz" instead of the number.
For each multiple of 5, print "Buzz" instead of the number.
For numbers that are multiples of both 3 and 5, print "FizzBuzz" instead of the number.The input contains a single integer N.
<b>Constraints:-</b>
1 ≤ N ≤ 1000Print N space separated number or Fizz buzz according to the condition.Sample Input:-
3
Sample Output:-
1 2 Fizz
Sample Input:-
5
Sample Output:-
1 2 Fizz 4 Buzz, 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 x= sc.nextInt();
fizzbuzz(x);
}
static void fizzbuzz(int n){
for(int i=1;i<=n;i++){
if(i%3==0 && i%5==0){System.out.print("FizzBuzz ");}
else if(i%5==0){System.out.print("Buzz ");}
else if(i%3==0){System.out.print("Fizz ");}
else{System.out.print(i+" ");}
}
}
}, In this Programming Language: Java, 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 "Fizz" instead of the number.
For each multiple of 5, print "Buzz" instead of the number.
For numbers that are multiples of both 3 and 5, print "FizzBuzz" instead of the number.The input contains a single integer N.
<b>Constraints:-</b>
1 ≤ N ≤ 1000Print N space separated number or Fizz buzz according to the condition.Sample Input:-
3
Sample Output:-
1 2 Fizz
Sample Input:-
5
Sample Output:-
1 2 Fizz 4 Buzz, I have written this Solution Code: def fizzbuzz(n):
for i in range (1,n+1):
if (i%3==0 and i%5==0):
print("FizzBuzz",end=' ')
elif i%3==0:
print("Fizz",end=' ')
elif i%5==0:
print("Buzz",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 "Fizz" instead of the number.
For each multiple of 5, print "Buzz" instead of the number.
For numbers that are multiples of both 3 and 5, print "FizzBuzz" instead of the number.The input contains a single integer N.
<b>Constraints:-</b>
1 ≤ N ≤ 1000Print N space separated number or Fizz buzz according to the condition.Sample Input:-
3
Sample Output:-
1 2 Fizz
Sample Input:-
5
Sample Output:-
1 2 Fizz 4 Buzz, I have written this Solution Code: void fizzbuzz(int n){
for(int i=1;i<=n;i++){
if(i%3==0 && i%5==0){printf("FizzBuzz ");}
else if(i%5==0){printf("Buzz ");}
else if(i%3==0){printf("Fizz ");}
else{printf("%d ",i);}
}
}, In this Programming Language: C, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a string <i>s</i> consisting of lowercase English letters. Find the number of different palindromes you can make by changing <b>exactly</b> one charecter from the string to some other lowercase English letter.The first and single line contains string <i>s</i> (1 ≤ |<i>s</i>| ≤ 10).Print the number of different palindromes you can make by changing <b>exactly</b> one charecter from the string to some other lowercase English letter. Sample Input 1
abbb
Sample Output 1
2
Explanation:
The possible palindromes are:
1. abba
2. bbbb
========================================================================
Sample Input 2
abba
Sample Output 2
0
, 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));
char str[] = br.readLine().toCharArray();
int ans = 0;
char arr[] = {'a','b','c','d','e','f','g','h','i','j','k','l','m','n','o','p','q','r','s','t','u','v','w','x','y','z'};
Set<String> set = new HashSet<>();
for(int i=0;i<str.length;i++){
char p = str[i];
for(char ch:arr){
if(ch==p) continue;
str[i] = ch;
if(isPallindrome(str)){
if(set.contains(String.valueOf(str))==false){
set.add(String.valueOf(str));
ans++;
}
}
str[i] = p;
}
}
System.out.println(ans);
}
static boolean isPallindrome(char[] str){
int i = 0;
int j = str.length-1;
while(i<j){
if(str[i]!=str[j]) return false;
i++;
j--;
}
return true;
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a string <i>s</i> consisting of lowercase English letters. Find the number of different palindromes you can make by changing <b>exactly</b> one charecter from the string to some other lowercase English letter.The first and single line contains string <i>s</i> (1 ≤ |<i>s</i>| ≤ 10).Print the number of different palindromes you can make by changing <b>exactly</b> one charecter from the string to some other lowercase English letter. Sample Input 1
abbb
Sample Output 1
2
Explanation:
The possible palindromes are:
1. abba
2. bbbb
========================================================================
Sample Input 2
abba
Sample Output 2
0
, I have written this Solution Code: n=input()
n=list(n)
ln=len(n)
cnt=0
for i in range(ln//2):
if not(n[i]==n[ln-i-1]):
cnt+=1
if(cnt==1):
print(2)
elif(cnt==0 and ln%2==1):
print(25)
else:
print(0), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a string <i>s</i> consisting of lowercase English letters. Find the number of different palindromes you can make by changing <b>exactly</b> one charecter from the string to some other lowercase English letter.The first and single line contains string <i>s</i> (1 ≤ |<i>s</i>| ≤ 10).Print the number of different palindromes you can make by changing <b>exactly</b> one charecter from the string to some other lowercase English letter. Sample Input 1
abbb
Sample Output 1
2
Explanation:
The possible palindromes are:
1. abba
2. bbbb
========================================================================
Sample Input 2
abba
Sample Output 2
0
, I have written this Solution Code: #include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
using namespace std;
using namespace __gnu_pbds;
template <typename T> using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
#define endl '\n'
#define pb push_back
#define ub upper_bound
#define lb lower_bound
#define fi first
#define se second
#define int long long
typedef long long ll;
typedef long double ld;
#define pii pair<int,int>
#define sz(x) ((ll)x.size())
#define fr(a,b,c) for(int a=b; a<=c; a++)
#define frev(a,b,c) for(int a=c; a>=b; a--)
#define rep(a,b,c) for(int a=b; a<c; a++)
#define trav(a,x) for(auto &a:x)
#define all(con) con.begin(),con.end()
#define done(x) {cout << x << endl;return;}
#define mini(x,y) x = min(x,y)
#define maxi(x,y) x = max(x,y)
const ll infl = 0x3f3f3f3f3f3f3f3fLL;
const int infi = 0x3f3f3f3f;
mt19937_64 mt(chrono::steady_clock::now().time_since_epoch().count());
//const int mod = 998244353;
const int mod = 1e9 + 7;
typedef vector<int> vi;
typedef vector<string> vs;
typedef vector<vector<int>> vvi;
typedef vector<pair<int, int>> vpii;
typedef map<int, int> mii;
typedef set<int> si;
typedef set<pair<int,int>> spii;
typedef queue<int> qi;
uniform_int_distribution<int> rng(0, 1e9);
// DEBUG FUNCTIONS START
void __print(int x) {cerr << x;}
void __print(double x) {cerr << x;}
void __print(long double x) {cerr << x;}
void __print(char x) {cerr << '\'' << x << '\'';}
void __print(const char *x) {cerr << '\"' << x << '\"';}
void __print(const string &x) {cerr << '\"' << x << '\"';}
void __print(bool x) {cerr << (x ? "true" : "false");}
template<typename T, typename V> void __print(const pair<T, V> &x) {cerr << '{'; __print(x.first); cerr << ','; __print(x.second); cerr << '}';}
template<typename T> void __print(const T &x) {int f = 0; cerr << '{'; for (auto &i: x) cerr << (f++ ? "," : ""), __print(i); cerr << "}";}
void deb() {cerr << "\n";}
template <typename T, typename... V> void deb(T t, V... v) {__print(t); if (sizeof...(v)) cerr << ", "; deb(v...);}
// DEBUG FUNCTIONS END
const int N = 2e5 + 5;
void solve(){
string s;
cin >> s;
int n = sz(s);
int x = 0;
rep(i, 0, n / 2){
x += s[i] != s[n - 1 - i];
}
if(x == 1){
cout << 2 << endl;
}
else if(x > 1){
cout << 0 << endl;
}
else{
if(n & 1){
cout << 25 << endl;
}
else{
cout << 0 << endl;
}
}
}
signed main(){
ios_base::sync_with_stdio(0), cin.tie(0);
cout << fixed << setprecision(15);
int t = 1;
//cin >> t;
while (t--)
solve();
return 0;
}
int powm(int a, int b){
int res = 1;
while (b) {
if (b & 1)
res = res * a % mod;
a = a * a % mod;
b >>= 1;
}
return res;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A of size N, you need to find its maximum, 2<sup>nd</sup> maximum and 3<sup>rd</sup> maximum element.
Try solving it in O(N) per test caseThe first line of the input contains the number of test cases T.
For each test case, the first line of the input contains an integer N denoting the number of elements in the array A. The following line contains N (space-separated) elements of A.
<b>Constraints:</b>
1 <= T <= 100
3 <= N <= 10<sup>6</sup>
1 <= A[i] <= 10<sup>9</sup>
<b>Note</b>:-It is guaranteed that the sum of N over all text cases does not exceed 10<sup>6</sup>For each test case, output the first, second and third maximum elements in the array.Sample Input:
3
5
1 4 2 4 5
6
1 3 5 7 9 8
7
11 22 33 44 55 66 77
Sample Output:
5 4 4
9 8 7
77 66 55
<b>Explanation:</b>
Testcase 1:
[1 4 2 4 5]
First max = 5
Second max = 4
Third max = 4, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) {
Scanner sc =new Scanner(System.in);
int T= sc.nextInt();
for(int i=0;i<T;i++){
int arrsize=sc.nextInt();
int max=0,secmax=0,thirdmax=0,j;
for(int k=0;k<arrsize;k++){
j=sc.nextInt();
if(j>max){
thirdmax=secmax;
secmax=max;
max=j;
}
else if(j>secmax){
thirdmax=secmax;
secmax=j;
}
else if(j>thirdmax){
thirdmax=j;
}
if(k%10000==0){
System.gc();
}
}
System.out.println(max+" "+secmax+" "+thirdmax+" ");
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A of size N, you need to find its maximum, 2<sup>nd</sup> maximum and 3<sup>rd</sup> maximum element.
Try solving it in O(N) per test caseThe first line of the input contains the number of test cases T.
For each test case, the first line of the input contains an integer N denoting the number of elements in the array A. The following line contains N (space-separated) elements of A.
<b>Constraints:</b>
1 <= T <= 100
3 <= N <= 10<sup>6</sup>
1 <= A[i] <= 10<sup>9</sup>
<b>Note</b>:-It is guaranteed that the sum of N over all text cases does not exceed 10<sup>6</sup>For each test case, output the first, second and third maximum elements in the array.Sample Input:
3
5
1 4 2 4 5
6
1 3 5 7 9 8
7
11 22 33 44 55 66 77
Sample Output:
5 4 4
9 8 7
77 66 55
<b>Explanation:</b>
Testcase 1:
[1 4 2 4 5]
First max = 5
Second max = 4
Third max = 4, I have written this Solution Code: t=int(input())
while t>0:
t-=1
n=int(input())
l=list(map(int,input().strip().split()))
li=[0,0,0]
for i in l:
x=i
for j in range(0,3):
y=min(x,li[j])
li[j]=max(x,li[j])
x=y
print(li[0],end=" ")
print(li[1],end=" ")
print(li[2]), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A of size N, you need to find its maximum, 2<sup>nd</sup> maximum and 3<sup>rd</sup> maximum element.
Try solving it in O(N) per test caseThe first line of the input contains the number of test cases T.
For each test case, the first line of the input contains an integer N denoting the number of elements in the array A. The following line contains N (space-separated) elements of A.
<b>Constraints:</b>
1 <= T <= 100
3 <= N <= 10<sup>6</sup>
1 <= A[i] <= 10<sup>9</sup>
<b>Note</b>:-It is guaranteed that the sum of N over all text cases does not exceed 10<sup>6</sup>For each test case, output the first, second and third maximum elements in the array.Sample Input:
3
5
1 4 2 4 5
6
1 3 5 7 9 8
7
11 22 33 44 55 66 77
Sample Output:
5 4 4
9 8 7
77 66 55
<b>Explanation:</b>
Testcase 1:
[1 4 2 4 5]
First max = 5
Second max = 4
Third max = 4, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main(){
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
int t;
cin>>t;
while(t--){
long long n;
cin>>n;
vector<long> a(n);
long ans[3]={0};
long x,y;
for(int i=0;i<n;i++){
cin>>a[i];
x=a[i];
for(int j=0;j<3;j++){
y=min(x,ans[j]);
ans[j]=max(x,ans[j]);
// cout<<ans[j]<<" ";
x=y;
}
}
if(ans[1]<ans[0]){
swap(ans[1],ans[0]);
}
if(ans[2]<ans[1]){
swap(ans[1],ans[2]);
}
if(ans[1]<ans[0]){
swap(ans[1],ans[0]);
}
cout<<ans[2]<<" "<<ans[1]<<" "<<ans[0]<<endl;
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A of size N, you need to find its maximum, 2<sup>nd</sup> maximum and 3<sup>rd</sup> maximum element.
Try solving it in O(N) per test caseThe first line of the input contains the number of test cases T.
For each test case, the first line of the input contains an integer N denoting the number of elements in the array A. The following line contains N (space-separated) elements of A.
<b>Constraints:</b>
1 <= T <= 100
3 <= N <= 10<sup>6</sup>
1 <= A[i] <= 10<sup>9</sup>
<b>Note</b>:-It is guaranteed that the sum of N over all text cases does not exceed 10<sup>6</sup>For each test case, output the first, second and third maximum elements in the array.Sample Input:
3
5
1 4 2 4 5
6
1 3 5 7 9 8
7
11 22 33 44 55 66 77
Sample Output:
5 4 4
9 8 7
77 66 55
<b>Explanation:</b>
Testcase 1:
[1 4 2 4 5]
First max = 5
Second max = 4
Third max = 4, I have written this Solution Code: function maxNumbers(arr,n) {
// write code here
// do not console.log the answer
// return the answer as an array of 3 numbers
return arr.sort((a,b)=>b-a).slice(0,3)
};
, In this Programming Language: JavaScript, 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: Write a Program to duplicate an array N number of times.The first line of the input contains two space separated integers N and M - The size of an array and Number of times to be repeated.
The second line contains N space separated integers A<sub>1</sub>, A<sub>2</sub>,. , A<sub>n</sub>.
<b>Constraints</b>
1 ≤ N,M ≤ 100
1 ≤ A[i] ≤ 1000Print the required output.Sample Input
3 2
1 2 3
Sample Output
1 2 3 1 2 3, 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));
StringTokenizer st = new StringTokenizer(br.readLine());
int n = Integer.parseInt(st.nextToken());
int m = Integer.parseInt(st.nextToken());
st = new StringTokenizer(br.readLine());
int[] arr = new int[n*m];
for(int i=0; i<n; i++){
arr[i] = Integer.parseInt(st.nextToken());
}
for(int i=n, j=0; i<m*n; i++, j++){
arr[i] = arr[j];
if(j==n-1) j=-1;
}
for(int i=0; i<n*m; 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: Write a Program to duplicate an array N number of times.The first line of the input contains two space separated integers N and M - The size of an array and Number of times to be repeated.
The second line contains N space separated integers A<sub>1</sub>, A<sub>2</sub>,. , A<sub>n</sub>.
<b>Constraints</b>
1 ≤ N,M ≤ 100
1 ≤ A[i] ≤ 1000Print the required output.Sample Input
3 2
1 2 3
Sample Output
1 2 3 1 2 3, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#ifdef LOCAL
#define debug(...) cerr << "[" << #__VA_ARGS__ << "]:", debug_out(__VA_ARGS__)
#else
#define debug(...) 2351
#endif
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, k;
cin >> n >> k;
vector<int> a(n);
for (int i = 0; i < n; i++) {
cin >> a[i];
}
while (k--) {
for (auto &i : a) {
cout << i << " ";
}
}
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: Given an array Arr, of N integers find the sum of max(A[i], A[j]) for all i, j such that i < j.The first line of the input contains an integer N, the size of the array.
The second line of the input contains N integers, the elements of the array Arr.
<b>Constraints:</b>
1 <= N <= 100000
1 <= Arr[i] <= 100000000Print a single integer which is the sum of min(A[i], A[j]) for all i, j such that i < j.Sample Input 1
4
5 3 3 1
Sample Output 1
24
Sample Input 2
2
1 10
Sample Output 2
10
<b>Explanation 1</b>
max(5,3) + max(5,3) + max(5,1) + max(3,3) + max(3,1) + max(3,1) = 24
, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws IOException {
BufferedReader read = new BufferedReader(new InputStreamReader(System.in));
int N = Integer.parseInt(read.readLine());
long[] arr = new long[N];
StringTokenizer st = new StringTokenizer(read.readLine());
for(int i=0; i<N; i++) {
arr[i] = Long.parseLong(st.nextToken());
}
Arrays.sort(arr);
long res = 0;
for(int i=N-1;i >-1; i--) {
res += arr[i]*i;
}
System.out.println(res);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array Arr, of N integers find the sum of max(A[i], A[j]) for all i, j such that i < j.The first line of the input contains an integer N, the size of the array.
The second line of the input contains N integers, the elements of the array Arr.
<b>Constraints:</b>
1 <= N <= 100000
1 <= Arr[i] <= 100000000Print a single integer which is the sum of min(A[i], A[j]) for all i, j such that i < j.Sample Input 1
4
5 3 3 1
Sample Output 1
24
Sample Input 2
2
1 10
Sample Output 2
10
<b>Explanation 1</b>
max(5,3) + max(5,3) + max(5,1) + max(3,3) + max(3,1) + max(3,1) = 24
, I have written this Solution Code: n=int(input())
arr=list(map(int,input().split()))
arr.sort(reverse=True)
l=[]
for i in range(n-1):
l.append(arr[i]*((n-1)-i))
print(sum(l)), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array Arr, of N integers find the sum of max(A[i], A[j]) for all i, j such that i < j.The first line of the input contains an integer N, the size of the array.
The second line of the input contains N integers, the elements of the array Arr.
<b>Constraints:</b>
1 <= N <= 100000
1 <= Arr[i] <= 100000000Print a single integer which is the sum of min(A[i], A[j]) for all i, j such that i < j.Sample Input 1
4
5 3 3 1
Sample Output 1
24
Sample Input 2
2
1 10
Sample Output 2
10
<b>Explanation 1</b>
max(5,3) + max(5,3) + max(5,1) + max(3,3) + max(3,1) + max(3,1) = 24
, I have written this Solution Code: #pragma GCC optimize ("Ofast")
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define VV vector
#define pb push_back
#define bitc __builtin_popcountll
#define m_p make_pair
#define infi 1e18+1
#define eps 0.000000000001
#define fastio ios_base::sync_with_stdio(false);cin.tie(NULL);
string char_to_str(char c){string tem(1,c);return tem;}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
template<class T>//usage rand<long long>()
T rand() {
return uniform_int_distribution<T>()(rng);
}
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
template<class T>
using oset = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
// string to integer stoi()
// string to long long stoll()
// string.substr(position,length);
// integer to string to_string();
//////////////
auto clk=clock();
#define all(x) x.begin(),x.end()
#define S second
#define F first
#define sz(x) ((long long)x.size())
#define int long long
#define f80 __float128
#define pii pair<int,int>
/////////////
signed main()
{
fastio;
#ifdef ANIKET_GOYAL
freopen("inputf.in","r",stdin);
freopen("outputf.in","w",stdout);
#endif
int n;
cin>>n;
int a[n+1];
for(int i=1;i<=n;++i){
cin>>a[i];
}
sort(a+1,a+n+1);
int ans=0;
for(int i=1;i<=n;++i)
ans+=(a[i]*(i-1));
cout<<ans;
#ifdef ANIKET_GOYAL
// cout<<endl<<endl<<endl<<endl<<"Time elapsed: "<<(double)(clock()-clk)/CLOCKS_PER_SEC<<endl;
#endif
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: The universe contains a magic number <b>z</b>. Thor's power is known to <b>x</b> and Loki's power to be <b>y</b>.
One's strength is defined to be <b>z - a</b>, if his power is <b>a</b>. Your task is to find out who among Thor and Loki has the highest strength, and print that strength.
<b>Note:</b> The input and answer may not fit in a 32-bit integer type. In particular, if you are using C++ consider using <em>long long int</em> over <em>int</em>.The first line contains one integer t — the number of test cases.
Each test case consists of one line containing three space-separated integers x, y and z.
<b> Constraints: </b>
1 ≤ t ≤ 10<sup>4</sup>
1 ≤ x, y ≤ 10<sup>15</sup>
max(x, y) < z ≤ 10<sup>15</sup>For each test case, print a single value - the largest strength among Thor and Loki.Sample Input
2
2 3 4
1 1 5
Sample Output
2
4, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) {
long z=0,x=0,y=0;
int choice;
Scanner in = new Scanner(System.in);
choice = in.nextInt();
String s="";
int f = 1;
while(f<=choice){
x = in.nextLong();
y = in.nextLong();
z = in.nextLong();
System.out.println((long)(Math.max((z-x),(z-y))));
f++;
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: The universe contains a magic number <b>z</b>. Thor's power is known to <b>x</b> and Loki's power to be <b>y</b>.
One's strength is defined to be <b>z - a</b>, if his power is <b>a</b>. Your task is to find out who among Thor and Loki has the highest strength, and print that strength.
<b>Note:</b> The input and answer may not fit in a 32-bit integer type. In particular, if you are using C++ consider using <em>long long int</em> over <em>int</em>.The first line contains one integer t — the number of test cases.
Each test case consists of one line containing three space-separated integers x, y and z.
<b> Constraints: </b>
1 ≤ t ≤ 10<sup>4</sup>
1 ≤ x, y ≤ 10<sup>15</sup>
max(x, y) < z ≤ 10<sup>15</sup>For each test case, print a single value - the largest strength among Thor and Loki.Sample Input
2
2 3 4
1 1 5
Sample Output
2
4, I have written this Solution Code: n = int(input())
for i in range(n):
l = list(map(int,input().split()))
print(l[2]-min(l)), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: The universe contains a magic number <b>z</b>. Thor's power is known to <b>x</b> and Loki's power to be <b>y</b>.
One's strength is defined to be <b>z - a</b>, if his power is <b>a</b>. Your task is to find out who among Thor and Loki has the highest strength, and print that strength.
<b>Note:</b> The input and answer may not fit in a 32-bit integer type. In particular, if you are using C++ consider using <em>long long int</em> over <em>int</em>.The first line contains one integer t — the number of test cases.
Each test case consists of one line containing three space-separated integers x, y and z.
<b> Constraints: </b>
1 ≤ t ≤ 10<sup>4</sup>
1 ≤ x, y ≤ 10<sup>15</sup>
max(x, y) < z ≤ 10<sup>15</sup>For each test case, print a single value - the largest strength among Thor and Loki.Sample Input
2
2 3 4
1 1 5
Sample Output
2
4, 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 mod2 = 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;
}
signed main()
{
read(t);
assert(1 <= t && t <= ll(1e4));
while (t--)
{
readc(x, y, z);
assert(1 <= x && x <= ll(1e15));
assert(1 <= y && y <= ll(1e15));
assert(max(x, y) < z && z <= ll(1e15));
int r = 2*z - x - y - 1;
int l = z - max(x, y);
print(r - l + 1);
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: The universe contains a magic number <b>z</b>. Thor's power is known to <b>x</b> and Loki's power to be <b>y</b>.
One's strength is defined to be <b>z - a</b>, if his power is <b>a</b>. Your task is to find out who among Thor and Loki has the highest strength, and print that strength.
<b>Note:</b> The input and answer may not fit in a 32-bit integer type. In particular, if you are using C++ consider using <em>long long int</em> over <em>int</em>.The first line contains one integer t — the number of test cases.
Each test case consists of one line containing three space-separated integers x, y and z.
<b> Constraints: </b>
1 ≤ t ≤ 10<sup>4</sup>
1 ≤ x, y ≤ 10<sup>15</sup>
max(x, y) < z ≤ 10<sup>15</sup>For each test case, print a single value - the largest strength among Thor and Loki.Sample Input
2
2 3 4
1 1 5
Sample Output
2
4, I have written this Solution Code: #include<bits/stdc++.h>
using namespace std;
#define int long long
void solve()
{
int t;
cin>>t;
while(t--)
{
int x, y, z;
cin>>x>>y>>z;
cout<<max(z - y, z- x)<<endl;
}
}
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: Given an array, A of length N, find the contiguous subarray within A which has the largest sum.First line of each test case contain the number of test cases.
The first line of each test case contains an integer n, the length of the array A
and the next line contains n integers.
Constraints:
1<=T<=100
1 <= N <= 10^5
-10^6 <= A[i] <= 10^6Output an integer representing the maximum possible sum of the contiguous subarray.Input:
1
5
1 2 3 4 -10
Output:
10
Explanation:-
1+2+3+4=10, I have written this Solution Code: t=int(input())
while t>0:
n=int(input())
a=map(int,input().split())
m=0
c=0
for i in a:
c+=i
if c>m:m=c
elif c<0:c=0
print(m)
t-=1, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array, A of length N, find the contiguous subarray within A which has the largest sum.First line of each test case contain the number of test cases.
The first line of each test case contains an integer n, the length of the array A
and the next line contains n integers.
Constraints:
1<=T<=100
1 <= N <= 10^5
-10^6 <= A[i] <= 10^6Output an integer representing the maximum possible sum of the contiguous subarray.Input:
1
5
1 2 3 4 -10
Output:
10
Explanation:-
1+2+3+4=10, I have written this Solution Code:
import java.util.*;
import java.lang.*;
import java.io.*;
class Main
{
static long Sum(int a[], int size)
{
long max_so_far = -1000000007, max_ending_here = 0;
for (int i = 0; i < size; i++)
{
max_ending_here = max_ending_here + a[i];
if (max_so_far < max_ending_here)
max_so_far = max_ending_here;
if (max_ending_here < 0)
max_ending_here = 0;
}
return max_so_far;
}
public static void main (String[] args) throws java.lang.Exception
{
BufferedReader br =
new BufferedReader(new InputStreamReader(System.in));
int t = Integer.parseInt(br.readLine().trim());
while(t-->0){
int n = Integer.parseInt(br.readLine().trim());
int arr[] = new int[n];
String inputLine[] = br.readLine().trim().split(" ");
for (int i = 0; i < n; i++) {
arr[i] = Integer.parseInt(inputLine[i]);
}
System.out.println(Sum(arr,n));
}
}
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array, A of length N, find the contiguous subarray within A which has the largest sum.First line of each test case contain the number of test cases.
The first line of each test case contains an integer n, the length of the array A
and the next line contains n integers.
Constraints:
1<=T<=100
1 <= N <= 10^5
-10^6 <= A[i] <= 10^6Output an integer representing the maximum possible sum of the contiguous subarray.Input:
1
5
1 2 3 4 -10
Output:
10
Explanation:-
1+2+3+4=10, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
long long Sum(long long a[], int size)
{
long long max_so_far = INT_MIN, max_ending_here = 0;
for (int i = 0; i < size; i++)
{
max_ending_here = max_ending_here + a[i];
if (max_so_far < max_ending_here)
max_so_far = max_ending_here;
if (max_ending_here < 0)
max_ending_here = 0;
}
return max_so_far;
}
int main(){
int t;
cin>>t;
while(t--){
int n;
cin>>n;
long long a[n];
for(int i=0;i<n;i++){
cin>>a[i];
}
cout<<Sum(a,n)<<endl;
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A[ ] of size N containing positive integers, find maximum and minimum elements from the array.The first line of input contains an integer T, denoting the number of testcases. The description of T testcases follows. The first line of each testcase contains a single integer N denoting the size of array. The second line contains N space-separated integers denoting the elements of the array.
Constraints:
1 <= T <= 100
1 <= N <= 10^5
1 <= A[i] <= 10^7For each testcase you need to print the maximum and minimum element found separated by space.Sample Input:
2
5
7 3 4 5 6
4
1 2 3 4
Sample Output:
7 3
4 1
, I have written this Solution Code: def solve(a):
maxi = 0
mini = 1e7+1
for i in a:
if(i < mini):
mini = i
if(i > maxi):
maxi = i
return mini,maxi, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A[ ] of size N containing positive integers, find maximum and minimum elements from the array.The first line of input contains an integer T, denoting the number of testcases. The description of T testcases follows. The first line of each testcase contains a single integer N denoting the size of array. The second line contains N space-separated integers denoting the elements of the array.
Constraints:
1 <= T <= 100
1 <= N <= 10^5
1 <= A[i] <= 10^7For each testcase you need to print the maximum and minimum element found separated by space.Sample Input:
2
5
7 3 4 5 6
4
1 2 3 4
Sample Output:
7 3
4 1
, I have written this Solution Code:
import java.io.*;
import java.util.*;
class Main
{
public static void main(String[] args)throws IOException {
BufferedReader read = new BufferedReader(new InputStreamReader(System.in));
int t = Integer.parseInt(read.readLine());
while (t-- > 0) {
int n = Integer.parseInt(read.readLine());
int[] arr = new int[n];
String str[] = read.readLine().trim().split(" ");
for(int i = 0; i < n; i++)
arr[i] = Integer.parseInt(str[i]);
findMinMax(arr, n);
System.out.println();
}
}
public static void findMinMax(int arr[], int n)
{
int min = Integer.MAX_VALUE;
int max = Integer.MIN_VALUE;
for(int i = 0; i < n; i++)
{
min = Math.min(arr[i], min);
max = Math.max(arr[i], max);
}
System.out.print(max + " " + min);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given an array of integers of length N, where N is even.
You need to split the array into two parts of equal size in such a way that the difference between the sums of the parts is maximised. The sum of a part is defined as the sum of the elements it contains.
Note that each element of the original array <b>must be in exactly one of the parts</b>.
Print the maximum possible difference.The first line of the input contains a single integer N (2 <= N <= 10<sup>4</sup>, N is even).
The second line of the input contains N space separated integers, the elements of the original array. Each element is within the range 0 to 10<sup>4</sup> inclusive.Print a single integer, the required maximum difference.Sample Input:
4
3 5 1 4
Sample Output:
5
Explanation:
It is optimal to make 2 parts as -- {4, 5} and {3, 1}., I have written this Solution Code: n = int(input())
elements = input()
array = str.split(elements)
for i in range(len(array)):
array[i] = int(array[i])
array.sort()
sum1 = 0
sum2 = 0
for x in range(int(n/2)):
sum1 += array[x]
for x in range(int(n/2)):
sum2 += array[x+int((n/2))]
diff = sum2 - sum1
print(diff), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given an array of integers of length N, where N is even.
You need to split the array into two parts of equal size in such a way that the difference between the sums of the parts is maximised. The sum of a part is defined as the sum of the elements it contains.
Note that each element of the original array <b>must be in exactly one of the parts</b>.
Print the maximum possible difference.The first line of the input contains a single integer N (2 <= N <= 10<sup>4</sup>, N is even).
The second line of the input contains N space separated integers, the elements of the original array. Each element is within the range 0 to 10<sup>4</sup> inclusive.Print a single integer, the required maximum difference.Sample Input:
4
3 5 1 4
Sample Output:
5
Explanation:
It is optimal to make 2 parts as -- {4, 5} and {3, 1}., 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 infty = 1e18;
#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 pob pop_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) freopen("input.txt", "r", stdin); freopen(x, "w", stdout);
#define out(x) cout << ((x) ? "Yes\n" : "No\n")
#define sz(x) x.size()
#define start_clock() auto start_time = std::chrono::high_resolution_clock::now();
#define measure() auto end_time = std::chrono::high_resolution_clock::now(); cerr << (end_time - start_time)/std::chrono::milliseconds(1) << "ms" << endl;
typedef long long ll;
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 mymod(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++; //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.resize(n+1);
}
void add_edge(int a, int b, bool directed = false)
{
adj[a].pb(b);
if(!directed) adj[b].pb(a);
}
};
int main()
{
ios_base::sync_with_stdio(false);
cin.tie(NULL);
ll n;
cin >> n;
assert(n%2 == 0);
vll A(n);
REP(i, 0, n)
{
cin >> A[i];
}
sort(all(A));
ll big = 0, small = 0;
REP(i, 0, n)
{
if(i<n/2) small += A[i];
else big += A[i];
}
cout << big - small << "\n";
return 0;
}
/*
1. Check borderline constraints. Can a variable you are dividing by be 0?
2. Use ll while using bitshifts
3. Do not erase from set while iterating it
4. Initialise everything
5. Read the task carefully, is something unique, sorted, adjacent, guaranteed??
6. DO NOT use if(!mp[x]) if you want to iterate the map later
7. Are you using i in all loops? Are the i's conflicting?
*/
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given an array of integers of length N, where N is even.
You need to split the array into two parts of equal size in such a way that the difference between the sums of the parts is maximised. The sum of a part is defined as the sum of the elements it contains.
Note that each element of the original array <b>must be in exactly one of the parts</b>.
Print the maximum possible difference.The first line of the input contains a single integer N (2 <= N <= 10<sup>4</sup>, N is even).
The second line of the input contains N space separated integers, the elements of the original array. Each element is within the range 0 to 10<sup>4</sup> inclusive.Print a single integer, the required maximum difference.Sample Input:
4
3 5 1 4
Sample Output:
5
Explanation:
It is optimal to make 2 parts as -- {4, 5} and {3, 1}., 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();
long arr[] = sc.readArrayLong(n);
ruffleSort(arr);
long a = 0;
for(int i = 0; i<n/2; i++) {
a+=arr[i];
}
long b = 0;
for(int i = n/2; i<n; i++) {
b+=arr[i];
}
System.out.println(abs(b-a));
}
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;
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: Modulo exponentiation is super awesome. But can you still think of a solution to the following problem?
Given three integers {a, b, c}, find the value of a<sup>b<sup>c</sup></sup> % 1000000007.
Here a<sup>b</sup> means a raised to the power b or pow(a, b). Expression evaluates to pow(a, pow(b, c)) % 1000000007.
(Read Euler's Theorem before solving this problem)The first input line has an integer t: the number of test cases.
After this, there are n lines, each containing three integers a, b and c.
Constraints
1≤ t ≤ 100
0 ≤ a, b, c ≤ 1000000000For each test case, output the value corresponding to the expression.Sample Input
3
3 7 1
15 2 2
3 4 5
Sample Output
2187
50625
763327764
Explaination:
In the first test, a = 3, b = 7, c = 1
b<sup>c</sup> = 7<sup>1</sup> = 7
a<sup>b<sup>c</sup></sup> = 3<sup>7</sup> = 2187, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args)throws IOException{
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
int T=Integer.parseInt(br.readLine());
for(int k=0;k<T;k++)
{
String[] str= br.readLine().split(" ");
int a=Integer.parseInt(str[0]);
int b=Integer.parseInt(str[1]);
int c=Integer.parseInt(str[2]);
int M=1000000007;
System.out.print(superExponentation(a,superExponentation(b,c,M-1),M));
System.out.println();
}
}
public static long superExponentation(long a,long b,int m)
{
long res=1;
while(b>0)
{
if(b%2!=0)
res=(long)(res%m*a%m)%m;
a=((a%m)*(a%m))%m;
b=b>>1;
}
return res;
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Modulo exponentiation is super awesome. But can you still think of a solution to the following problem?
Given three integers {a, b, c}, find the value of a<sup>b<sup>c</sup></sup> % 1000000007.
Here a<sup>b</sup> means a raised to the power b or pow(a, b). Expression evaluates to pow(a, pow(b, c)) % 1000000007.
(Read Euler's Theorem before solving this problem)The first input line has an integer t: the number of test cases.
After this, there are n lines, each containing three integers a, b and c.
Constraints
1≤ t ≤ 100
0 ≤ a, b, c ≤ 1000000000For each test case, output the value corresponding to the expression.Sample Input
3
3 7 1
15 2 2
3 4 5
Sample Output
2187
50625
763327764
Explaination:
In the first test, a = 3, b = 7, c = 1
b<sup>c</sup> = 7<sup>1</sup> = 7
a<sup>b<sup>c</sup></sup> = 3<sup>7</sup> = 2187, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define mem(a, b) memset(a, (b), sizeof(a))
#define fore(i,a) for(int i=0;i<a;i++)
#define fore1(i,j,a) for(int i=j;i<a;i++)
#define print(ar) for(int i=0;i<ar.size();i++)cout<<ar[i]<<" ";
#define END cout<<'\n'
const double pi=acos(-1.0);
typedef pair<int, int> PII;
typedef vector<long long> 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;
ll fastexp (ll a, ll b, ll n) {
ll res = 1;
while (b) {
if (b & 1) res = res*a%n;
a = a*a%n;
b >>= 1;
}
return res;
}
void fast(){
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
}
int main()
{
fast();
ll a,b,c;
int t, n, k;
cin >> t;
while(t--) {
cin >> a >>b >>c;
ll mod = 1e9+7;
ll k = fastexp(b,c,mod-1);
ll ans= fastexp(a,k,mod);
cout<<ans<<endl;
}}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: There are Q queries and for each query you are given a positive integer N. Find whether N is a perfect square or not.
Expected Space Complexity: O(1)First line contains an integer Q - the number of queries.
Q lines follow, each line containing a single integer N.
Constraints
1 <= Q <= 10^5
1 <= Ai <= 10^9Print Q lines. Each line should contain either "YES" or "NO".Sample Input 1:
3
1
2
4
Output
YES
NO
YES
Explanation:
1 and 4 are perfect squares.
Sample Input 2:
2
144
63
Output
YES
NO
Explanation:
12 * 12 = 144, I have written this Solution Code: import math
def checkperfectsquare(x):
if (math.ceil(math.sqrt(n)) ==
math.floor(math.sqrt(n))):
print("YES")
else:
print("NO")
t=int(input())
for _ in range(t):
n=int(input())
checkperfectsquare(n), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: There are Q queries and for each query you are given a positive integer N. Find whether N is a perfect square or not.
Expected Space Complexity: O(1)First line contains an integer Q - the number of queries.
Q lines follow, each line containing a single integer N.
Constraints
1 <= Q <= 10^5
1 <= Ai <= 10^9Print Q lines. Each line should contain either "YES" or "NO".Sample Input 1:
3
1
2
4
Output
YES
NO
YES
Explanation:
1 and 4 are perfect squares.
Sample Input 2:
2
144
63
Output
YES
NO
Explanation:
12 * 12 = 144, I have written this Solution Code: import java.io.*; // for handling input/output
import java.util.*; // contains Collections framework
// don't change the name of this class
// you can add inner classes if needed
class Main {
public static void main (String[] args) {
Scanner sc = new Scanner(System.in);
long Q = sc.nextLong();
while(Q-- > 0){
long n = sc.nextLong();
long sqrt = (long)Math.sqrt(n);
if(sqrt*sqrt == n){
System.out.println("YES");
}else{
System.out.println("NO");
}
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: There are Q queries and for each query you are given a positive integer N. Find whether N is a perfect square or not.
Expected Space Complexity: O(1)First line contains an integer Q - the number of queries.
Q lines follow, each line containing a single integer N.
Constraints
1 <= Q <= 10^5
1 <= Ai <= 10^9Print Q lines. Each line should contain either "YES" or "NO".Sample Input 1:
3
1
2
4
Output
YES
NO
YES
Explanation:
1 and 4 are perfect squares.
Sample Input 2:
2
144
63
Output
YES
NO
Explanation:
12 * 12 = 144, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main() {
int q;
cin >> q;
while (q--) {
int n;
cin >> n;
int x = sqrt(n);
if (x * x == n) {
cout << "YES\n";
} else {
cout << "NO\n";
}
}
return 0;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a temperature F in Farenheit, your task is to convert it into Celsius using the following equation:-
T(°c) = (T(°f) - 32)*5/9You don't have to worry about the input, you just have to complete the function <b>fahrenheitToCelsius</b>
Constraints:-
-10^3 <= F <= 10^3
<b>Note:-</b> It is guaranteed that F - 32 will be a multiple of 9.Print an integer containing converted temperature in Fahrenheit.Sample Input 1:
77
Sample Output 1:
25
Sample Input 2:-
-40
Sample Output 2:-
-40
<b>Explanation 1</b>:
T(°c) = (T(°f) - 32)*5/9
T(°c) = (77-32)*5/9
T(°c) =25, I have written this Solution Code: void farhenheitToCelsius(int n){
n-=32;
n/=9;
n*=5;
cout<<n;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a temperature F in Farenheit, your task is to convert it into Celsius using the following equation:-
T(°c) = (T(°f) - 32)*5/9You don't have to worry about the input, you just have to complete the function <b>fahrenheitToCelsius</b>
Constraints:-
-10^3 <= F <= 10^3
<b>Note:-</b> It is guaranteed that F - 32 will be a multiple of 9.Print an integer containing converted temperature in Fahrenheit.Sample Input 1:
77
Sample Output 1:
25
Sample Input 2:-
-40
Sample Output 2:-
-40
<b>Explanation 1</b>:
T(°c) = (T(°f) - 32)*5/9
T(°c) = (77-32)*5/9
T(°c) =25, I have written this Solution Code: Fahrenheit= int(input())
Celsius = int(((Fahrenheit-32)*5)/9 )
print(Celsius), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a temperature F in Farenheit, your task is to convert it into Celsius using the following equation:-
T(°c) = (T(°f) - 32)*5/9You don't have to worry about the input, you just have to complete the function <b>fahrenheitToCelsius</b>
Constraints:-
-10^3 <= F <= 10^3
<b>Note:-</b> It is guaranteed that F - 32 will be a multiple of 9.Print an integer containing converted temperature in Fahrenheit.Sample Input 1:
77
Sample Output 1:
25
Sample Input 2:-
-40
Sample Output 2:-
-40
<b>Explanation 1</b>:
T(°c) = (T(°f) - 32)*5/9
T(°c) = (77-32)*5/9
T(°c) =25, I have written this Solution Code: static void farhrenheitToCelsius(int farhrenheit)
{
int celsius = ((farhrenheit-32)*5)/9;
System.out.println(celsius);
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a string S check if it is Pangram or not. A pangram is a sentence containing all 26 letters of the English Alphabet.First line of input contains of an integer T denoting number od test cases then T test cases follow. Each testcase contains a String S.
Constraints:
1 <= T <= 100
1 <= |S| <= 1000
Note:- String will not contain any spacesFor each test case print in a new line 1 if its a pangram else print 0.Input:
2
Bawdsjogflickquartzvenymph
sdfs
Output:
0
0
Explanation :
Testcase 1: In the given input, the letter 'x' of the english alphabet is not present. Hence, the output is 0.
Testcase 2: In the given input, there aren't all the letters present in the english alphabet. Hence, the output is 0., I have written this Solution Code: def ispangram(str):
alphabet = "abcdefghijklmnopqrstuvwxyz"
for char in alphabet:
if char not in str.lower():
return False
return True
N = int(input())
arr = []
for i in range(N):
arr.append(input())
for i in range(N):
if(ispangram(arr[i]) == True):
print(1)
else:
print(0), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a string S check if it is Pangram or not. A pangram is a sentence containing all 26 letters of the English Alphabet.First line of input contains of an integer T denoting number od test cases then T test cases follow. Each testcase contains a String S.
Constraints:
1 <= T <= 100
1 <= |S| <= 1000
Note:- String will not contain any spacesFor each test case print in a new line 1 if its a pangram else print 0.Input:
2
Bawdsjogflickquartzvenymph
sdfs
Output:
0
0
Explanation :
Testcase 1: In the given input, the letter 'x' of the english alphabet is not present. Hence, the output is 0.
Testcase 2: In the given input, there aren't all the letters present in the english alphabet. Hence, the output is 0., I have written this Solution Code: function pangrams(s) {
// write code here
// do not console.log it
// return 1 or 0
let alphabet = "abcdefghijklmnopqrstuvwxyz";
let regex = /\s/g;
let lowercase = s.toLowerCase().replace(regex, "");
for(let i = 0; i < alphabet.length; i++){
if(lowercase.indexOf(alphabet[i]) === -1){
return 0;
}
}
return 1;
}, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a string S check if it is Pangram or not. A pangram is a sentence containing all 26 letters of the English Alphabet.First line of input contains of an integer T denoting number od test cases then T test cases follow. Each testcase contains a String S.
Constraints:
1 <= T <= 100
1 <= |S| <= 1000
Note:- String will not contain any spacesFor each test case print in a new line 1 if its a pangram else print 0.Input:
2
Bawdsjogflickquartzvenymph
sdfs
Output:
0
0
Explanation :
Testcase 1: In the given input, the letter 'x' of the english alphabet is not present. Hence, the output is 0.
Testcase 2: In the given input, there aren't all the letters present in the english alphabet. Hence, the output is 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 t = sc.nextInt();
while(t-->0){
String s = sc.next();
int check = 1;
int p =0;
for(char ch = 'a';ch<='z';ch++){
p=0;
for(int i = 0;i<s.length();i++){
if(s.charAt(i)==ch){p=1;}
}
if(p==0){check=0;}
}
System.out.println(check);
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a string S check if it is Pangram or not. A pangram is a sentence containing all 26 letters of the English Alphabet.First line of input contains of an integer T denoting number od test cases then T test cases follow. Each testcase contains a String S.
Constraints:
1 <= T <= 100
1 <= |S| <= 1000
Note:- String will not contain any spacesFor each test case print in a new line 1 if its a pangram else print 0.Input:
2
Bawdsjogflickquartzvenymph
sdfs
Output:
0
0
Explanation :
Testcase 1: In the given input, the letter 'x' of the english alphabet is not present. Hence, the output is 0.
Testcase 2: In the given input, there aren't all the letters present in the english alphabet. Hence, the output is 0., I have written this Solution Code: #include<bits/stdc++.h>
using namespace std;
#define pu push_back
#define fi first
#define se second
#define mp make_pair
#define int long long
#define pii pair<int,int>
#define mm (s+e)/2
#define all(x) x.begin(), x.end()
#define For(i, st, en) for(int i=st; i<en; i++)
#define tr(x) for(auto it=x.begin(); it!=x.end(); it++)
#define fast std::ios::sync_with_stdio(false);cin.tie(NULL);
#define sz 200000
int A[26],B[26];
signed main()
{
int t;
cin>>t;
string p;
getline(cin,p);
while(t>0)
{
t--;
string s;
getline(cin,s);
int n=s.size();
memset(A,0,sizeof(A));
int ch=1;
for(int i=0;i<n;i++)
{
int x=s[i]-'a';
if(x>=0 && x<26)
{
A[x]++;
}
x=s[i]-'A';
if(x>=0 && x<26)
{
A[x]++;
}
}
for(int i=0;i<26;i++)
{
if(A[i]==0) ch=0;
}
cout<<ch<<endl;
}
}, 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: Write a program to find the roots of a quadratic equation.
<b>Note</b>: Try to do it using Switch Case.
<b>Quadratic equations</b> are the polynomial equations of degree 2 in one variable of type f(x) = ax<sup>2</sup> + bx + c = 0 where a, b, c, ∈ R and a ≠ 0. It is the general form of a quadratic equation where 'a' is called the leading coefficient and 'c' is called the absolute term of f (x).The first line of the input contains the three integer values a, b, and c of equation ax^2 + bx + c.
<b>Constraints</b>
1 ≤ a, b, c ≤ 50Print the two roots in two different lines and for imaginary roots print real and imaginary part of one root with (+/- and i )sign in between in one line and other in next line. For clarity see sample Output 2.
<b>Note</b> Imaginary roots can also be there and roots are considered upto 2 decimal places.Sample Input 1 :
4 -2 -10
Sample Output 2 :
1.85
-1.35
Sample Input 2 :
2 1 10
Sample Output 2:
-0.25+i2.22
-0.25-i2.22, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
String s=br.readLine().toString();
String[] str = s.split(" ");
float a=Float.parseFloat(str[0]);
float b=Float.parseFloat(str[1]);
float c=Float.parseFloat(str[2]);
float div = (float)(b*b-4*a*c);
if(div>0.0){
float alpha= (-b+(float)Math.sqrt(div))/(2*a);
float beta= (-b-(float)Math.sqrt(div))/(2*a);
System.out.printf("%.2f\n",alpha);
System.out.printf("%.2f",beta);
}
else{
float rp=-b/(2*a);
float ip=(float)Math.sqrt(-div)/(2*a);
System.out.printf("%.2f+i%.2f\n",rp,ip);
System.out.printf("%.2f-i%.2f\n",rp,ip);
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Write a program to find the roots of a quadratic equation.
<b>Note</b>: Try to do it using Switch Case.
<b>Quadratic equations</b> are the polynomial equations of degree 2 in one variable of type f(x) = ax<sup>2</sup> + bx + c = 0 where a, b, c, ∈ R and a ≠ 0. It is the general form of a quadratic equation where 'a' is called the leading coefficient and 'c' is called the absolute term of f (x).The first line of the input contains the three integer values a, b, and c of equation ax^2 + bx + c.
<b>Constraints</b>
1 ≤ a, b, c ≤ 50Print the two roots in two different lines and for imaginary roots print real and imaginary part of one root with (+/- and i )sign in between in one line and other in next line. For clarity see sample Output 2.
<b>Note</b> Imaginary roots can also be there and roots are considered upto 2 decimal places.Sample Input 1 :
4 -2 -10
Sample Output 2 :
1.85
-1.35
Sample Input 2 :
2 1 10
Sample Output 2:
-0.25+i2.22
-0.25-i2.22, I have written this Solution Code: import math
a,b,c = map(int, input().split(' '))
disc = (b ** 2) - (4*a*c)
sq = disc ** 0.5
if disc > 0:
print("{:.2f}".format((-b + sq)/(2*a)))
print("{:.2f}".format((-b - sq)/(2*a)))
elif disc == 0:
print("{:.2f}".format(-b/(2*a)))
elif disc < 0:
r1 = complex((-b + sq)/(2*a))
r2 = complex((-b - sq)/(2*a))
print("{:.2f}+i{:.2f}".format(r1.real, abs(r1.imag)))
print("{:.2f}-i{:.2f}".format(r2.real, abs(r2.imag))), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Write a program to find the roots of a quadratic equation.
<b>Note</b>: Try to do it using Switch Case.
<b>Quadratic equations</b> are the polynomial equations of degree 2 in one variable of type f(x) = ax<sup>2</sup> + bx + c = 0 where a, b, c, ∈ R and a ≠ 0. It is the general form of a quadratic equation where 'a' is called the leading coefficient and 'c' is called the absolute term of f (x).The first line of the input contains the three integer values a, b, and c of equation ax^2 + bx + c.
<b>Constraints</b>
1 ≤ a, b, c ≤ 50Print the two roots in two different lines and for imaginary roots print real and imaginary part of one root with (+/- and i )sign in between in one line and other in next line. For clarity see sample Output 2.
<b>Note</b> Imaginary roots can also be there and roots are considered upto 2 decimal places.Sample Input 1 :
4 -2 -10
Sample Output 2 :
1.85
-1.35
Sample Input 2 :
2 1 10
Sample Output 2:
-0.25+i2.22
-0.25-i2.22, I have written this Solution Code: /**
* Author : tourist1256
* Time : 2022-01-19 02:44:22
**/
#include <bits/stdc++.h>
using namespace std;
#ifdef LOCAL
#define debug(...) cerr << "[" << #__VA_ARGS__ << "]:", debug_out(__VA_ARGS__)
#else
#define debug(...) 2351
#endif
int main() {
float a, b, c;
float root1, root2, imaginary;
float discriminant;
scanf("%f%f%f", &a, &b, &c);
discriminant = (b * b) - (4 * a * c);
switch (discriminant > 0) {
case 1:
root1 = (-b + sqrt(discriminant)) / (2 * a);
root2 = (-b - sqrt(discriminant)) / (2 * a);
printf("%.2f\n%.2f", root1, root2);
break;
case 0:
switch (discriminant < 0) {
case 1:
root1 = root2 = -b / (2 * a);
imaginary = sqrt(-discriminant) / (2 * a);
printf("%.2f + i%.2f\n%.2f - i%.2f", root1, imaginary, root2, imaginary);
break;
case 0:
root1 = root2 = -b / (2 * a);
printf("%.2f\n%.2f", root1, root2);
break;
}
}
return 0;
}, 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 |
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: function calculateCI(P, R, T)
{
let interest = P * (Math.pow(1.0 + R/100.0, T) - 1);
return interest.toFixed(2);
}, In this Programming Language: JavaScript, 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: #include <bits/stdc++.h>
using namespace std;
int main(){
int p,r,t;
cin>>p>>r>>t;
double rate= (float)r/100;
double amt = (float)p*(pow(1+rate,t));
cout << fixed << setprecision(2) << (amt - p);
return 0;
}, 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: import java.io.*;
import java.util.*;
import java.lang.Math;
class Main {
public static void main (String[] args) throws Exception{
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
String[] s= br.readLine().split(" ");
double[] darr = new double[s.length];
for(int i=0;i<s.length;i++){
darr[i] = Double.parseDouble(s[i]);
}
double ans = darr[0]*Math.pow(1+darr[1]/100,darr[2])-darr[0];
System.out.printf("%.2f",ans);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a string <em>S</em> and an integer <em>A</em>. You have to check whether or not the given integer A is smaller than 3200 or not. If it's smaller, print "red", otherwise print the string S.The first line of the input contains an integer A
The second line of the input contains a string S
<b>Constraints:</b>
1) 2800 ≤ A ≤ 5000Print the output string.<b>Sample Input 1:</b>
3400
lol
<b>Sample Output 1:</b>
lol
<b>Sample Input 2:</b>
3000
lol
<b>Sample Output 2:</b>
red, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main() {
int A;
string S;
cin >> A >> S;
if(A < 3200){
cout << "red" << endl;
}
else{
cout << S << endl;
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of N elements where each element is either 1 or 0. You have to divide the array into maximum number of subarrays such that each element of the array is in exactly one subarray such that each subarray has equal number of 1's and 0's.First line of input contains N.
Second line of input contains N space separated elements of the array.
Constraints:
1 <= N <= 100000
0 <= elements of the array <= 1Print the single integer which is the maximum number of subarrays the array can be divided into. If it is not possible then print -1.Sample input 1
4
1 0 1 0
Sample output 1
2
Sample input 2
4
1 1 0 0
Sample output 2
1, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws Exception {
BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
int N = Integer.parseInt(reader.readLine());
StringTokenizer tokens = new StringTokenizer(reader.readLine());
int partitions = 0;
int c0 = 0, c1 = 0;
for(int i = 0; i < N; i++) {
int x = Integer.parseInt(tokens.nextToken());
if(x == 0) {
c0++;
} else {
c1++;
}
if(c0 > 0 || c1 > 0) {
if(c0 == c1) {
partitions++;
c0 = c1 = 0;
}
}
}
if(c1 > 0 || c0 > 0) {
partitions = -1;
}
System.out.println(partitions);
}
}, 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 where each element is either 1 or 0. You have to divide the array into maximum number of subarrays such that each element of the array is in exactly one subarray such that each subarray has equal number of 1's and 0's.First line of input contains N.
Second line of input contains N space separated elements of the array.
Constraints:
1 <= N <= 100000
0 <= elements of the array <= 1Print the single integer which is the maximum number of subarrays the array can be divided into. If it is not possible then print -1.Sample input 1
4
1 0 1 0
Sample output 1
2
Sample input 2
4
1 1 0 0
Sample output 2
1, I have written this Solution Code: n = int(input())
a = input().split()
o,z = 0,0
cnt = 0
for i in a:
if i == '1':
o += 1
else:
z +=1
if o == z:
cnt += 1
if o == z:
print(cnt)
else:
print(-1), 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 where each element is either 1 or 0. You have to divide the array into maximum number of subarrays such that each element of the array is in exactly one subarray such that each subarray has equal number of 1's and 0's.First line of input contains N.
Second line of input contains N space separated elements of the array.
Constraints:
1 <= N <= 100000
0 <= elements of the array <= 1Print the single integer which is the maximum number of subarrays the array can be divided into. If it is not possible then print -1.Sample input 1
4
1 0 1 0
Sample output 1
2
Sample input 2
4
1 1 0 0
Sample output 2
1, I have written this Solution Code: #include<bits/stdc++.h>
using namespace std;
#define ll long long
#define VV vector
#define pb push_back
#define bitc __builtin_popcountll
#define m_p make_pair
#define infi 1e18+1
#define eps 0.000000000001
#define fastio ios_base::sync_with_stdio(false);cin.tie(NULL);
string char_to_str(char c){string tem(1,c);return tem;}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
template<class T>//usage rand<long long>()
T rand() {
return uniform_int_distribution<T>()(rng);
}
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
typedef tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> oset;
// string to integer stoi()
// string to long long stoll()
// string.substr(position,length);
// integer to string to_string();
//////////////
auto clk=clock();
#define all(x) x.begin(),x.end()
#define S second
#define F first
#define sz(x) ((long long)x.size())
#define int long long
#define f80 __float128
#define pii pair<int,int>
/////////////
signed main()
{
fastio;
#ifdef ANIKET_GOYAL
freopen("inputf.in","r",stdin);
freopen("outputf.in","w",stdout);
#endif
int n;
cin>>n;
int c=0;
int ans=0;
for(int i=0;i<n;++i){
int x;
cin>>x;
if(x==0)
++c;
else
--c;
if(c==0)
++ans;
}
if(c==0){
cout<<ans;
}else
{
cout<<"-1";
}
#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: 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: You are given a matrix Mat of m rows and n columns. The matrix is boolean so the elements of the matrix can only be either 0 or 1.
Now, if any row of the matrix contains a 1, then you need to fill that whole row with 1. After doing the mentioned operation, you need to print the modified matrix.The first line of input contains T denoting the number of test cases. T test cases follow.
The first line of each test case contains m and n denotes the number of rows and a number of columns.
Then next m lines contain n elements denoting the elements of the matrix.
Constraints:
1 ≤ T ≤ 20
1 ≤ m, n ≤ 700
Mat[I][j] ∈ {0,1}For each testcase, in a new line, print the modified matrix.Input:
1
5 4
1 0 0 0
0 0 0 0
0 1 0 0
0 0 0 0
0 0 0 1
Output:
1 1 1 1
0 0 0 0
1 1 1 1
0 0 0 0
1 1 1 1
Explanation:
Rows = 5 and columns = 4
The given matrix is
1 0 0 0
0 0 0 0
0 1 0 0
0 0 0 0
0 0 0 1
Evidently, the first row contains a 1 so fill the whole row with 1. The third row also contains a 1 so that row will be filled too. Finally, the last row contains a 1 and therefore it needs to be filled with 1 too.
The final matrix is
1 1 1 1
0 0 0 0
1 1 1 1
0 0 0 0
1 1 1 1, I have written this Solution Code: t=int(input())
while t!=0:
m,n=input().split()
m,n=int(m),int(n)
for i in range(m):
arr=input().strip()
if '1' in arr:
arr='1 '*n
else:
arr='0 '*n
print(arr)
t-=1, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a matrix Mat of m rows and n columns. The matrix is boolean so the elements of the matrix can only be either 0 or 1.
Now, if any row of the matrix contains a 1, then you need to fill that whole row with 1. After doing the mentioned operation, you need to print the modified matrix.The first line of input contains T denoting the number of test cases. T test cases follow.
The first line of each test case contains m and n denotes the number of rows and a number of columns.
Then next m lines contain n elements denoting the elements of the matrix.
Constraints:
1 ≤ T ≤ 20
1 ≤ m, n ≤ 700
Mat[I][j] ∈ {0,1}For each testcase, in a new line, print the modified matrix.Input:
1
5 4
1 0 0 0
0 0 0 0
0 1 0 0
0 0 0 0
0 0 0 1
Output:
1 1 1 1
0 0 0 0
1 1 1 1
0 0 0 0
1 1 1 1
Explanation:
Rows = 5 and columns = 4
The given matrix is
1 0 0 0
0 0 0 0
0 1 0 0
0 0 0 0
0 0 0 1
Evidently, the first row contains a 1 so fill the whole row with 1. The third row also contains a 1 so that row will be filled too. Finally, the last row contains a 1 and therefore it needs to be filled with 1 too.
The final matrix is
1 1 1 1
0 0 0 0
1 1 1 1
0 0 0 0
1 1 1 1, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define N 1000
int a[N][N];
// Driver code
int main()
{
int t;
cin>>t;
while(t--){
int n,m;
cin>>n>>m;
bool b[n];
for(int i=0;i<n;i++){
b[i]=false;
}
for(int i=0;i<n;i++){
for(int j=0;j<m;j++){
cin>>a[i][j];
if(a[i][j]==1){
b[i]=true;
}
}
}
for(int i=0;i<n;i++){
if(b[i]){
for(int j=0;j<m;j++){
cout<<1<<" ";
}}
else{
for(int j=0;j<m;j++){
cout<<0<<" ";
}
}
cout<<endl;
}
}}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a matrix Mat of m rows and n columns. The matrix is boolean so the elements of the matrix can only be either 0 or 1.
Now, if any row of the matrix contains a 1, then you need to fill that whole row with 1. After doing the mentioned operation, you need to print the modified matrix.The first line of input contains T denoting the number of test cases. T test cases follow.
The first line of each test case contains m and n denotes the number of rows and a number of columns.
Then next m lines contain n elements denoting the elements of the matrix.
Constraints:
1 ≤ T ≤ 20
1 ≤ m, n ≤ 700
Mat[I][j] ∈ {0,1}For each testcase, in a new line, print the modified matrix.Input:
1
5 4
1 0 0 0
0 0 0 0
0 1 0 0
0 0 0 0
0 0 0 1
Output:
1 1 1 1
0 0 0 0
1 1 1 1
0 0 0 0
1 1 1 1
Explanation:
Rows = 5 and columns = 4
The given matrix is
1 0 0 0
0 0 0 0
0 1 0 0
0 0 0 0
0 0 0 1
Evidently, the first row contains a 1 so fill the whole row with 1. The third row also contains a 1 so that row will be filled too. Finally, the last row contains a 1 and therefore it needs to be filled with 1 too.
The final matrix is
1 1 1 1
0 0 0 0
1 1 1 1
0 0 0 0
1 1 1 1, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main(String[] args) throws Exception{
InputStreamReader isr = new InputStreamReader(System.in);
BufferedReader bf = new BufferedReader(isr);
int t = Integer.parseInt(bf.readLine());
while (t-- > 0){
String inputs[] = bf.readLine().split(" ");
int m = Integer.parseInt(inputs[0]);
int n = Integer.parseInt(inputs[1]);
String[] matrix = new String[m];
for(int i=0; i<m; i++){
matrix[i] = bf.readLine();
}
StringBuffer ones = new StringBuffer("");
StringBuffer zeros = new StringBuffer("");
for(int i=0; i<n; i++){
ones.append("1 ");
zeros.append("0 ");
}
for(int i=0; i<m; i++){
if(matrix[i].contains("1")){
System.out.println(ones);
}else{
System.out.println(zeros);
}
}
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a positive integer N, your task is to check if the given number prime or not.<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 <i>Ifprime()</i> which contains the given number N.
<b>Constraints:</b>
1 <= N <= 10<sup>9</sup>
<b>Note:</b>
<i>But there is a catch here given user function has already code in it which may or may not be correct, now you need to figure out these and correct if it is required</i>Print "Yes" if the given number is prime else print "No"Sample Input:-
6
Sample Output:-
No
Sample Input:-
3
Sample Output:-
Yes, I have written this Solution Code: def IFprime(N):
is_prime = 'Yes'
if N == 1: is_prime = 'No'
else:
for i in range(2, N//2+1):
if N % i == 0:
is_prime = 'No'
break
print(is_prime)
IFprime(int(input())), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a positive integer N, your task is to check if the given number prime or not.<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 <i>Ifprime()</i> which contains the given number N.
<b>Constraints:</b>
1 <= N <= 10<sup>9</sup>
<b>Note:</b>
<i>But there is a catch here given user function has already code in it which may or may not be correct, now you need to figure out these and correct if it is required</i>Print "Yes" if the given number is prime else print "No"Sample Input:-
6
Sample Output:-
No
Sample Input:-
3
Sample Output:-
Yes, I have written this Solution Code: static void Ifprime(int N)
{
if(N==1){
System.out.print("No");
return;
}
boolean prime = true;
for(int i=2; i*i<=N ;i++){
if(N%i==0){prime=false;break;}
}
if(prime==true){
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: You are given an array of integers. Consider absolute difference between all the pairs of the the elements. You need to find Kth smallest absolute difference. If the size of the array is N then value of K will be less than N and more than or equal to 1.The first line of input contains number of test cases T.
The first line of each test case contains a two integers N and K denoting the number of elements in the array A and difference you need to output. The second line of each test case contains N space separated integers denoting the elements of the array A
Constraints:
1<= T <= 10
2 <= N <= 100000
1 <= K < N < 100000
0 <= A[i] <= 100000For each test case, output Kth smallest absolute difference.Input :
1
6 2
1 3 4 1 3 8
Output :
0
Explanation :
Test case 1: First smallest difference is 0, between the pair (1, 1) and second smallest absolute difference difference is also 0 between the pairs (3, 3)., 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 read = new BufferedReader(new InputStreamReader(System.in));
int t = Integer.parseInt(read.readLine().trim());
while (t-- > 0) {
String str[] = read.readLine().trim().split(" ");
int n = Integer.parseInt(str[0]);
int k = Integer.parseInt(str[1]);
int arr[] = new int[n];
str = read.readLine().trim().split(" ");
for (int i = 0; i < n; i++)
arr[i] = Integer.parseInt(str[i]);
System.out.println(Math.abs(small(arr, k)));
}
}
public static int small(int arr[], int k) {
Arrays.sort(arr);
int l = 0, r = arr[arr.length - 1] - arr[0];
while (r > l) {
int mid = l + (r - l) / 2;
if (count(arr, mid) < k) {
l = mid + 1;
} else {
r = mid;
}
}
return r;
}
public static int count(int arr[], int mid) {
int ans = 0, j = 0;
for (int i = 1; i < arr.length; ++i) {
while (j < i && arr[i] - arr[j] > mid) {
++j;
}
ans += i - j;
}
return ans;
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a natural number N, your task is to print all the digits of the number in English words. The words have to separate by space and in lowercase English letters.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>Print_Digit()</b> that takes integer N as a parameter.
<b>Constraints:-</b>
1 ≤ N ≤ 10<sup>7</sup>Print the digits of the number as shown in the example.
<b>Note:-</b>
Print all digits in lowercase English lettersSample Input:-
1024
Sample Output:-
one zero two four
Sample Input:-
2
Sample Output:-
two, I have written this Solution Code: def Print_Digit(n):
dc = {1: "one", 2: "two", 3: "three", 4: "four",
5: "five", 6: "six", 7: "seven", 8: "eight", 9: "nine", 0: "zero"}
final_list = []
while (n > 0):
final_list.append(dc[int(n%10)])
n = int(n / 10)
for val in final_list[::-1]:
print(val, end=' '), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a natural number N, your task is to print all the digits of the number in English words. The words have to separate by space and in lowercase English letters.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>Print_Digit()</b> that takes integer N as a parameter.
<b>Constraints:-</b>
1 ≤ N ≤ 10<sup>7</sup>Print the digits of the number as shown in the example.
<b>Note:-</b>
Print all digits in lowercase English lettersSample Input:-
1024
Sample Output:-
one zero two four
Sample Input:-
2
Sample Output:-
two, I have written this Solution Code: class Solution {
public static void Print_Digits(int N){
if(N==0){return;}
Print_Digits(N/10);
int x=N%10;
if(x==1){System.out.print("one ");}
else if(x==2){System.out.print("two ");}
else if(x==3){System.out.print("three ");}
else if(x==4){System.out.print("four ");}
else if(x==5){System.out.print("five ");}
else if(x==6){System.out.print("six ");}
else if(x==7){System.out.print("seven ");}
else if(x==8){System.out.print("eight ");}
else if(x==9){System.out.print("nine ");}
else if(x==0){System.out.print("zero ");}
}
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
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