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For this Question: Gian and Suneo want their heights to be equal so they asked Doraemon's help. Doraemon gave a big light to both of them but the both big lights have different speed of magnifying. Let's assume the big light given to Gian can increase height of a person by v1 m/s and that of Suneo's big light is v2 m/s.
At the end of each second Doraemon check if their heights are equal or not.
Given initial height of Gian and Suneo, your task is to check whether the height of Gian and Suneo will become equal at some point or not, assuming they both started at the same time.First line takes the input of integer h1(height of gian), h2(height of suneo), v1(speed of Gian's big light) and v2(speed of Suneo's big light) as parameter.
<b>Constraints:-</b>
1 <b>≤</b> h2 < h1<b>≤</b> 10<sup>4</sup>
1 <b>≤</b> v1 <b>≤</b> 10<sup>4</sup>
1 <b>≤</b> v2 <b>≤</b> 10<sup>4</sup>complete the function EqualOrNot and return a boolean True if their height will become equal at some point (as seen by Doraemon) else print False
Sample input:-
4 2 2 4
Sample output:-
Yes
Explanation:-
height of Gian goes as- 4 6 8 10. .
height of Suneo goes as:- 2 6 10..
at the end of 1 second their height will become equal.
Sample Input:-
5 4 1 6
Sample Output:
No, I have written this Solution Code: static boolean EqualOrNot(int h1, int h2, int v1,int v2){
if (v2>v1&&(h1-h2)%(v2-v1)==0){
return true;
}
return false;
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given two arrays - value and frequency both containing N elements.
There is also a third array C which is currently empty. Then you perform N insertion operation in the array. For ith operation you insert value[i] to the end of the array frequency[i] number of times.
Finally you have to tell the kth smallest element in the array C.First line of input contains N.
Second line contains N integers denoting array - value
Third line contains N integers denoting array - frequency
Fourth line contains single integer K.
Constraints
1 <= N, value[i], frequency[i] <= 100000
1 <= k <= frequency[1] + frequency[2] +frequency[3] +........ + frequency[N]
Output a single integer which is the kth smallest element of the array C.Sample input 1
5
1 2 3 4 5
1 1 1 2 2
3
Sample output 1
3
Explanation 1:
Array C constructed is 1 2 3 4 4 5 5
Third smallest element is 3
Sample input 2
3
2 1 3
3 3 2
2
sample output 2
1
Explanation 2:
Array C constructed is 2 2 2 1 1 1 3 3
Second smallest element is 1, 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));
int n = Integer.parseInt(br.readLine());
StringTokenizer st = new StringTokenizer(br.readLine());
int[] val = new int[n];
for(int i=0; i<n; i++){
val[i] = Integer.parseInt(st.nextToken());
}
st = new StringTokenizer(br.readLine());
int[] freq = new int[n];
for(int i=0; i<n; i++){
freq[i] = Integer.parseInt(st.nextToken());
}
int k = Integer.parseInt(br.readLine());
for (int i=0; i<n; i++) {
for (int j=i+1; j<n; j++) {
if (val[j] < val[i]) {
int temp = val[i];
val[i] = val[j];
val[j] = temp;
int temp1 = freq[i];
freq[i] = freq[j];
freq[j] = temp1;
}
}
}
int element=0;
for(int i=0; i<n; i++){
for(int j=0; j<freq[i]; j++){
element++;
int value = val[i];
if(element==k){
System.out.print(value);
break;
}
}
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given two arrays - value and frequency both containing N elements.
There is also a third array C which is currently empty. Then you perform N insertion operation in the array. For ith operation you insert value[i] to the end of the array frequency[i] number of times.
Finally you have to tell the kth smallest element in the array C.First line of input contains N.
Second line contains N integers denoting array - value
Third line contains N integers denoting array - frequency
Fourth line contains single integer K.
Constraints
1 <= N, value[i], frequency[i] <= 100000
1 <= k <= frequency[1] + frequency[2] +frequency[3] +........ + frequency[N]
Output a single integer which is the kth smallest element of the array C.Sample input 1
5
1 2 3 4 5
1 1 1 2 2
3
Sample output 1
3
Explanation 1:
Array C constructed is 1 2 3 4 4 5 5
Third smallest element is 3
Sample input 2
3
2 1 3
3 3 2
2
sample output 2
1
Explanation 2:
Array C constructed is 2 2 2 1 1 1 3 3
Second smallest element is 1, I have written this Solution Code: def myFun():
n = int(input())
arr1 = list(map(int,input().strip().split()))
arr2 = list(map(int,input().strip().split()))
k = int(input())
arr = []
for i in range(n):
arr.append((arr1[i], arr2[i]))
arr.sort()
c = 0
for i in arr:
k -= i[1]
if k <= 0:
print(i[0])
return
myFun()
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given two arrays - value and frequency both containing N elements.
There is also a third array C which is currently empty. Then you perform N insertion operation in the array. For ith operation you insert value[i] to the end of the array frequency[i] number of times.
Finally you have to tell the kth smallest element in the array C.First line of input contains N.
Second line contains N integers denoting array - value
Third line contains N integers denoting array - frequency
Fourth line contains single integer K.
Constraints
1 <= N, value[i], frequency[i] <= 100000
1 <= k <= frequency[1] + frequency[2] +frequency[3] +........ + frequency[N]
Output a single integer which is the kth smallest element of the array C.Sample input 1
5
1 2 3 4 5
1 1 1 2 2
3
Sample output 1
3
Explanation 1:
Array C constructed is 1 2 3 4 4 5 5
Third smallest element is 3
Sample input 2
3
2 1 3
3 3 2
2
sample output 2
1
Explanation 2:
Array C constructed is 2 2 2 1 1 1 3 3
Second smallest element is 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 inf 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;ll K;
cin>>N;
int c=0;
pair<int, ll> A[N];
for(int i=0;i<N;++i){
cin >> A[i].first ;
}
for(int i=0;i<N;++i){
cin >> A[i].second ;
}
cin>>K;
sort(A, A+N);
for(int i=0;i<N;++i){
K -= A[i].second;
if(K <= 0){
cout << A[i].first << endl;;
break;
}
}
#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: You must be familiar with various types of operators. One of the most commonly used operators in any language are increment and decrement operators. Given two numbers X and Y. Your task is to print the value of X decremented by 1 and value of Y after incremented by 1.The first line of the input contains two integers X and Y.
<b>Constraints:</b>
1 ≤ X, Y ≤ 10000You need to perform the task as mentioned in the question, and finally, print the result separated by a space.Sample Input 1:
4 5
Sample Output 1:
3 6
Sample Input 2:
5 6
Sample Output 2:
4 7, 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 x = sc.nextInt();
int y = sc.nextInt();
x--;
y++;
System.out.print(x);
System.out.print(" ");
System.out.print(y);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You must be familiar with various types of operators. One of the most commonly used operators in any language are increment and decrement operators. Given two numbers X and Y. Your task is to print the value of X decremented by 1 and value of Y after incremented by 1.The first line of the input contains two integers X and Y.
<b>Constraints:</b>
1 ≤ X, Y ≤ 10000You need to perform the task as mentioned in the question, and finally, print the result separated by a space.Sample Input 1:
4 5
Sample Output 1:
3 6
Sample Input 2:
5 6
Sample Output 2:
4 7, I have written this Solution Code: def incremental_decremental(x, y):
x -= 1
y += 1
print(x, y, end =' ')
def main():
input1 = input().split()
x = int(input1[0])
y = int(input1[1])
#z = int(input1[2])
incremental_decremental(x, y)
if __name__ == '__main__':
main(), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Gian and Suneo want their heights to be equal so they asked Doraemon's help. Doraemon gave a big light to both of them but the both big lights have different speed of magnifying. Let's assume the big light given to Gian can increase height of a person by v1 m/s and that of Suneo's big light is v2 m/s.
At the end of each second Doraemon check if their heights are equal or not.
Given initial height of Gian and Suneo, your task is to check whether the height of Gian and Suneo will become equal at some point or not, assuming they both started at the same time.First line takes the input of integer h1(height of gian), h2(height of suneo), v1(speed of Gian's big light) and v2(speed of Suneo's big light) as parameter.
<b>Constraints:-</b>
1 <b>≤</b> h2 < h1<b>≤</b> 10<sup>4</sup>
1 <b>≤</b> v1 <b>≤</b> 10<sup>4</sup>
1 <b>≤</b> v2 <b>≤</b> 10<sup>4</sup>complete the function EqualOrNot and return a boolean True if their height will become equal at some point (as seen by Doraemon) else print False
Sample input:-
4 2 2 4
Sample output:-
Yes
Explanation:-
height of Gian goes as- 4 6 8 10. .
height of Suneo goes as:- 2 6 10..
at the end of 1 second their height will become equal.
Sample Input:-
5 4 1 6
Sample Output:
No, I have written this Solution Code:
#include <bits/stdc++.h>
using namespace std;
bool EqualOrNot(int h1, int h2, int v1,int v2){
if (v2>v1&&(h1-h2)%(v2-v1)==0){
return true;
}
return false;
}
int main(){
int n1,n2,v1,v2;
cin>>n1>>n2>>v1>>v2;
if(EqualOrNot(n1,n2,v1,v2)){
cout<<"Yes";}
else{
cout<<"No";
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Gian and Suneo want their heights to be equal so they asked Doraemon's help. Doraemon gave a big light to both of them but the both big lights have different speed of magnifying. Let's assume the big light given to Gian can increase height of a person by v1 m/s and that of Suneo's big light is v2 m/s.
At the end of each second Doraemon check if their heights are equal or not.
Given initial height of Gian and Suneo, your task is to check whether the height of Gian and Suneo will become equal at some point or not, assuming they both started at the same time.First line takes the input of integer h1(height of gian), h2(height of suneo), v1(speed of Gian's big light) and v2(speed of Suneo's big light) as parameter.
<b>Constraints:-</b>
1 <b>≤</b> h2 < h1<b>≤</b> 10<sup>4</sup>
1 <b>≤</b> v1 <b>≤</b> 10<sup>4</sup>
1 <b>≤</b> v2 <b>≤</b> 10<sup>4</sup>complete the function EqualOrNot and return a boolean True if their height will become equal at some point (as seen by Doraemon) else print False
Sample input:-
4 2 2 4
Sample output:-
Yes
Explanation:-
height of Gian goes as- 4 6 8 10. .
height of Suneo goes as:- 2 6 10..
at the end of 1 second their height will become equal.
Sample Input:-
5 4 1 6
Sample Output:
No, I have written this Solution Code: def EqualOrNot(h1,h2,v1,v2):
if (v2>v1 and (h1-h2)%(v2-v1)==0):
return True
else:
return False
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Gian and Suneo want their heights to be equal so they asked Doraemon's help. Doraemon gave a big light to both of them but the both big lights have different speed of magnifying. Let's assume the big light given to Gian can increase height of a person by v1 m/s and that of Suneo's big light is v2 m/s.
At the end of each second Doraemon check if their heights are equal or not.
Given initial height of Gian and Suneo, your task is to check whether the height of Gian and Suneo will become equal at some point or not, assuming they both started at the same time.First line takes the input of integer h1(height of gian), h2(height of suneo), v1(speed of Gian's big light) and v2(speed of Suneo's big light) as parameter.
<b>Constraints:-</b>
1 <b>≤</b> h2 < h1<b>≤</b> 10<sup>4</sup>
1 <b>≤</b> v1 <b>≤</b> 10<sup>4</sup>
1 <b>≤</b> v2 <b>≤</b> 10<sup>4</sup>complete the function EqualOrNot and return a boolean True if their height will become equal at some point (as seen by Doraemon) else print False
Sample input:-
4 2 2 4
Sample output:-
Yes
Explanation:-
height of Gian goes as- 4 6 8 10. .
height of Suneo goes as:- 2 6 10..
at the end of 1 second their height will become equal.
Sample Input:-
5 4 1 6
Sample Output:
No, I have written this Solution Code: static boolean EqualOrNot(int h1, int h2, int v1,int v2){
if (v2>v1&&(h1-h2)%(v2-v1)==0){
return true;
}
return false;
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a matrix of N*N dimensions (Mat). Print the matrix rotated by 90 degrees and 180 degrees.The first line contains N.
N lines follow each containing N space-separated integers.
<b>Constraints</b>
2 <= N <= 100
1 <= Mat[i][j] <= 10000Output 2*N+1 lines.
First N lines should contain the Matrix rotated by 90 degrees.
Then print a blank line.
Then N lines should contain the Matrix rotated by 180 degrees.Sample Input 1:
2
3 4
7 6
Sample Output 1:
7 3
6 4
6 7
4 3
Sample Input 2:
2
1 2
3 4
Sample Output 2:
3 1
4 2
4 3
2 1, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define N 1000
// Function to rotate the matrix 90 degree clockwise
void rotate90Clockwise(int a[][N],int n)
{
// Traverse each cycle
for (int i = 0; i < n / 2; i++) {
for (int j = i; j < n - i - 1; j++) {
// Swap elements of each cycle
// in clockwise direction
int temp = a[i][j];
a[i][j] = a[n - 1 - j][i];
a[n - 1 - j][i] = a[n - 1 - i][n - 1 - j];
a[n - 1 - i][n - 1 - j] = a[j][n - 1 - i];
a[j][n - 1 - i] = temp;
}
}
}
// Function for print matrix
void printMatrix(int arr[][N],int n)
{
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++)
cout << arr[i][j] << " ";
cout << '\n';
}
}
// Driver code
int main()
{
int n;
cin>>n;
int arr[n][N];
for(int i=0;i<n;i++){
for(int j=0;j<n;j++){
cin>>arr[i][j];
}}
rotate90Clockwise(arr,n);
printMatrix(arr,n);
cout<<endl;
rotate90Clockwise(arr,n);
printMatrix(arr,n);
return 0;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a matrix of N*N dimensions (Mat). Print the matrix rotated by 90 degrees and 180 degrees.The first line contains N.
N lines follow each containing N space-separated integers.
<b>Constraints</b>
2 <= N <= 100
1 <= Mat[i][j] <= 10000Output 2*N+1 lines.
First N lines should contain the Matrix rotated by 90 degrees.
Then print a blank line.
Then N lines should contain the Matrix rotated by 180 degrees.Sample Input 1:
2
3 4
7 6
Sample Output 1:
7 3
6 4
6 7
4 3
Sample Input 2:
2
1 2
3 4
Sample Output 2:
3 1
4 2
4 3
2 1, I have written this Solution Code:
import java.util.*;
import java.lang.*;
import java.io.*;
class Main
{
public static void main (String[] args) throws java.lang.Exception
{
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
int a[][] = new int[n][n];
for(int i=0;i<n;i++){
for(int j=0;j<n;j++){
a[i][j]= sc.nextInt();
}
}
//rotating by 90 degree
for (int i = 0; i < n / 2; i++) {
for (int j = i; j < n - i - 1; j++) {
int temp = a[i][j];
a[i][j] = a[n - 1 - j][i];
a[n - 1 - j][i] = a[n - 1 - i][n - 1 - j];
a[n - 1 - i][n - 1 - j] = a[j][n - 1 - i];
a[j][n - 1 - i] = temp;
}
}
for(int i =0;i<n;i++){
for(int j=0;j<n;j++){
System.out.print(a[i][j]+" ");
}
System.out.println();
}
//rotating by 90 degree
for (int i = 0; i < n / 2; i++) {
for (int j = i; j < n - i - 1; j++) {
int temp = a[i][j];
a[i][j] = a[n - 1 - j][i];
a[n - 1 - j][i] = a[n - 1 - i][n - 1 - j];
a[n - 1 - i][n - 1 - j] = a[j][n - 1 - i];
a[j][n - 1 - i] = temp;
}
}
System.out.println();
for(int i =0;i<n;i++){
for(int j=0;j<n;j++){
System.out.print(a[i][j]+" ");
}
System.out.println();
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a matrix of N*N dimensions (Mat). Print the matrix rotated by 90 degrees and 180 degrees.The first line contains N.
N lines follow each containing N space-separated integers.
<b>Constraints</b>
2 <= N <= 100
1 <= Mat[i][j] <= 10000Output 2*N+1 lines.
First N lines should contain the Matrix rotated by 90 degrees.
Then print a blank line.
Then N lines should contain the Matrix rotated by 180 degrees.Sample Input 1:
2
3 4
7 6
Sample Output 1:
7 3
6 4
6 7
4 3
Sample Input 2:
2
1 2
3 4
Sample Output 2:
3 1
4 2
4 3
2 1, I have written this Solution Code: x=int(input())
l1=[]
for i in range(x):
a1=list(map(int,input().split()))
l1.append(a1)
for j in range(x):
for i in range(1,x+1):
print(l1[-i][j], end=" ")
print()
print()
for i in range(1,x+1):
for j in range(1,x+1):
print(l1[-i][-j], end=" ")
print(), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You think that integers divisible by K are nice. Given L and R (L<=R), find the number of nice integers from L to R (both inclusive).First Line of input contains three integers L R K
Constraints :
0 <= L <= R <= 1000000000000000000(10^18)
1 <= K <= 1000000000000000000(10^18)Output a single integer which is the number of nice integers from L to R (both inclusive).Sample input 1
1 10 2
Sample output 1
5
Sample intput 2
4 5 3
Sample output 2
0
Explanation:-
Required divisors = 2 4 6 8 10, I have written this Solution Code:
import java.io.*;
import java.util.*;
class Main{
public static void main(String[] args)throws IOException{
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
String str[]=br.readLine().split(" ");
long l=Long.parseLong(str[0]);
long r=Long.parseLong(str[1]);
long k=Long.parseLong(str[2]);
long count=(r/k)-((l-1)/k);
System.out.print(count);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You think that integers divisible by K are nice. Given L and R (L<=R), find the number of nice integers from L to R (both inclusive).First Line of input contains three integers L R K
Constraints :
0 <= L <= R <= 1000000000000000000(10^18)
1 <= K <= 1000000000000000000(10^18)Output a single integer which is the number of nice integers from L to R (both inclusive).Sample input 1
1 10 2
Sample output 1
5
Sample intput 2
4 5 3
Sample output 2
0
Explanation:-
Required divisors = 2 4 6 8 10, I have written this Solution Code: l,r,k=[int(x)for x in input().split()]
print(r//k - (l-1)//k), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You think that integers divisible by K are nice. Given L and R (L<=R), find the number of nice integers from L to R (both inclusive).First Line of input contains three integers L R K
Constraints :
0 <= L <= R <= 1000000000000000000(10^18)
1 <= K <= 1000000000000000000(10^18)Output a single integer which is the number of nice integers from L to R (both inclusive).Sample input 1
1 10 2
Sample output 1
5
Sample intput 2
4 5 3
Sample output 2
0
Explanation:-
Required divisors = 2 4 6 8 10, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main(){
unsigned long long x,l,r,k;
cin>>l>>r>>k;
x=l/k;
if(l%k==0){x--;}
cout<<r/k-x;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer N, find the closest prime number to N. If there are multiple print the smaller one.The input contains a single integer N.
Constraints:
1 <= N <= 1000000000Print the closest prime to N.Sample Input 1
12
Sample Output 1
11
Explanation: Closest prime to 12 is 11 and 13 smaller of which is 11.
Sample Input 2
17
Sample Output 2
17
Explanation: Closest prime to 17 is 17., I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws IOException{
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int n = Integer.parseInt(br.readLine());
if(n==1){
System.out.println(2);
}else{
int after = afterPrime(n);
int before = beforePrime(n);
if(before>after){
System.out.println(n+after);
}
else{System.out.println(n-before);}
}
}
public static boolean isPrime(int n)
{
int count=0;
for(int i=2;i*i<n;i++)
{
if(n%i==0)
count++;
}
if(count==0)
return true;
else
return false;
}
public static int beforePrime(int n)
{
int c=0;
while(true)
{
if(isPrime(n))
return c;
else
{
n=n-1;
c++;
}
}
}
public static int afterPrime(int n)
{
int c=0;
while(true)
{
if(isPrime(n))
return c;
else
{
n=n+1;
c++;
}
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer N, find the closest prime number to N. If there are multiple print the smaller one.The input contains a single integer N.
Constraints:
1 <= N <= 1000000000Print the closest prime to N.Sample Input 1
12
Sample Output 1
11
Explanation: Closest prime to 12 is 11 and 13 smaller of which is 11.
Sample Input 2
17
Sample Output 2
17
Explanation: Closest prime to 17 is 17., I have written this Solution Code: from math import sqrt
def NearPrime(N):
if N >1:
for i in range(2,int(sqrt(N))+1):
if N%i ==0:
return False
break
else: return True
else: return False
N=int(input())
i =0
while NearPrime(N-i)==False and NearPrime(N+i)==False:
i+=1
if NearPrime(N-i) and NearPrime(N+i):print(N-i)
elif NearPrime(N-i):print(N-i)
elif NearPrime(N+i): print(N+i), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer N, find the closest prime number to N. If there are multiple print the smaller one.The input contains a single integer N.
Constraints:
1 <= N <= 1000000000Print the closest prime to N.Sample Input 1
12
Sample Output 1
11
Explanation: Closest prime to 12 is 11 and 13 smaller of which is 11.
Sample Input 2
17
Sample Output 2
17
Explanation: Closest prime to 17 is 17., 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>
/////////////
bool isPrime(int n){
if(n<=1)
return false;
for(int i=2;i*i<=n;++i)
if(n%i==0)
return false;
return true;
}
signed main()
{
fastio;
#ifdef ANIKET_GOYAL
freopen("inputf.in","r",stdin);
freopen("outputf.in","w",stdout);
#endif
int n;
cin>>n;
if(n==1)
cout<<"2";
else{
int v1=n,v2=n;
while(isPrime(v1)==false)
--v1;
while(isPrime(v2)==false)
++v2;
if(v2-n==n-v1)
cout<<v1;
else{
if(v2-n<n-v1)
cout<<v2;
else
cout<<v1;
}
}
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer K and an array height[] of size N, where height[i] denotes the height of the ith tree in a forest. The task is to make a cut of height X from the ground such that at max K units wood is collected. Find the minimum value of X
<b>If you make a cut of height X from the ground then every tree with a height greater than X will be reduced to X and the remaining part of the wood can be collected</b>The first line contains two integers N and K.
The next line contains N integers denoting the elements of the array height[]
<b>Constraints</b>
1 ≤ N ≤ 10<sup>5</sup>
1 ≤ arr[i] ≤ 10<sup>5</sup>
1 ≤ K ≤ 10<sup>7</sup>Print a single integer with the value of X.Sample Input:
4 2
1 2 1 2
Sample Output:
1
<b>Explanation:</b>
Make a cut at height 1, the updated array will be {1, 1, 1, 1} and
the collected wood will be {0, 1, 0, 1} i. e. 0 + 1 + 0 + 1 = 2., I have written this Solution Code: import java.io.*; // for handling input/output
import java.util.*; // contains Collections framework
// don't change the name of this class
// you can add inner classes if needed
class Main {
public static void main (String[] args) {
// Your code here
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
int k = sc.nextInt();
int arr[] = new int[n];
for(int i = 0; i < n; i++)
arr[i] = sc.nextInt();
System.out.println(minValue(arr, n, k));
}
static int minValue(int arr[], int N, int k)
{
int l = 0, h = N;
while(l+1 < h){
int m = (l + h) >> 1;
if(f(arr, m, N) <= k)
h = m;
else
l = m;
}
return h;
}
static int f(int a[], int x, int n)
{
int sum = 0;
for(int i = 0; i < n; i++)
sum += Math.max(a[i]-x, 0);
return sum;
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer K and an array height[] of size N, where height[i] denotes the height of the ith tree in a forest. The task is to make a cut of height X from the ground such that at max K units wood is collected. Find the minimum value of X
<b>If you make a cut of height X from the ground then every tree with a height greater than X will be reduced to X and the remaining part of the wood can be collected</b>The first line contains two integers N and K.
The next line contains N integers denoting the elements of the array height[]
<b>Constraints</b>
1 ≤ N ≤ 10<sup>5</sup>
1 ≤ arr[i] ≤ 10<sup>5</sup>
1 ≤ K ≤ 10<sup>7</sup>Print a single integer with the value of X.Sample Input:
4 2
1 2 1 2
Sample Output:
1
<b>Explanation:</b>
Make a cut at height 1, the updated array will be {1, 1, 1, 1} and
the collected wood will be {0, 1, 0, 1} i. e. 0 + 1 + 0 + 1 = 2., I have written this Solution Code: def minCut(arr,k):
mini = 1
maxi = k
mid = 0
while mini<=maxi:
mid = mini + int((maxi - mini)/2)
wood = 0
for j in range(n):
if(arr[j]-mid>=0):
wood += arr[j] - mid
if wood == k:
break;
elif wood > k:
mini = mid+1
else:
maxi = mid-1
print(mini)
n,k = list(map(int, input().split()))
arr = list(map(int, input().split()))
minCut(arr,k), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer K and an array height[] of size N, where height[i] denotes the height of the ith tree in a forest. The task is to make a cut of height X from the ground such that at max K units wood is collected. Find the minimum value of X
<b>If you make a cut of height X from the ground then every tree with a height greater than X will be reduced to X and the remaining part of the wood can be collected</b>The first line contains two integers N and K.
The next line contains N integers denoting the elements of the array height[]
<b>Constraints</b>
1 ≤ N ≤ 10<sup>5</sup>
1 ≤ arr[i] ≤ 10<sup>5</sup>
1 ≤ K ≤ 10<sup>7</sup>Print a single integer with the value of X.Sample Input:
4 2
1 2 1 2
Sample Output:
1
<b>Explanation:</b>
Make a cut at height 1, the updated array will be {1, 1, 1, 1} and
the collected wood will be {0, 1, 0, 1} i. e. 0 + 1 + 0 + 1 = 2., I have written this Solution Code: #include "bits/stdc++.h"
using namespace std;
#define ll 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 = 1e5 + 5;
const int mod = 1e9 + 7;
const int inf = 1e9 + 9;
int a[N], n, k;
int f(int x){
int sum = 0;
for(int i = 1; i <= n; i++)
sum += max(a[i]-x, 0);
return sum;
}
void solve(){
cin >> n >> k;
for(int i = 1; i <= n; i++)
cin >> a[i];
int l = 0, h = N;
while(l+1 < h){
int m = (l + h) >> 1;
if(f(m) <= k)
h = m;
else
l = m;
}
cout << h;
}
void testcases(){
int tt = 1;
//cin >> tt;
while(tt--){
solve();
}
}
signed main() {
IOS;
testcases();
return 0;
} , In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: There is a tree given with N nodes.
You need to answer Q queries. In each query you are given a tree node. You need to choose some nodes such that any two nodes is at most D distance apart. Distance between the two nodes is the number of minimum number of edges on the path between the two nodes. You have to choose the node given in the query as one of its chosen node. You need to maximise the total distance between all the pair of selected nodes.First line contains two intengers N, D
Next N-1 lines contains two integers A, B (1 ≤ A,B ≤ N) which denotes an edge between that.
Next line contains Q the number of queries
Next Q lines contains a single integer X (1 ≤ X ≤ N) the node that should be chosen
<b> Constraints: </b>
1 ≤ N ≤ 500000
1 ≤ D ≤ N -1
1 ≤ Q ≤ 10Output Q lines each line contains a single integer denoting the maximum total distance between all the pair of selected nodes.INPUT
15 3
5 1
3 2
13 3
2 4
3 5
7 6
3 7
3 8
14 9
5 10
1 11
13 12
1 14
14 15
10
1
2
5
6
7
8
10
12
13
15
OUTPUT
57
57
57
40
57
57
57
40
57
29
, I have written this Solution Code: #include <bits/stdc++.h>
#define For(i,a,b) for(int i=(a);i<=(b);++i)
#define Rep(i,a,b) for(int i=(a);i>=(b);--i)
#define ll unsigned long long
using namespace std;
inline int read() {
char c = getchar();
int x = 0;
bool f = 0;
for (; !isdigit(c); c = getchar())
f ^= !(c ^ 45);
for (; isdigit(c); c = getchar())
x = (x << 1) + (x << 3) + (c ^ 48);
if (f)
x = -x;
return x;
}
#define fi first
#define se second
#define pb push_back
#define mkp make_pair
typedef pair<int, int>pii;
typedef vector<int>vi;
#define maxn 1000005
#define inf 0x3f3f3f3f
bool mbe;
struct info {
int u, v, d, c;
ll su, sv, res;
} emptyinfo;
bool isempty(info o) {
return !o.u && !o.v;
}
info MR(info x, info y) {
if (y.v == x.u)
swap(y.u, y.v), swap(y.su, y.sv);
if (y.u == x.v)
swap(x.u, x.v), swap(x.su, x.sv);
if (x.v == y.v)
swap(y.u, y.v), swap(y.su, y.sv), swap(x.u, x.v), swap(x.su, x.sv);
info z = x;
z.c = x.c + y.c - 1;
z.su = x.su + y.su;
z.sv = x.sv + y.su + 1ll * (y.c - 1) * x.d;
z.res = x.res + y.res + x.su * (y.c - 1) + y.su * (x.c - 1);
return z;
}
info MC(info x, info y) {
if (y.v == x.u)
swap(y.u, y.v), swap(y.su, y.sv);
if (y.u == x.v)
swap(x.u, x.v), swap(x.su, x.sv);
if (x.v == y.v)
swap(y.u, y.v), swap(y.su, y.sv), swap(x.u, x.v), swap(x.su, x.sv);
info z;
z.u = x.v, z.v = y.v, z.d = x.d + y.d;
z.c = x.c + y.c - 1;
z.su = x.sv + y.su + 1ll * (y.c - 1) * x.d;
z.sv = y.sv + x.su + 1ll * (x.c - 1) * y.d;
z.res = x.res + y.res + x.su * (y.c - 1) + y.su * (x.c - 1);
return z;
}
info R(info x, info y) {
return isempty(x) ? y : (isempty(y) ? x : MR(x, y));
}
info C(info x, info y) {
return isempty(x) ? y : (isempty(y) ? x : MC(x, y));
}
struct cluster {
int u, v, sz, len, op;
info w;
} t[maxn];
int cnt, up[maxn], ls[maxn], rs[maxn];
#define mxn 500005
int n, d, dd;
int fa[mxn], son[mxn], siz[mxn], dep[mxn], top[mxn], dfn[mxn], idx;
vi e[mxn];
void dfs1(int u) {
siz[u] = 1, dep[u] = dep[fa[u]] + 1;
if (fa[u]) {
int i = u;
t[i].sz = 1, t[i].u = fa[i], t[i].v = i, t[i].len = 1;
t[i].w = (info) {
u, fa[u], 1, 2, 1, 1, 1
};
}
for (auto v : e[u]) {
if (v == fa[u])
continue;
fa[v] = u, dfs1(v), siz[u] += siz[v];
if (siz[v] > siz[son[u]])
son[u] = v;
}
}
void dfs2(int u, int tp) {
top[u] = tp;
dfn[u] = ++idx;
if (son[u])
dfs2(son[u], tp);
for (auto v : e[u])
if (v != fa[u] && v != son[u])
dfs2(v, v);
}
int lca(int u, int v) {
for (; top[u] != top[v]; u = fa[top[u]])
if (dep[top[u]] < dep[top[v]])
swap(u, v);
return dep[u] < dep[v] ? u : v;
}
int dist(int u, int v) {
return dep[u] + dep[v] - 2 * dep[lca(u, v)];
}
bool isanc(int u, int v) {
return dfn[v] >= dfn[u] && dfn[v] <= dfn[u] + siz[u] - 1;
}
int rake(int a, int b) {
if (!a || !b)
return a | b;
int c = ++cnt;
up[a] = up[b] = c, ls[c] = a, rs[c] = b;
t[c].sz = t[a].sz + t[b].sz, t[c].op = 0;
t[c].len = t[a].len;
t[c].w = R(t[a].w, t[b].w);
t[c].u = t[a].u, t[c].v = t[a].v;
return c;
}
int com(int a, int b) {
if (!a || !b)
return a | b;
int c = ++cnt;
up[a] = up[b] = c, ls[c] = a, rs[c] = b;
t[c].sz = t[a].sz + t[b].sz, t[c].op = 1;
t[c].w = C(t[a].w, t[b].w);
t[c].len = t[a].len + t[b].len;
vi o = {t[a].u, t[a].v, t[b].u, t[b].v};
sort(o.begin(), o.end());
t[c].u = t[a].u, t[c].v = t[b].v;
return c;
}
struct cmp {
bool operator()(int x, int y) {
return t[x].sz > t[y].sz;
}
};
int lid[mxn], zid[mxn], rt;
int sid[mxn], su[mxn], len;
int get(int l, int r) {
if (l == r)
return lid[sid[l]];
int sum = 0, nw = 0, mid = 0;
For(i, l, r)sum += t[lid[sid[i]]].sz;
For(i, l, r - 1) {
nw += t[lid[sid[i]]].sz;
if (nw * 2 >= sum || i == r - 1) {
mid = i;
break;
}
}
return com(get(l, mid), get(mid + 1, r));
}
void solve(int u) {
priority_queue<int, vector<int>, cmp>q;
for (auto v : e[u])
if (v != fa[u] && v != son[u])
solve(v), q.push(zid[v]);
while (q.size() > 1) {
int x = q.top();
q.pop();
int y = q.top();
q.pop();
q.push(rake(x, y));
}
if (q.size())
lid[u] = q.top();
if (u != 1)
lid[u] = rake(u, lid[u]);
if (son[u])
solve(son[u]);
if (u != top[u])
return;
len = 0;
for (int x = u; x; x = son[x])
sid[++len] = x;
if (u != 1)
zid[u] = get(1, len);
else
zid[u] = rake(lid[u], get(2, len));
}
vector<info>f[maxn][2], g[maxn][2];
bool pd(int x, int y) {
return t[x].v == t[y].u || t[x].v == t[y].v;
}
int P(int x, int y) {
return y ? t[x].v : t[x].u;
}
info F(int x, int y, int z) {
if (z < 0)
return emptyinfo;
if (x <= n)
return z < 0 ? emptyinfo : t[x].w;
return f[x][y][min(max(z, 0), (int)f[x][y].size() - 1)];
}
info G(int x, int y, int z) {
// --z;
if (z < 0)
return emptyinfo;
return g[x][y][min(max(z, 0), (int)g[x][y].size() - 1)];
}
bool med;
void dfs3(int u) {
if (u <= n)
return;
For(i, 0, 1)f[u][i].resize(min(t[u].sz + 1, dd + 2) - 1, emptyinfo);
int ls =::ls[u], rs =::rs[u];
dfs3(ls), dfs3(rs);
int l = pd(ls, rs), r = pd(rs, ls), o = (t[u].v == P(ls, !l));
int lim = f[u][0].size();
if (t[u].op) {
For(i, 0, lim - 1) {
f[u][o][i] = C(F(ls, !l, i), F(rs, r, i - t[ls].len));
f[u][!o][i] = C(F(rs, !r, i), F(ls, l, i - t[rs].len));
}
} else {
For(i, 0, lim - 1) {
f[u][o][i] = R(F(ls, !l, i), F(rs, r, i - t[ls].len));
f[u][!o][i] = R(F(ls, l, i), F(rs, r, i));
}
}
if (t[u].sz <= d / 2) {
for (int x : {
ls, rs
})
For(y, 0, 1)
f[x][y].clear(), f[x][y].shrink_to_fit();
}
}
bool need[maxn];
void jump(int u) {
while (up[u] && t[up[u]].sz <= d / 2)
u = up[u];
if (u != rt)
need[u] = 1;
}
void dfs4(int u) {
if (u <= n || t[u].sz <= d / 2)
return;
int ls =::ls[u], rs =::rs[u], l = pd(ls, rs), r = pd(rs, ls), o = (t[u].v == P(ls, !l));
for (int x : {
ls, rs
})
For(y, 0, 1)g[x][y].resize(dd + 2, emptyinfo);
if (t[u].op) {
For(i, 1, dd + 1) {
g[ls][!l][i] = G(u, o, i);
g[ls][l][i] = C(F(rs, r, i - 1), G(u, !o, i - t[rs].len));
g[rs][!r][i] = G(u, !o, i);
g[rs][r][i] = C(F(ls, l, i - 1), G(u, o, i - t[ls].len));
}
} else {
For(i, 1, dd + 1) {
g[ls][!l][i] = G(u, o, i);
g[ls][l][i] = R(G(u, !o, i), F(rs, r, i - 1));
g[rs][r][i] = R(R(F(ls, l, i - 1), G(u, !o, i)), G(u, o, i - t[ls].len));
}
}
if (!need[u]) {
For(o, 0, 1)
g[u][o].clear(), g[u][o].shrink_to_fit();
}
dfs4(ls), dfs4(rs);
}
void init() {
cnt = n;
dfs1(1), dfs2(1, 1);
solve(1);
rt = zid[1];
dfs3(rt);
bool MED;
// printf("%.6lf %.6lf\n",(&mbe-&MED)/1024576.0,(&med-&MED)/1024576.0);
For(i, 1, n) jump(i);
For(i, 0, 1)g[rt][i].resize(dd + 1, emptyinfo);
dfs4(rt);
// if(n>=400000)exit(0);
For(i, 2, cnt) for (auto &tmp : g[i][0])
tmp = C(tmp, t[i].w);
}
info ask(int u) {
int x = u;
if (u == 1)
u = 2;
if (d % 2 == 0) {
while (up[u] && t[up[u]].sz <= d / 2)
u = up[u];
if (u == rt)
return t[rt].w;
return C(G(u, 0, dd - dist(x, t[u].u)), G(u, 1, dd - dist(x, t[u].v)));
}
while (up[u] && t[up[u]].sz <= d / 2)
u = up[u];
if (u == rt)
return t[rt].w;
// t[u].u is upon t[u].v
// cout<<"u: "<<x<<" "<<u<<" "<<t[up[u]].sz<<"\n";
if (isanc(t[u].u, fa[x]) && isanc(x, t[u].v))
return C(G(u, 0, d / 2 + 1 - dist(x, t[u].u)), G(u, 1, d / 2 - dist(x, t[u].v)));
return C(G(u, 0, d / 2 + 1 - dist(x, t[u].u)), G(u, 1, d / 2 + 1 - dist(x, t[u].v)));
}
ll ans[mxn];
int D[mxn];
void getd(int u, int pa) {
if (pa)
D[u] = D[pa] + 1;
for (int v : e[u])
if (v != pa)
getd(v, u);
}
signed main() {
// freopen("17.1.in","r",stdin);
// freopen("my.out","w",stdout);
// printf("%.6lf\n",(&mbe-&med)/1024576.0);
n = read(), d = read();
dd = d / 2;
For(i, 2, n) {
int u = read(), v = read();
e[u].pb(v), e[v].pb(u);
}
init();
For(i, 1 + d % 2, n) ans[i] = ask(i).res;
int q = read();
For(_, 1, q) {
int u = read();
if (d == 0) {
puts("0");
continue;
}
D[u] = 0, getd(u, 0);
ll res = 0;
if (d % 2 == 0) {
For(i, 1, n) if (D[i] <= d / 2)
res = max(res, ans[i]);
} else {
For(i, 2, n) if (D[i] <= d / 2 || D[fa[i]] <= d / 2)
res = max(res, ans[i]);
}
printf("%lld\n", res);
}
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: This is a time of conflict in Westeros as Viserys Targaryen, the king of all seven kingdoms, rejected the hand of Lady Laena Velaryon. So, the Velaryon soldiers are not deemed trustworthy anymore. The Targaryen soldiers have to keep an eye on them.
You are given a sequential order of N soldiers standing in a line. The order is provided as a binary string, with 0 representing Velaryon soldiers, and 1 representing the Targaryen soldiers. Viserys wants each contiguous segment of N/2 soldiers to contain an even number of Targaryen soldiers.
Formally you are given a binary string of length N, where N is an even natural number. Each character of the string is either '0' or '1'. You want to rearrange the elements of the string in such a way that the final string contains an even number of 1s in each contiguous subsegment of length N/2.
Your task is to find out whether there exists a rearrangement of the soldiers that satisfies the above conditions.The first line contains an integer T, the number of test cases. Then, T test cases follow.
The first line of each test case contains an even positive integer N, the length of the line.
The second line of each test case contains a binary string of length N, representing the current arrangement of soldiers.
<b> Constraints: </b>
1 ≤ T ≤ 10
2 ≤ N ≤ 10<sup>4</sup>
N is evenPrint a single character in a new line for each test case. Print '1' (without quotes) if a required rearrangement exists, and '0' (without quotes) otherwise.Sample Input:
3
2
10
2
00
4
0011
Sample Output:
0
1
0
(In the last case, no matter how you rearrange the string, there will always be a single one in at least one subsegment of length 2 of the string), I have written this Solution Code: import java.io.*;
import java.util.*;
class Main{
public static void main(String[] args)throws IOException
{
StringBuilder out=new StringBuilder();
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
int test=Integer.parseInt(br.readLine());
while(test-->0)
{
int n=Integer.parseInt(br.readLine());
String s=br.readLine();
int c1=0;
for(int i=0;i<s.length();i++)
{
if(s.charAt(i)=='1') c1++;
}
if(c1%4==0) out.append("1\n");
else if(c1==s.length() && (c1/2)%2==0) out.append("1\n");
else
out.append("0\n");
}
System.out.print(out);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: This is a time of conflict in Westeros as Viserys Targaryen, the king of all seven kingdoms, rejected the hand of Lady Laena Velaryon. So, the Velaryon soldiers are not deemed trustworthy anymore. The Targaryen soldiers have to keep an eye on them.
You are given a sequential order of N soldiers standing in a line. The order is provided as a binary string, with 0 representing Velaryon soldiers, and 1 representing the Targaryen soldiers. Viserys wants each contiguous segment of N/2 soldiers to contain an even number of Targaryen soldiers.
Formally you are given a binary string of length N, where N is an even natural number. Each character of the string is either '0' or '1'. You want to rearrange the elements of the string in such a way that the final string contains an even number of 1s in each contiguous subsegment of length N/2.
Your task is to find out whether there exists a rearrangement of the soldiers that satisfies the above conditions.The first line contains an integer T, the number of test cases. Then, T test cases follow.
The first line of each test case contains an even positive integer N, the length of the line.
The second line of each test case contains a binary string of length N, representing the current arrangement of soldiers.
<b> Constraints: </b>
1 ≤ T ≤ 10
2 ≤ N ≤ 10<sup>4</sup>
N is evenPrint a single character in a new line for each test case. Print '1' (without quotes) if a required rearrangement exists, and '0' (without quotes) otherwise.Sample Input:
3
2
10
2
00
4
0011
Sample Output:
0
1
0
(In the last case, no matter how you rearrange the string, there will always be a single one in at least one subsegment of length 2 of the string), I have written this Solution Code: T=int(input())
for i in range(T):
n=int(input())
a=input()
count_1=0
for i in a:
if i=='1':
count_1+=1
if count_1%2==0 and ((count_1)//2)%2==0:
print('1')
else:
print('0'), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: This is a time of conflict in Westeros as Viserys Targaryen, the king of all seven kingdoms, rejected the hand of Lady Laena Velaryon. So, the Velaryon soldiers are not deemed trustworthy anymore. The Targaryen soldiers have to keep an eye on them.
You are given a sequential order of N soldiers standing in a line. The order is provided as a binary string, with 0 representing Velaryon soldiers, and 1 representing the Targaryen soldiers. Viserys wants each contiguous segment of N/2 soldiers to contain an even number of Targaryen soldiers.
Formally you are given a binary string of length N, where N is an even natural number. Each character of the string is either '0' or '1'. You want to rearrange the elements of the string in such a way that the final string contains an even number of 1s in each contiguous subsegment of length N/2.
Your task is to find out whether there exists a rearrangement of the soldiers that satisfies the above conditions.The first line contains an integer T, the number of test cases. Then, T test cases follow.
The first line of each test case contains an even positive integer N, the length of the line.
The second line of each test case contains a binary string of length N, representing the current arrangement of soldiers.
<b> Constraints: </b>
1 ≤ T ≤ 10
2 ≤ N ≤ 10<sup>4</sup>
N is evenPrint a single character in a new line for each test case. Print '1' (without quotes) if a required rearrangement exists, and '0' (without quotes) otherwise.Sample Input:
3
2
10
2
00
4
0011
Sample Output:
0
1
0
(In the last case, no matter how you rearrange the string, there will always be a single one in at least one subsegment of length 2 of the string), I have written this Solution Code: #include<bits/stdc++.h>
using namespace std;
signed main() {
int t;
cin >> t;
for(int i=0; i<t; i++) {
int n;
cin >> n;
string s;
cin >> s;
int cnt = 0;
for(int j=0; j<n; j++) {
if(s[j] == '1') cnt++;
}
if(cnt % 4 == 0) cout << 1 << "\n";
else cout << 0 << "\n";
}
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Find factorial of a given number where n! = n * n-1 * n-2 .....* 1First line consists of a single integer denoting n
Constraints:-
0 <= n <= 20Output is a single line containing factorial(n)Sample Input
5
Sample Output
120
Explanation:-
5!= 5 * 4 * 3 * 2 * 1 = 120
Sample Input
10
Sample Output
3628800, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args)throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int n=Integer.parseInt(br.readLine());
long fact=1;
for(int i=1; i<=n;i++){
fact*=i;
}
System.out.print(fact);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Find factorial of a given number where n! = n * n-1 * n-2 .....* 1First line consists of a single integer denoting n
Constraints:-
0 <= n <= 20Output is a single line containing factorial(n)Sample Input
5
Sample Output
120
Explanation:-
5!= 5 * 4 * 3 * 2 * 1 = 120
Sample Input
10
Sample Output
3628800, I have written this Solution Code: def fact(n):
if( n==0 or n==1):
return 1
return n*fact(n-1);
n=int(input())
print(fact(n)), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Find factorial of a given number where n! = n * n-1 * n-2 .....* 1First line consists of a single integer denoting n
Constraints:-
0 <= n <= 20Output is a single line containing factorial(n)Sample Input
5
Sample Output
120
Explanation:-
5!= 5 * 4 * 3 * 2 * 1 = 120
Sample Input
10
Sample Output
3628800, I have written this Solution Code: #include <bits/stdc++.h>
// #define ll long long
using namespace std;
int main(){
int t;
t=1;
while(t--){
int n;
cin>>n;
unsigned long long sum=1;
for(int i=1;i<=n;i++){
sum*=i;
}
cout<<sum<<endl;
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number N. The task is to find the Nth Catalan number.
The first few Catalan numbers for N = 1, 2, 3, … are 1, 2, 5, 14, 42, 132, 429, 1430, 4862, …
You can read more about Catalan numbers <a href="https://en.wikipedia.org/wiki/Catalan_number">here</a>.The first line of input contains a single integer T which denotes the number of test cases. The first line of each test case contains a single integer N.
Constraints:
1 <= T <= 100000
1 <= N <= 1000000For each test case, in a new line print the Catalan number at position N.
Since the answer can be large, print answer modulo (10^9 + 7)Sample Input:
3
5
4
10
Sample Output:
42
14
16796, I have written this Solution Code: import java.io.*;
class Main {
public static void main(String[] args) throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int t = Integer.parseInt(br.readLine());
dp[0]=0;
dp[1]=1;
for (int i = 1; i <=1000000 ; i++) {
dp[i+1]=(((4*i+2)%mod)*dp[i]%mod*modPow(i+2,mod-2))%mod;
}
while(t-->0){
int n = Integer.parseInt(br.readLine());
System.out.println(dp[n]);
}
}
static long [] dp= new long [1000002];
static long mod = 1000000007;
static long modPow(long x, long n){
if(n==0) return 1;
if(n==1) return x;
if(n%2==0) return modPow((x*x)%mod,n/2)%mod;
return (x*modPow((x*x)%mod,n/2)%mod);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number N. The task is to find the Nth Catalan number.
The first few Catalan numbers for N = 1, 2, 3, … are 1, 2, 5, 14, 42, 132, 429, 1430, 4862, …
You can read more about Catalan numbers <a href="https://en.wikipedia.org/wiki/Catalan_number">here</a>.The first line of input contains a single integer T which denotes the number of test cases. The first line of each test case contains a single integer N.
Constraints:
1 <= T <= 100000
1 <= N <= 1000000For each test case, in a new line print the Catalan number at position N.
Since the answer can be large, print answer modulo (10^9 + 7)Sample Input:
3
5
4
10
Sample Output:
42
14
16796, I have written this Solution Code: #include<bits/stdc++.h>
#define int long long
#define ld long double
#define ll long long
#define pb push_back
#define endl '\n'
#define pi pair<int,int>
#define vi vector<int>
#define all(a) (a).begin(),(a).end()
#define fi first
#define se second
#define sz(x) (int)x.size()
#define hell 1000000007
#define rep(i,a,b) for(int i=a;i<b;i++)
#define dep(i,a,b) for(int i=a;i>=b;i--)
#define lbnd lower_bound
#define ubnd upper_bound
#define bs binary_search
#define mp make_pair
using namespace std;
const int N = 2e6 + 5;
const int mod = 1e9 + 7;
const int inf = 1e9 + 9;
int f[N], inv[N], res[N];
int power(int a, int b){
int ans = 1;
while(b){
if(b&1)
ans = (ans*a) % mod;
b >>= 1;
a = (a*a) % mod;
}
return ans;
}
void solve(){
int n; cin >> n;
cout << res[n] << endl;
}
void testcases(){
int tt = 1;
f[0] = 1;
for(int i = 1; i < N; i++)
f[i] = (i*f[i-1]) % mod;
inv[N-1] = power(f[N-1], mod-2);
for(int i = N-2; i >= 1; i--)
inv[i] = ((i+1)*inv[i+1]) % mod;
for(int i = 1; i < N/2; i++){
res[i] = f[2*i];
res[i] = (res[i]*inv[i]) % mod;
res[i] = (res[i]*inv[i]) % mod;
res[i] = (res[i]*power(i+1, mod-2)) % mod;
}
cin >> tt;
while(tt--){
solve();
}
}
signed main()
{
ios_base::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
clock_t start = clock();
testcases();
cerr << (double)(clock() - start)*1000/CLOCKS_PER_SEC << " ms: ";
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Naruto and Sasuke are on a S- Rank mission. Now They got bored and thought of competing in a race against each other in a horizontal plane. They decides a common spot C at which both of them will try to reach. Whoever reaches first wins the race, both of them run at the same speed.
Given initial positions of Naruto and Sasuke as A and B recpectively.
you need to tell which of them will win the race. If Naruto wins print "N" ( without the quotes ), if Sasuke wins print "S" ( without the quotes ).
if both of them reach the common spot at the same time, print "D" (for draw, without the quotes ).<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>Race</b> that takes the integer A (initial position of Naruto) and B (initial position of Sasuke) and C (position of common spot) as parameter.
Constraints
1 <= A, B, C <= 100Return the character according to the given scenario.Sample Input
1 2 3
Sample Output
S
Sample Input
1 3 2
Sample Output
D, I have written this Solution Code:
char Race(int A, int B, int C){
if(abs(C-A)==abs(C-B)){return 'D';}
if(abs(C-A)>abs(C-B)){return 'S';}
else{
return 'N';}
}
, In this Programming Language: C, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Naruto and Sasuke are on a S- Rank mission. Now They got bored and thought of competing in a race against each other in a horizontal plane. They decides a common spot C at which both of them will try to reach. Whoever reaches first wins the race, both of them run at the same speed.
Given initial positions of Naruto and Sasuke as A and B recpectively.
you need to tell which of them will win the race. If Naruto wins print "N" ( without the quotes ), if Sasuke wins print "S" ( without the quotes ).
if both of them reach the common spot at the same time, print "D" (for draw, without the quotes ).<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>Race</b> that takes the integer A (initial position of Naruto) and B (initial position of Sasuke) and C (position of common spot) as parameter.
Constraints
1 <= A, B, C <= 100Return the character according to the given scenario.Sample Input
1 2 3
Sample Output
S
Sample Input
1 3 2
Sample Output
D, I have written this Solution Code: def Race(A,B,C):
if abs(C-A) ==abs(C-B):
return 'D'
if abs(C-A)>abs(C-B):
return 'S'
return 'N'
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Naruto and Sasuke are on a S- Rank mission. Now They got bored and thought of competing in a race against each other in a horizontal plane. They decides a common spot C at which both of them will try to reach. Whoever reaches first wins the race, both of them run at the same speed.
Given initial positions of Naruto and Sasuke as A and B recpectively.
you need to tell which of them will win the race. If Naruto wins print "N" ( without the quotes ), if Sasuke wins print "S" ( without the quotes ).
if both of them reach the common spot at the same time, print "D" (for draw, without the quotes ).<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>Race</b> that takes the integer A (initial position of Naruto) and B (initial position of Sasuke) and C (position of common spot) as parameter.
Constraints
1 <= A, B, C <= 100Return the character according to the given scenario.Sample Input
1 2 3
Sample Output
S
Sample Input
1 3 2
Sample Output
D, I have written this Solution Code:
char Race(int A, int B, int C){
if(abs(C-A)==abs(C-B)){return 'D';}
if(abs(C-A)>abs(C-B)){return 'S';}
else{
return 'N';}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Naruto and Sasuke are on a S- Rank mission. Now They got bored and thought of competing in a race against each other in a horizontal plane. They decides a common spot C at which both of them will try to reach. Whoever reaches first wins the race, both of them run at the same speed.
Given initial positions of Naruto and Sasuke as A and B recpectively.
you need to tell which of them will win the race. If Naruto wins print "N" ( without the quotes ), if Sasuke wins print "S" ( without the quotes ).
if both of them reach the common spot at the same time, print "D" (for draw, without the quotes ).<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>Race</b> that takes the integer A (initial position of Naruto) and B (initial position of Sasuke) and C (position of common spot) as parameter.
Constraints
1 <= A, B, C <= 100Return the character according to the given scenario.Sample Input
1 2 3
Sample Output
S
Sample Input
1 3 2
Sample Output
D, I have written this Solution Code: static char Race(int A,int B,int C){
if(Math.abs(C-A)==Math.abs(C-B)){return 'D';}
if(Math.abs(C-A)>Math.abs(C-B)){return 'S';}
else{
return 'N';}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a grid having N*M elements. Find the index of the row (1-base indexing) which has the maximum sum of elements.
Note: if more than two rows have the same sum of elements, then return the smallest index.The first line of the input contains two integers N and M.
The next N lines each contain M space separated integers.
<b>Constraints:</b>
1 <= N, M <= 10<sup>3</sup>
1 <= A<sub>i, j</sub> <= 10<sup>9Print the index of the row (1-base indexing) which has the maximum sum of elements.Sample Input:
4 3
3 4 2
5 1 7
2 8 1
2 3 3
Sample Output:
2
<b>Explaination:</b>
Row number 2 has sum = 5 + 1 + 7 = 13 which is the maximum among all rows., I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
static class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader()
{
br = new BufferedReader(
new InputStreamReader(System.in));
}
String next()
{
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
}
catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() { return Integer.parseInt(next()); }
long nextLong() { return Long.parseLong(next()); }
double nextDouble()
{
return Double.parseDouble(next());
}
String nextLine()
{
String str = "";
try {
if(st.hasMoreTokens()){
str = st.nextToken("\n");
}
else{
str = br.readLine();
}
}
catch (IOException e) {
e.printStackTrace();
}
return str;
}
}
public static void main (String[] args) {
FastReader sc=new FastReader();
int n=sc.nextInt();
int m=sc.nextInt();
int arr[][]=new int[n][m];
for(int i=0; i<n; i++)
{
for(int j=0; j<m; j++)
{
arr[i][j]= sc.nextInt();
}
}
long maxsum=0;
int index=0;
for(int i=0; i<n; i++)
{
long sum=0;
for(int j=0; j<m; j++)
{
sum += arr[i][j];
}
if(sum > maxsum)
{
maxsum=sum;
index=i+1;
}
}
System.out.print(index);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a grid having N*M elements. Find the index of the row (1-base indexing) which has the maximum sum of elements.
Note: if more than two rows have the same sum of elements, then return the smallest index.The first line of the input contains two integers N and M.
The next N lines each contain M space separated integers.
<b>Constraints:</b>
1 <= N, M <= 10<sup>3</sup>
1 <= A<sub>i, j</sub> <= 10<sup>9Print the index of the row (1-base indexing) which has the maximum sum of elements.Sample Input:
4 3
3 4 2
5 1 7
2 8 1
2 3 3
Sample Output:
2
<b>Explaination:</b>
Row number 2 has sum = 5 + 1 + 7 = 13 which is the maximum among all rows., I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define int long long
signed main(){
int n, m;
cin >> n >> m;
vector<vector<int>> a(n, vector<int> (m));
vector<int> sum(n);
for(int i = 0; i < n; i++){
for(int j = 0; j < m; j++){
cin >> a[i][j];
sum[i] += a[i][j];
}
}
int mx = 0, ans = 0;
for(int i = 0; i < n; i++){
if(mx < sum[i]){
mx = sum[i];
ans = i;
}
}
cout << ans + 1;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sheldon and Leonard are gone for lunch but none of them have money so they decided to wash dishes. In total, they washed T dishes from which N dishes are washed by Leonard. Now Leonard wants to know the number of dishes Sheldon washed. Help him to find it.The first line of the input contains N and T
Constraints:-
1 <= N <= T <= 1000Return the number of dishes Sheldon washed.Sample Input:-
3 6
Sample Output:-
3
Sample Input:-
2 4
Sample Output:-
2, I have written this Solution Code: nan, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sheldon and Leonard are gone for lunch but none of them have money so they decided to wash dishes. In total, they washed T dishes from which N dishes are washed by Leonard. Now Leonard wants to know the number of dishes Sheldon washed. Help him to find it.The first line of the input contains N and T
Constraints:-
1 <= N <= T <= 1000Return the number of dishes Sheldon washed.Sample Input:-
3 6
Sample Output:-
3
Sample Input:-
2 4
Sample Output:-
2, I have written this Solution Code:
#include <iostream>
using namespace std;
int Dishes(int N, int T){
return T-N;
}
int main(){
int n,k;
scanf("%d%d",&n,&k);
printf("%d",Dishes(n,k));
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sheldon and Leonard are gone for lunch but none of them have money so they decided to wash dishes. In total, they washed T dishes from which N dishes are washed by Leonard. Now Leonard wants to know the number of dishes Sheldon washed. Help him to find it.The first line of the input contains N and T
Constraints:-
1 <= N <= T <= 1000Return the number of dishes Sheldon washed.Sample Input:-
3 6
Sample Output:-
3
Sample Input:-
2 4
Sample Output:-
2, I have written this Solution Code: nan, In this Programming Language: C, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sheldon and Leonard are gone for lunch but none of them have money so they decided to wash dishes. In total, they washed T dishes from which N dishes are washed by Leonard. Now Leonard wants to know the number of dishes Sheldon washed. Help him to find it.The first line of the input contains N and T
Constraints:-
1 <= N <= T <= 1000Return the number of dishes Sheldon washed.Sample Input:-
3 6
Sample Output:-
3
Sample Input:-
2 4
Sample Output:-
2, I have written this Solution Code: nan, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Saurabh has a paper of size N*M and some horizontal and vertical lines in the form of arrays. Saurabh wants to know the maximum area which is trapped inside the lines.
Note:- Consider the boundary of the paper to be vertical and horizontal lines. Also consider 0 indexingThe first line of input contains 4 space separated integers depicting N, M, size of array contains horizontal lines(H), size of array containing vertical lines(V). The second line contains H space separated integers depicting horizontal lines. Last lines contains V space separated integers depicting vertical lines.
Constraints:-
1 <= N, M <= 10<sup>9</sup>
1 <= H, V <= 100000
0 <= horizontal lines <= N
0 <= vertical lines <= MPrint the maximum area trapped between the lines.
<b>Note:</b>Area to be printed might be large print area as (area%(10<sup>9</sup> +7)).Sample Input:-
5 4 3 2
1 2 4
1 3
Sample Output:-
4
Explanation:-
The area is- (2,1), (2,3) (4,1) (4,3)
Sample Input:-
5 4 2 1
3 1
1
Sample Output:-
6, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args)
{
Scanner scn = new Scanner(System.in);
int n = scn.nextInt();
int m = scn.nextInt();
int h = scn.nextInt();
int w = scn.nextInt();
int harr[] = new int[h];
int warr[] = new int[w];
for(int i=0; i<h; i++)
{
harr[i] = scn.nextInt();
}
for(int i=0; i<w; i++)
{
warr[i] = scn.nextInt();
}
System.out.println(maxArea( n, m, harr, warr));
}
public static int maxArea(int n, int m, int[] hc, int[] vc) {
Arrays.sort(hc);
Arrays.sort(vc);
int maxh = Math.max(hc[0], n - hc[hc.length-1]),
maxv = Math.max(vc[0], m - vc[vc.length-1]);
for (int i = 1; i < hc.length; i++)
maxh = Math.max(maxh, hc[i] - hc[i-1]);
for (int i = 1; i < vc.length; i++)
maxv = Math.max(maxv, vc[i] - vc[i-1]);
return (int)((long)maxh * maxv % 1000000007);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Saurabh has a paper of size N*M and some horizontal and vertical lines in the form of arrays. Saurabh wants to know the maximum area which is trapped inside the lines.
Note:- Consider the boundary of the paper to be vertical and horizontal lines. Also consider 0 indexingThe first line of input contains 4 space separated integers depicting N, M, size of array contains horizontal lines(H), size of array containing vertical lines(V). The second line contains H space separated integers depicting horizontal lines. Last lines contains V space separated integers depicting vertical lines.
Constraints:-
1 <= N, M <= 10<sup>9</sup>
1 <= H, V <= 100000
0 <= horizontal lines <= N
0 <= vertical lines <= MPrint the maximum area trapped between the lines.
<b>Note:</b>Area to be printed might be large print area as (area%(10<sup>9</sup> +7)).Sample Input:-
5 4 3 2
1 2 4
1 3
Sample Output:-
4
Explanation:-
The area is- (2,1), (2,3) (4,1) (4,3)
Sample Input:-
5 4 2 1
3 1
1
Sample Output:-
6, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define MEM(a, b) memset(a, (b), sizeof(a))
#define FOREACH(it, l) for (auto it = l.begin(); it != l.end(); it++)
#define IN(A, B, C) assert( B <= A && A <= C)
#define MP make_pair
#define FOR(i,a) for(int i=0;i<a;i++)
#define FOR1(i,j,a) for(int i=j;i<a;i++)
#define EB emplace_back
#define INF (int)1e9
#define EPS 1e-9
#define PI 3.1415926535897932384626433832795
#define max1 101
#define MOD 1000000007
#define read(type) readInt<type>()
#define out(x) cout<<x<<'\n'
#define out1(x) cout<<x<<" "
#define END cout<<'\n'
#define int long long
#define sz(v) ((int)(v).size())
#define all(v) (v).begin(), (v).end()
void fast(){
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
}
int solve(int ha, int w, vector<int>h, vector<int> v){
sort(h.begin(),h.end());
sort(v.begin(),v.end());
h.push_back(ha);
v.push_back(w);
long long hh=h[0],vv=v[0];;
for(int i=1;i<h.size();i++){
hh=max(hh,(1LL)*(h[i]-h[i-1]));
}
for(int i=1;i<v.size();i++){
vv=max(vv,(1LL)*(v[i]-v[i-1]));
}
return (hh*vv*1LL)%(1000000007);
}
signed main(){
fast();
int n;
cin>>n;
int m;
cin>>m;
int x,y;
cin>>x>>y;
vector<int> h(x),v(y);
FOR(i,x){
cin>>h[i];}
FOR(i,y){
cin>>v[i];}
cout<<solve(n,m,h,v);
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array, A of N integers. Find the product of maximum values for every subarray of size K. Print the answer modulo 10<sup>9</sup>+7.
A subarray is any contiguous sequence of elements in an array.The first line contains two integers N and K, denoting the size of the array and the size of the subarray respectively.
The next line contains N integers denoting the elements of the array.
<b>Constarints</b>
1 <= K <= N <= 1000000
1 <= A[i] <= 1000000Print a single integer denoting the product of maximums for every subarray of size K modulo 1000000007Sample Input 1:
6 4
1 5 2 3 6 4
Sample Output 1:
180
<b>Explanation:</b>
For subarray [1, 5, 2, 3], maximum = 5
For subarray [5, 2, 3, 6], maximum = 6
For subarray [2, 3, 6, 4], maximum = 6
Therefore, ans = 5*6*6 = 180, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args)throws IOException {
Reader sc=new Reader();
int n=sc.nextInt();
int b=sc.nextInt();
int[] a=new int[n];
for(int i=0;i<n;i++){
a[i]=sc.nextInt();
}
int j, max;
long mul=1;
Deque<Integer> dq= new LinkedList<Integer>();
for (int i = 0;i<b;i++)
{
while (!dq.isEmpty() && a[i] >=a[dq.peekLast()])
dq.removeLast();
dq.addLast(i);
}
mul=((mul%1000000007)*(a[dq.peek()]%1000000007))%1000000007;
for (int i=b; i < n;i++)
{
while ((!dq.isEmpty()) && dq.peek() <=i-b)
dq.removeFirst();
while ((!dq.isEmpty()) && a[i]>=a[dq.peekLast()])
dq.removeLast();
dq.addLast(i);
mul=((mul%1000000007)*(a[dq.peek()]%1000000007))%1000000007;
}
System.out.println(mul%1000000007);
}
static class Reader
{
final private int BUFFER_SIZE = 1 << 16;
private DataInputStream din;
private byte[] buffer;
private int bufferPointer, bytesRead;
public Reader()
{
din = new DataInputStream(System.in);
buffer = new byte[BUFFER_SIZE];
bufferPointer = bytesRead = 0;
}
public Reader(String file_name) throws IOException
{
din = new DataInputStream(new FileInputStream(file_name));
buffer = new byte[BUFFER_SIZE];
bufferPointer = bytesRead = 0;
}
public String readLine() throws IOException
{
byte[] buf = new byte[64];
int cnt = 0, c;
while ((c = read()) != -1)
{
if (c == '\n')
break;
buf[cnt++] = (byte) c;
}
return new String(buf, 0, cnt);
}
public int nextInt() throws IOException
{
int ret = 0;
byte c = read();
while (c <= ' ')
c = read();
boolean neg = (c == '-');
if (neg)
c = read();
do
{
ret = ret * 10 + c - '0';
} while ((c = read()) >= '0' && c <= '9');
if (neg)
return -ret;
return ret;
}
public long nextLong() throws IOException
{
long ret = 0;
byte c = read();
while (c <= ' ')
c = read();
boolean neg = (c == '-');
if (neg)
c = read();
do {
ret = ret * 10 + c - '0';
}
while ((c = read()) >= '0' && c <= '9');
if (neg)
return -ret;
return ret;
}
public double nextDouble() throws IOException
{
double ret = 0, div = 1;
byte c = read();
while (c <= ' ')
c = read();
boolean neg = (c == '-');
if (neg)
c = read();
do {
ret = ret * 10 + c - '0';
}
while ((c = read()) >= '0' && c <= '9');
if (c == '.')
{
while ((c = read()) >= '0' && c <= '9')
{
ret += (c - '0') / (div *= 10);
}
}
if (neg)
return -ret;
return ret;
}
private void fillBuffer() throws IOException
{
bytesRead = din.read(buffer, bufferPointer = 0, BUFFER_SIZE);
if (bytesRead == -1)
buffer[0] = -1;
}
private byte read() throws IOException
{
if (bufferPointer == bytesRead)
fillBuffer();
return buffer[bufferPointer++];
}
public void close() throws IOException
{
if (din == null)
return;
din.close();
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array, A of N integers. Find the product of maximum values for every subarray of size K. Print the answer modulo 10<sup>9</sup>+7.
A subarray is any contiguous sequence of elements in an array.The first line contains two integers N and K, denoting the size of the array and the size of the subarray respectively.
The next line contains N integers denoting the elements of the array.
<b>Constarints</b>
1 <= K <= N <= 1000000
1 <= A[i] <= 1000000Print a single integer denoting the product of maximums for every subarray of size K modulo 1000000007Sample Input 1:
6 4
1 5 2 3 6 4
Sample Output 1:
180
<b>Explanation:</b>
For subarray [1, 5, 2, 3], maximum = 5
For subarray [5, 2, 3, 6], maximum = 6
For subarray [2, 3, 6, 4], maximum = 6
Therefore, ans = 5*6*6 = 180, I have written this Solution Code: from collections import deque
deq=deque()
n,k=list(map(int,input().split()))
array=list(map(int,input().split()))
for i in range(k):
while(deq and array[deq[-1]]<=array[i]):
deq.pop()
deq.append(i)
ans=1
for j in range(k,n):
ans=ans*array[deq[0]]
ans=(ans)%1000000007
while(deq and deq[0]<=j-k):
deq.popleft()
while(deq and array[deq[-1]]<=array[j]):
deq.pop()
deq.append(j)
ans=ans*array[deq[0]]
ans=(ans)%1000000007
print(ans), 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. Find the product of maximum values for every subarray of size K. Print the answer modulo 10<sup>9</sup>+7.
A subarray is any contiguous sequence of elements in an array.The first line contains two integers N and K, denoting the size of the array and the size of the subarray respectively.
The next line contains N integers denoting the elements of the array.
<b>Constarints</b>
1 <= K <= N <= 1000000
1 <= A[i] <= 1000000Print a single integer denoting the product of maximums for every subarray of size K modulo 1000000007Sample Input 1:
6 4
1 5 2 3 6 4
Sample Output 1:
180
<b>Explanation:</b>
For subarray [1, 5, 2, 3], maximum = 5
For subarray [5, 2, 3, 6], maximum = 6
For subarray [2, 3, 6, 4], maximum = 6
Therefore, ans = 5*6*6 = 180, I have written this Solution Code: #include<bits/stdc++.h>
#define int long long
#define ll long long
#define pb push_back
#define endl '\n'
#define pii pair<int,int>
#define vi vector<int>
#define all(a) (a).begin(),(a).end()
#define F first
#define S second
#define sz(x) (int)x.size()
#define hell 1000000007
#define rep(i,a,b) for(int i=a;i<b;i++)
#define dep(i,a,b) for(int i=a;i>=b;i--)
#define lbnd lower_bound
#define ubnd upper_bound
#define bs binary_search
#define mp make_pair
using namespace std;
const int N = 1e6 + 5;
const int mod = 1e9 + 7;
const int inf = 1e9 + 9;
int a[N];
void solve(){
int n, k;
cin >> n >> k;
for(int i = 1; i <= n; i++)
cin >> a[i];
deque<int> q;
for(int i = 1; i <= k; i++){
while(!q.empty() && a[q.back()] <= a[i])
q.pop_back();
q.push_back(i);
}
int ans = a[q.front()];
for(int i = k+1; i <= n; i++){
if(q.front() == i-k)
q.pop_front();
while(!q.empty() && a[q.back()] <= a[i])
q.pop_back();
q.push_back(i);
ans = (ans*a[q.front()]) % mod;
}
cout << ans;
}
void testcases(){
int tt = 1;
//cin >> tt;
while(tt--){
solve();
}
}
signed main()
{
ios_base::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
clock_t start = clock();
testcases();
cerr << (double)(clock() - start)*1000/CLOCKS_PER_SEC << " ms: ";
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a circular linked list consisting of N nodes and an integer K, your task is to add the integer K at the end of the list.
<b>Note:
Sample Input and Output just show how a linked list will look depending on the questions. Do not copy-paste as it is in custom input</b><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>Insertion()</b> that takes head node of circular linked list and the integer K as parameter.
Constraints:
1 <=N <= 1000
1 <= Node.data, K<= 1000Return the head node of the modified circular linked list.Sample Input 1:-
3
1- >2- >3
4
Sample Output 1:-
1- >2- >3- >4
Sample Input 2:-
3
1- >3- >2
1
Sample Output 2:-
1- >3- >2- >1, I have written this Solution Code: public static Node Insertion(Node head, int K){
Node node=head;
while ( node.next != head)
{node = node.next; }
Node temp = new Node(K);
node.next=temp;
temp.next=head;
return head;}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer N, your task is to check whether the given number is prime or notThe input contains a single integer N.
Constraints:-
1 <= N <= 100000000000Print "YES" If the given number is prime else print "NO".Sample Input:-
2
Sample Output:-
YES
Sample Input:-
4
Sample Output:-
NO, 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);
long n = sc.nextLong();
int p=(int)Math.sqrt(n);
for(int i=2;i<=p;i++){
if(n%i==0){System.out.print("NO");return;}
}
System.out.print("YES");
}
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer N, your task is to check whether the given number is prime or notThe input contains a single integer N.
Constraints:-
1 <= N <= 100000000000Print "YES" If the given number is prime else print "NO".Sample Input:-
2
Sample Output:-
YES
Sample Input:-
4
Sample Output:-
NO, I have written this Solution Code: import math
def isprime(A):
if A == 1:
return False
sqrt = int(math.sqrt(A))
for i in range(2,sqrt+1):
if A%i == 0:
return False
return True
inp = int(input())
if isprime(inp):
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 N, your task is to check whether the given number is prime or notThe input contains a single integer N.
Constraints:-
1 <= N <= 100000000000Print "YES" If the given number is prime else print "NO".Sample Input:-
2
Sample Output:-
YES
Sample Input:-
4
Sample Output:-
NO, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main(){
long n;
cin>>n;
long x = sqrt(n);
for(int i=2;i<=x;i++){
if(n%i==0){cout<<"NO";return 0;}
}
cout<<"YES";
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array <b>arr[]</b> of <b>N</b> elements. The task is to <b>count maximum number of distinct smaller elements on right side of any array element</b>. For example, in {10, 6, 9, 7, 20, 19, 21, 18, 17, 16}, the result is 4.
<b>Note</b> that 20 has a maximum of 4 smaller elements on the right side. Other elements have less count, for example 10 has 3 smaller elements on the right side.First line of input contains number of testcases T. For each testcase, first line of input contains number of elements in the array and next line contains array elements.
<b>Constraints:</b>
1 <= T <= 100
1 <= N <= 10^4
1 <= arr[i] <= 10^6
<b>Sum of N over all testcases does not exceed 10^5</b>For each testcase, print the maximum count of smaller elements on right side. Input:
4
10
10 6 9 7 20 19 21 18 17 16
5
5 4 3 2 1
5
1 2 3 4 5
5
1 2 3 2 1
Output:
4
4
0
2
Explanation:
Testcase 1: Number of smaller elements on the right side of every element (from left to right) in the array are 3 0 1 0 4 3 3 2 1 and 0 respectively. The maximum of all these counts is 4.
Testcase 2: Number of smaller elements on the right side of every element (from left to right) in the array are 4 3 2 1 and 0 respectively. The maximum of all these counts is 4., I have written this Solution Code: import java.util.*;
import java.io.*;
class Main{
public static void main(String args[])throws IOException{
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
int T=Integer.parseInt(br.readLine());
while(T-->0){
int N=Integer.parseInt(br.readLine());
String s[]=br.readLine().split(" ");
int [] arr=new int[N];
for(int i=0;i<N;i++){
arr[i]=Integer.parseInt(s[i]);
}
System.out.println(maxcountonrightside(arr,N));
}
}
public static int maxcountonrightside(int [] arr,int N){
TreeSet<Integer>tree=new TreeSet<>();
int count=0;
for(int i=N-1;i>=0;i--){
tree.add(arr[i]);
count=Integer.max(tree.headSet(arr[i]).size(),count);
}
return count;
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array <b>arr[]</b> of <b>N</b> elements. The task is to <b>count maximum number of distinct smaller elements on right side of any array element</b>. For example, in {10, 6, 9, 7, 20, 19, 21, 18, 17, 16}, the result is 4.
<b>Note</b> that 20 has a maximum of 4 smaller elements on the right side. Other elements have less count, for example 10 has 3 smaller elements on the right side.First line of input contains number of testcases T. For each testcase, first line of input contains number of elements in the array and next line contains array elements.
<b>Constraints:</b>
1 <= T <= 100
1 <= N <= 10^4
1 <= arr[i] <= 10^6
<b>Sum of N over all testcases does not exceed 10^5</b>For each testcase, print the maximum count of smaller elements on right side. Input:
4
10
10 6 9 7 20 19 21 18 17 16
5
5 4 3 2 1
5
1 2 3 4 5
5
1 2 3 2 1
Output:
4
4
0
2
Explanation:
Testcase 1: Number of smaller elements on the right side of every element (from left to right) in the array are 3 0 1 0 4 3 3 2 1 and 0 respectively. The maximum of all these counts is 4.
Testcase 2: Number of smaller elements on the right side of every element (from left to right) in the array are 4 3 2 1 and 0 respectively. The maximum of all these counts is 4., I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
void countSmallerRight(int A[], int len)
{
set<int> s;
int countSmaller[len];
for (int i = len - 1; i >= 0; i--) {
s.insert(A[i]);
auto it = s.lower_bound(A[i]);
countSmaller[i] = distance(s.begin(), it);
}
int x=0;
for (int i = 0; i < len; i++) {
x=max(x, countSmaller[i]); }
cout<<x<<endl;
}
// Driver code
int main()
{
int t;
cin>>t;
while(t--){
int n;
cin>>n;
int a[n];
for(int i=0;i<n;i++){cin>>a[i];}
countSmallerRight(a, n);}
return 0;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a positive integer <b>N</b> and <b>Q</b> queries. For every query, you will be given <b>start</b> and <b>end</b> point and a positive integer <b>K</b> which is going to be added to the numbers in the range(from start to end) as per query. Once you are done with all the queries, you need to print the maximum number obtained after updation.
Note:-Initially the array is-<b> 1,2,3,....N </b>First line of input contains number of testcases T. For each testcase, first line contains N and Q( number of queries). Q lines after this will contain three integers each line has start, end and K, where start is the starting index of the range, end is the ending index of the range and K is the value to add with the elements in the range.
<b>Note:</b> Array is 1-based index
<b>Constraints:</b>
1 <= T <= 100
1 <= N, K <= 10^5
1 <= Q <= 10^5
1 <= start <= end <= N
Sum of N, Q for every test case is less than or equal to 10^5For each testcase, you need to print the maximum.Input:
1
5 3
1 2 5
2 5 10
3 4 5
Output:
19
Explanation:
Testcase 1: After the queries, we have elements added in the given ranges, so updated numbers from 1 to 5 will be as 6, 17, 18, 19, 15. Thus maximum after updation comes as19., I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void abc(int[] a,int s,int e,int k)
{
for(int i=s;i<e;i++)
a[i]+=k;
}
public static void main (String[] args) throws IOException
{
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
int t=Integer.parseInt(br.readLine());
for(int z=0;z<t;z++)
{
String[] s=br.readLine().split(" ");
int n=Integer.parseInt(s[0]);
int q=Integer.parseInt(s[1]);
int[] arr=new int[n];
for(int i=0;i<n;i++)
arr[i]=i+1;
for(int i=0;i<q;i++)
{
String[] ss=br.readLine().split(" ");
int st=Integer.parseInt(ss[0]);
int en=Integer.parseInt(ss[1]);
int k=Integer.parseInt(ss[2]);
abc(arr,st-1,en,k);
}
int max=-9999999;
for(int i=0;i<n;i++)
{
if(arr[i]>max)
max=arr[i];
}
System.out.println(max);
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a positive integer <b>N</b> and <b>Q</b> queries. For every query, you will be given <b>start</b> and <b>end</b> point and a positive integer <b>K</b> which is going to be added to the numbers in the range(from start to end) as per query. Once you are done with all the queries, you need to print the maximum number obtained after updation.
Note:-Initially the array is-<b> 1,2,3,....N </b>First line of input contains number of testcases T. For each testcase, first line contains N and Q( number of queries). Q lines after this will contain three integers each line has start, end and K, where start is the starting index of the range, end is the ending index of the range and K is the value to add with the elements in the range.
<b>Note:</b> Array is 1-based index
<b>Constraints:</b>
1 <= T <= 100
1 <= N, K <= 10^5
1 <= Q <= 10^5
1 <= start <= end <= N
Sum of N, Q for every test case is less than or equal to 10^5For each testcase, you need to print the maximum.Input:
1
5 3
1 2 5
2 5 10
3 4 5
Output:
19
Explanation:
Testcase 1: After the queries, we have elements added in the given ranges, so updated numbers from 1 to 5 will be as 6, 17, 18, 19, 15. Thus maximum after updation comes as19., I have written this Solution Code: tc=int(input())
while(tc>0):
n,q=[int(i) for i in input().split()]
li=[0]*(n+1);
while(q>0):
a,b,k=[int(i) for i in input().split()]
li[a-1]+=k
li[b]-=k
q-=1
m=0
for i in range(1,n):
li[i]+=li[i-1]
for i in range(0,n):
if(li[i]+i+1>m):
m=li[i]+i+1
print(m)
tc-=1, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a positive integer <b>N</b> and <b>Q</b> queries. For every query, you will be given <b>start</b> and <b>end</b> point and a positive integer <b>K</b> which is going to be added to the numbers in the range(from start to end) as per query. Once you are done with all the queries, you need to print the maximum number obtained after updation.
Note:-Initially the array is-<b> 1,2,3,....N </b>First line of input contains number of testcases T. For each testcase, first line contains N and Q( number of queries). Q lines after this will contain three integers each line has start, end and K, where start is the starting index of the range, end is the ending index of the range and K is the value to add with the elements in the range.
<b>Note:</b> Array is 1-based index
<b>Constraints:</b>
1 <= T <= 100
1 <= N, K <= 10^5
1 <= Q <= 10^5
1 <= start <= end <= N
Sum of N, Q for every test case is less than or equal to 10^5For each testcase, you need to print the maximum.Input:
1
5 3
1 2 5
2 5 10
3 4 5
Output:
19
Explanation:
Testcase 1: After the queries, we have elements added in the given ranges, so updated numbers from 1 to 5 will be as 6, 17, 18, 19, 15. Thus maximum after updation comes as19., I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main(){
int t;
cin>>t;
while(t--){
int n,k;
cin>>n>>k;
int a[n+1];
for(int i=0;i<=n;i++){
a[i]=0;
}
int x,y,p;
while(k--){
cin>>x>>y>>p;
x--;
a[x]+=p;
a[y]-=p;
}
long long sum=0,ans=0;
for(int i=0;i<n;i++){
sum+=a[i];
ans=max(sum+i+1,ans);
}
cout<<ans<<endl;
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a integer N, your task is to calculate the number of divisors in factorial of N.Input contains a single integer depicting value of N.
Constraints:-
1 < = N < = 100Print the number of divisors in N!.Sample Input:-
3
Sample Output:-
4
Sample Input:-
5
Sample Output:-
16, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
static Vector<Integer> allPrimes=new Vector<Integer>();
public static void main (String[] args) throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int n = Integer.parseInt(br.readLine());
System.out.println(factorialDivisors(n));
}
static void sieve(int n){
boolean []prime=new boolean[n+1];
for(int i=0;i<=n;i++)
prime[i]=true;
for (int p=2; p*p<=n; p++)
{
if (prime[p] == true)
{
for (int i=p*2; i<=n; i += p)
prime[i] = false;
}
}
for (int p=2; p<=n; p++)
if (prime[p])
allPrimes.add(p);
}
static long factorialDivisors(int n)
{
sieve(n);
long result = 1;
for (int i=0; i < allPrimes.size(); i++)
{
long p = allPrimes.get(i);
long exp = 0;
while (p <= n)
{
exp = exp + (n/p);
p = p*allPrimes.get(i);
}
result = result*(exp+1);
}
return result;
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a integer N, your task is to calculate the number of divisors in factorial of N.Input contains a single integer depicting value of N.
Constraints:-
1 < = N < = 100Print the number of divisors in N!.Sample Input:-
3
Sample Output:-
4
Sample Input:-
5
Sample Output:-
16, I have written this Solution Code: n=int(input())
prime=[True for i in range(n+1)]
p=2
while(p*p<=n):
if prime[p]:
for i in range(p*p,n+1,p):
prime[i]=False
p+=1
ans=1
for i in range(2,n+1):
if prime[i]:
x=n
e=0
while x>0:
x=x//i
e+=x
ans*=(e+1)
print(ans), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a integer N, your task is to calculate the number of divisors in factorial of N.Input contains a single integer depicting value of N.
Constraints:-
1 < = N < = 100Print the number of divisors in N!.Sample Input:-
3
Sample Output:-
4
Sample Input:-
5
Sample Output:-
16, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define int long long
// Sieve of Eratosthenes to mark all prime number
// in array prime as 1
void sieve(int n, bool prime[])
{
// Initialize all numbers as prime
for (int i=1; i<=n; i++)
prime[i] = 1;
// Mark composites
prime[1] = 0;
for (int i=2; i*i<=n; i++)
{
if (prime[i])
{
for (int j=i*i; j<=n; j += i)
prime[j] = 0;
}
}
}
// Returns the highest exponent of p in n!
int expFactor(int n, int p)
{
int x = p;
int exponent = 0;
while ((n/x) > 0)
{
exponent += n/x;
x *= p;
}
return exponent;
}
// Returns the no of factors in n!
int countFactors(int n)
{
// ans stores the no of factors in n!
int ans = 1;
// Find all primes upto n
bool prime[n+1];
sieve(n, prime);
// Multiply exponent (of primes) added with 1
for (int p=1; p<=n; p++)
{
// if p is a prime then p is also a
// prime factor of n!
if (prime[p]==1)
ans *= (expFactor(n, p) + 1);
}
return ans;
}
// Driver code
signed main()
{
int t ;
t=1;
while(t--){
int n ;
cin>>n;
cout<<(countFactors(n))<<endl;}
return 0;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: For a given string S and a pattern P, your task is to print all the indexes where the pattern occur in the given string S.First line of input contains the string P, the next line of input contains the string S.
Constraints:-
1 <= |S| <= 500000
1 <= |P| <= 100000
Note:- String will contain only lowercase English lettersPrint all the indexes in ascending order separated by spaces where the pattern occurs in the string S. if none indexes are present then print -1.Sample Input:-
na
banana
Sample Output:-
2 4
Sample Input:-
Newton
School
Sample Output:-
-1, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) {
Scanner sc = new Scanner(System.in);
String a = sc.next();
String b = sc.next();
int alen = a.length();
int blen = b.length();
int len=0;
if(alen>blen)
System.out.println("-1");
else {
StringBuilder sb = new StringBuilder("");
boolean flag = false;
int s=0;
while(true){
int value = b.indexOf(a,s);
if(value!=-1)
{
sb.append(b.indexOf(a,s)).append(" ");
s=value+1;
flag=true;
}else {
break;
}
}
if(flag)
System.out.println(sb);
else
System.out.println("-1");
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: For a given string S and a pattern P, your task is to print all the indexes where the pattern occur in the given string S.First line of input contains the string P, the next line of input contains the string S.
Constraints:-
1 <= |S| <= 500000
1 <= |P| <= 100000
Note:- String will contain only lowercase English lettersPrint all the indexes in ascending order separated by spaces where the pattern occurs in the string S. if none indexes are present then print -1.Sample Input:-
na
banana
Sample Output:-
2 4
Sample Input:-
Newton
School
Sample Output:-
-1, I have written this Solution Code:
#include <bits/stdc++.h>
using namespace std;
#define MEM(a, b) memset(a, (b), sizeof(a))
#define FOREACH(it, l) for (auto it = l.begin(); it != l.end(); it++)
#define IN(A, B, C) assert( B <= A && A <= C)
#define MP make_pair
#define FOR(i,a) for(int i=0;i<a;i++)
#define FOR1(i,j,a) for(int i=j;i<a;i++)
#define EB emplace_back
#define INF (int)1e9
#define EPS 1e-9
#define PI 3.1415926535897932384626433832795
#define max1 1000001
#define MOD 100000000
#define read(type) readInt<type>()
#define out(x) cout<<x<<'\n'
#define out1(x) cout<<x<<" "
#define END cout<<'\n'
#define int long long
#define sz(v) ((int)(v).size())
#define all(v) (v).begin(), (v).end()
void fast(){
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
}
void solve(){
string p,s;
cin>>p>>s;
vector<int> ans;
int found = s.find(p);
while (found != string::npos) {
ans.EB(found);
found = s.find(p, found + 1);
}
if(ans.size()==0){out(-1);return;}
FOR(i,ans.size()){
out1(ans[i]);
}
}
signed main(){
fast();
solve();
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: The username field in `PROFILE` must have at least 3 characters. Create a table `PROFILE` having this constraint. Fields are (USERNAME VARCHAR (24), FULL_NAME (72), HEADLINE VARCHAR (72)).
( USE ONLY UPPERCASE LETTERS FOR CODE )
<schema>[{'name': 'PROFILE', 'columns': [{'name': 'USERNAME', 'type': 'VARCHAR (24)'}, {'name': 'FULL_NAME', 'type': 'VARCHAR (72)'}, {'name': 'HEADLINE', 'type': 'VARCHAR (72)'}]}]</schema>nannannan, I have written this Solution Code: CREATE TABLE PROFILE(
USERNAME VARCHAR(24) CHECK (LENGTH(USERNAME) >= 3) ,
FULL_NAME VARCHAR(72),
HEADLINE VARCHAR(72)
);, In this Programming Language: SQL, Now tell me if this Code is compilable or not? | Compilable |
For this Question: For a given number N, Consider a table of N X N in which elements at the intersection of ith row and jth column are the product of i and j, i. e i x j.
Also given a positive integer X. Your task is to count the number of elements in the table that contain the integer X.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>OccurenceOfX()</b> that takes the integer N and the integer X as parameter.
Constraints:-
1 <= N <= 10^5
1 <= X <= 10^9Return the count of X.Sample Input:-
5 5
Sample Output:-
2
Explanation:-
table :-
1 2 3 4 5
2 4 6 8 10
3 6 9 12 15
4 8 12 16 20
5 10 15 20 25
Count of occurrence of X is :- 2
Sample Input:-
10 13
Sample Output:-
0, I have written this Solution Code: def OccurenceOfX(N,X):
cnt=0
for i in range(1, N+1):
if(X%i==0 and X/i<=N):
cnt=cnt+1
return cnt, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: For a given number N, Consider a table of N X N in which elements at the intersection of ith row and jth column are the product of i and j, i. e i x j.
Also given a positive integer X. Your task is to count the number of elements in the table that contain the integer X.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>OccurenceOfX()</b> that takes the integer N and the integer X as parameter.
Constraints:-
1 <= N <= 10^5
1 <= X <= 10^9Return the count of X.Sample Input:-
5 5
Sample Output:-
2
Explanation:-
table :-
1 2 3 4 5
2 4 6 8 10
3 6 9 12 15
4 8 12 16 20
5 10 15 20 25
Count of occurrence of X is :- 2
Sample Input:-
10 13
Sample Output:-
0, I have written this Solution Code:
int OccurenceOfX(int N,long X){
int cnt=0,i;
for( i=1;i<=N;i++){
if(X%i==0 && X/i<=N){cnt++;}}
return cnt;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: For a given number N, Consider a table of N X N in which elements at the intersection of ith row and jth column are the product of i and j, i. e i x j.
Also given a positive integer X. Your task is to count the number of elements in the table that contain the integer X.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>OccurenceOfX()</b> that takes the integer N and the integer X as parameter.
Constraints:-
1 <= N <= 10^5
1 <= X <= 10^9Return the count of X.Sample Input:-
5 5
Sample Output:-
2
Explanation:-
table :-
1 2 3 4 5
2 4 6 8 10
3 6 9 12 15
4 8 12 16 20
5 10 15 20 25
Count of occurrence of X is :- 2
Sample Input:-
10 13
Sample Output:-
0, I have written this Solution Code:
int OccurenceOfX(int N,long X){
int cnt=0,i;
for( i=1;i<=N;i++){
if(X%i==0 && X/i<=N){cnt++;}}
return cnt;
}
int main()
{
, In this Programming Language: C, Now tell me if this Code is compilable or not? | Compilable |
For this Question: For a given number N, Consider a table of N X N in which elements at the intersection of ith row and jth column are the product of i and j, i. e i x j.
Also given a positive integer X. Your task is to count the number of elements in the table that contain the integer X.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>OccurenceOfX()</b> that takes the integer N and the integer X as parameter.
Constraints:-
1 <= N <= 10^5
1 <= X <= 10^9Return the count of X.Sample Input:-
5 5
Sample Output:-
2
Explanation:-
table :-
1 2 3 4 5
2 4 6 8 10
3 6 9 12 15
4 8 12 16 20
5 10 15 20 25
Count of occurrence of X is :- 2
Sample Input:-
10 13
Sample Output:-
0, I have written this Solution Code: public static int OccurenceOfX(int N,int X){
int cnt=0,i;
for( i=1;i<=N;i++){
if(X%i==0 && X/i<=N){cnt++;}}
return cnt;
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of positive integers. The task is to find inversion count of array.
Inversion Count : For an array, inversion count indicates how far (or close) the array is from being sorted. If array is already sorted then inversion count is 0. If array is sorted in reverse order that inversion count is the maximum.
Formally, two elements a[i] and a[j] form an inversion if a[i] > a[j] and i < j.
Asked in Adobe, Amazon, Microsoft.The first line of each test case is N, the size of the array. The second line of each test case contains N elements.
Constraints:-
1 ≤ N ≤ 10^5
1 ≤ a[i] ≤ 10^5Print the inversion count of array.Sample Input:
5
2 4 1 3 5
Sample Output:
3
Explanation:
Testcase 1: The sequence 2, 4, 1, 3, 5 has three inversions (2, 1), (4, 1), (4, 3)., I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws IOException {
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
int n=Integer.parseInt(br.readLine());
String input[]=br.readLine().split("\\s");
int a[]=new int[n];
for(int i=0;i<n;i++)
{
a[i]=Integer.parseInt(input[i]);
}
System.out.print(implementMergeSort(a,0,n-1));
}
public static long implementMergeSort(int arr[], int start, int end)
{
long count=0;
if(start<end)
{
int mid=start+(end-start)/2;
count +=implementMergeSort(arr,start,mid);
count +=implementMergeSort(arr,mid+1,end);
count +=merge(arr,start,end,mid);
}
return count;
}
public static long merge(int []a,int start,int end,int mid)
{
int i=start;
int j=mid+1;
int k=0;
int len=end-start+1;
int c[]=new int[len];
long inv_count=0;
while(i<=mid && j<=end)
{
if(a[i]<=a[j])
{
c[k++]=a[i];
i++;
}
else
{
c[k++]=a[j];
j++;
inv_count +=(mid-i)+1;
}
}
while(i<=mid)
{
c[k++]=a[i++];
}
while(j<=end)
{
c[k++]=a[j++];
}
for(int l=0;l<len;l++)
a[start+l]=c[l];
return inv_count;
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of positive integers. The task is to find inversion count of array.
Inversion Count : For an array, inversion count indicates how far (or close) the array is from being sorted. If array is already sorted then inversion count is 0. If array is sorted in reverse order that inversion count is the maximum.
Formally, two elements a[i] and a[j] form an inversion if a[i] > a[j] and i < j.
Asked in Adobe, Amazon, Microsoft.The first line of each test case is N, the size of the array. The second line of each test case contains N elements.
Constraints:-
1 ≤ N ≤ 10^5
1 ≤ a[i] ≤ 10^5Print the inversion count of array.Sample Input:
5
2 4 1 3 5
Sample Output:
3
Explanation:
Testcase 1: The sequence 2, 4, 1, 3, 5 has three inversions (2, 1), (4, 1), (4, 3)., I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define mod 1000000007
#define int long long
long long _mergeSort(int arr[], int temp[], int left, int right);
long long merge(int arr[], int temp[], int left, int mid, int right);
/* This function sorts the input array and returns the
number of inversions in the array */
long long mergeSort(int arr[], int array_size)
{
int temp[array_size];
return _mergeSort(arr, temp, 0, array_size - 1);
}
/* An auxiliary recursive function that sorts the input array and
returns the number of inversions in the array. */
long long _mergeSort(int arr[], int temp[], int left, int right)
{
int mid, inv_count = 0;
if (right > left) {
/* Divide the array into two parts and
call _mergeSortAndCountInv()
for each of the parts */
mid = (right + left) / 2;
/* Inversion count will be sum of
inversions in left-part, right-part
and number of inversions in merging */
inv_count += _mergeSort(arr, temp, left, mid);
inv_count += _mergeSort(arr, temp, mid + 1, right);
/*Merge the two parts*/
inv_count += merge(arr, temp, left, mid + 1, right);
}
return inv_count;
}
/* This funt merges two sorted arrays
and returns inversion count in the arrays.*/
long long merge(int arr[], int temp[], int left,
int mid, int right)
{
int i, j, k;
long long inv_count = 0;
i = left; /* i is index for left subarray*/
j = mid; /* j is index for right subarray*/
k = left; /* k is index for resultant merged subarray*/
while ((i <= mid - 1) && (j <= right)) {
if (arr[i] <= arr[j]) {
temp[k++] = arr[i++];
}
else {
temp[k++] = arr[j++];
/* this is tricky -- see above
explanation/diagram for merge()*/
inv_count = inv_count + (mid - i);
}
}
/* Copy the remaining elements of left subarray
(if there are any) to temp*/
while (i <= mid - 1)
temp[k++] = arr[i++];
/* Copy the remaining elements of right subarray
(if there are any) to temp*/
while (j <= right)
temp[k++] = arr[j++];
/*Copy back the merged elements to original array*/
for (i = left; i <= right; i++)
arr[i] = temp[i];
return inv_count;
}
signed main()
{
int n;
cin>>n;
int a[n];
unordered_map<int,int> m;
for(int i=0;i<n;i++){
cin>>a[i];
if(m.find(a[i])==m.end()){
m[a[i]]=i;
}
}
cout<<mergeSort(a,n);
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Bob is given an integer a. He wants to find the value of a + a^2 + a^3.The input consists of an integer a.
<b>Constraints</b>
1≤a≤10
a is an integer.Print the answer.<b>Sample Input 1</b>
2
<b>Sample Output 1</b>
14
<b>Sample Input 2</b>
10
<b>Sample Output 2</b>
1110, I have written this Solution Code: #include <iostream>
using namespace std;
int main(){
int a;
cin >> a;
cout << a + a * a + a * a * a << endl;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: The `POST` table should contain a field `ID` that can be used as the primary key. So make a table post with id as primary key ( ID INT, USERNAME VARCHAR(24), POST_TITLE VARCHAR(72), POST_DESCRIPTION TEXT, DATETIME_CREATED DATETIME, NUMBER_OF_LIKES INT, PHOTO BLOB )
( USE ONLY UPPERCASE LETTERS FOR CODE)
<schema>[{'name': 'POST', 'columns': [{'name': 'ID', 'type': 'INT'}, {'name': 'USERname', 'type': 'VARCHAR(24)'}, {'name': 'POST_TITLE', 'type': 'VARCHAR (72)'}, {'name': 'POST_DESCRIPTION', 'type': 'TEXT'}, {'name': 'DATETIME_CREATED', 'type': 'DATETIME'}, {'name': 'NUMBER_OF_LIKES', 'type': 'INT'}, {'name': 'PHOTO', 'type': 'BLOB'}]}]</schema>nannannan, I have written this Solution Code: CREATE TABLE POST(
ID INT PRIMARY KEY,
USERNAME VARCHAR(24),
POST_TITLE VARCHAR(72),
POST_DESCRIPTION TEXT,
DATETIME_CREATED DATETIME,
NUMBER_OF_LIKES INT,
PHOTO BLOB
);, In this Programming Language: SQL, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Alice has Q queries for you. If you can solve all her queries, she will invite you to participate in a fireworks display with her.
In each query, she gives you a positive integer N, and you have to find the number of positive integers M such that M × d(M) = N.
Here, d(M) refers to the sum of all digits of M. For example, d(1023) = 6.The first line consists of a single integer Q – the number of queries.
Then Q lines follow, each line containing a single integer N denoting a query.
<b> Constraints: </b>
1 ≤ Q ≤ 1000
1 ≤ N ≤ 10<sup>18</sup>Print Q lines, the i<sup>th</sup> line containing the answer to the i<sup>th</sup> query.Sample Input 1:
3
1
6
36
Sample Output 1:
1
0
2
Sample Explanation 1:
For the first query, the only possibility is 1.
For the third query, the only possibilities are 6 and 12., I have written this Solution Code: import java.io.*;
import java.util.*;
public class Main {
private static final int START_TEST_CASE = 1;
public static void solveCase(FastIO io, int testCase) {
final long N = io.nextLong();
int count = 0;
for (int i = 1; i <= 200; ++i) {
if (N % i == 0) {
long j = N / i;
if (digitSum(j) == i) {
++count;
}
}
}
io.println(count);
}
private static long digitSum(long x) {
long s = 0;
while (x > 0) {
s += x % 10;
x /= 10;
}
return s;
}
public static void solve(FastIO io) {
final int T = io.nextInt();
for (int t = 0; t < T; ++t) {
solveCase(io, START_TEST_CASE + t);
}
}
public static class FastIO {
private InputStream reader;
private PrintWriter writer;
private byte[] buf = new byte[1024];
private int curChar;
private int numChars;
public FastIO(InputStream r, OutputStream w) {
reader = r;
writer = new PrintWriter(new BufferedWriter(new OutputStreamWriter(w)));
}
public int read() {
if (numChars == -1)
throw new InputMismatchException();
if (curChar >= numChars) {
curChar = 0;
try {
numChars = reader.read(buf);
} catch (IOException e) {
throw new InputMismatchException();
}
if (numChars <= 0)
return -1;
}
return buf[curChar++];
}
public String nextLine() {
int c = read();
while (isSpaceChar(c))
c = read();
StringBuilder res = new StringBuilder();
do {
res.appendCodePoint(c);
c = read();
} while (!isEndOfLine(c));
return res.toString();
}
public String nextString() {
int c = read();
while (isSpaceChar(c))
c = read();
StringBuilder res = new StringBuilder();
do {
res.appendCodePoint(c);
c = read();
} while (!isSpaceChar(c));
return res.toString();
}
public long nextLong() {
int c = read();
while (isSpaceChar(c))
c = read();
int sgn = 1;
if (c == '-') {
sgn = -1;
c = read();
}
long res = 0;
do {
if (c < '0' || c > '9')
throw new InputMismatchException();
res *= 10;
res += c - '0';
c = read();
} while (!isSpaceChar(c));
return res * sgn;
}
public int nextInt() {
int c = read();
while (isSpaceChar(c))
c = read();
int sgn = 1;
if (c == '-') {
sgn = -1;
c = read();
}
int res = 0;
do {
if (c < '0' || c > '9')
throw new InputMismatchException();
res *= 10;
res += c - '0';
c = read();
} while (!isSpaceChar(c));
return res * sgn;
}
public double nextDouble() {
return Double.parseDouble(nextString());
}
public int[] nextIntArray(int n) {
return nextIntArray(n, 0);
}
public int[] nextIntArray(int n, int off) {
int[] arr = new int[n + off];
for (int i = 0; i < n; i++) {
arr[i + off] = nextInt();
}
return arr;
}
public long[] nextLongArray(int n) {
return nextLongArray(n, 0);
}
public long[] nextLongArray(int n, int off) {
long[] arr = new long[n + off];
for (int i = 0; i < n; i++) {
arr[i + off] = nextLong();
}
return arr;
}
private boolean isSpaceChar(int c) {
return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
}
private boolean isEndOfLine(int c) {
return c == '\n' || c == '\r' || c == -1;
}
public void print(Object... objects) {
for (int i = 0; i < objects.length; i++) {
if (i != 0) {
writer.print(' ');
}
writer.print(objects[i]);
}
}
public void println(Object... objects) {
print(objects);
writer.println();
}
public void printArray(int[] arr) {
for (int i = 0; i < arr.length; i++) {
if (i != 0) {
writer.print(' ');
}
writer.print(arr[i]);
}
}
public void printArray(long[] arr) {
for (int i = 0; i < arr.length; i++) {
if (i != 0) {
writer.print(' ');
}
writer.print(arr[i]);
}
}
public void printlnArray(int[] arr) {
printArray(arr);
writer.println();
}
public void printlnArray(long[] arr) {
printArray(arr);
writer.println();
}
public void printf(String format, Object... args) {
print(String.format(format, args));
}
public void flush() {
writer.flush();
}
}
private static class Solution implements Runnable {
@Override
public void run() {
FastIO io = new FastIO(System.in, System.out);
solve(io);
io.flush();
}
}
public static void main(String[] args) throws InterruptedException {
Thread t = new Thread(null, new Solution(), "Solution", 1 << 30);
t.start();
t.join();
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Alice has Q queries for you. If you can solve all her queries, she will invite you to participate in a fireworks display with her.
In each query, she gives you a positive integer N, and you have to find the number of positive integers M such that M × d(M) = N.
Here, d(M) refers to the sum of all digits of M. For example, d(1023) = 6.The first line consists of a single integer Q – the number of queries.
Then Q lines follow, each line containing a single integer N denoting a query.
<b> Constraints: </b>
1 ≤ Q ≤ 1000
1 ≤ N ≤ 10<sup>18</sup>Print Q lines, the i<sup>th</sup> line containing the answer to the i<sup>th</sup> query.Sample Input 1:
3
1
6
36
Sample Output 1:
1
0
2
Sample Explanation 1:
For the first query, the only possibility is 1.
For the third query, the only possibilities are 6 and 12., I have written this Solution Code:
q = int(input())
for _ in range(q):
n = int(input())
count = 0
for i in range(1,9*len(str(n))):
if not n % i:
dig = n//i
if sum(map(int,str(dig))) == i:
count += 1
print(count), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Alice has Q queries for you. If you can solve all her queries, she will invite you to participate in a fireworks display with her.
In each query, she gives you a positive integer N, and you have to find the number of positive integers M such that M × d(M) = N.
Here, d(M) refers to the sum of all digits of M. For example, d(1023) = 6.The first line consists of a single integer Q – the number of queries.
Then Q lines follow, each line containing a single integer N denoting a query.
<b> Constraints: </b>
1 ≤ Q ≤ 1000
1 ≤ N ≤ 10<sup>18</sup>Print Q lines, the i<sup>th</sup> line containing the answer to the i<sup>th</sup> query.Sample Input 1:
3
1
6
36
Sample Output 1:
1
0
2
Sample Explanation 1:
For the first query, the only possibility is 1.
For the third query, the only possibilities are 6 and 12., I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define int long long
// #pragma gcc optimize("ofast")
// #pragma gcc target("avx,avx2,fma")
#define all(x) (x).begin(), (x).end()
#define pb push_back
#define endl '\n'
#define fi first
#define se second
// const int mod = 1e9 + 7;
const int mod=998'244'353;
const long long INF = 2e18 + 10;
// const int INF=4e9+10;
#define readv(x, n) \
vector<int> x(n); \
for (auto &i : x) \
cin >> i;
template <typename t>
using v = vector<t>;
template <typename t>
using vv = vector<vector<t>>;
template <typename t>
using vvv = vector<vector<vector<t>>>;
typedef vector<int> vi;
typedef vector<double> vd;
typedef vector<vector<int>> vvi;
typedef vector<vector<vector<int>>> vvvi;
typedef vector<vector<vector<vector<int>>>> vvvvi;
typedef vector<vector<double>> vvd;
typedef pair<int, int> pii;
int multiply(int a, int b, int in_mod) { return (int)(1ll * a * b % in_mod); }
int mult_identity(int a) { return 1; }
const double pi = acosl(-1);
vector<vector<int> > multiply(vector<vector<int>> a, vector<vector<int>> b, int in_mod)
{
int n = a.size();
int l = b.size();
int m = b[0].size();
vector<vector<int> > result(n,vector<int>(n));
for(int i=0;i<n;i++)
{
for(int j=0;j<m;j++)
{
for(int k=0;k<l;k++)
{
result[i][j] = (result[i][j] + a[i][k]*b[k][j])%in_mod;
}
}
}
return result;
}
vector<vector<int>> operator%(vector<vector<int>> a, int in_mod)
{
for(auto &i:a)
for(auto &j:i)
j%=in_mod;
return a;
}
vector<vector<int>> mult_identity(vector<vector<int>> a)
{
int n=a.size();
vector<vector<int>> output(n, vector<int> (n));
for(int i=0;i<n;i++)
output[i][i]=1;
return output;
}
vector<int> mat_vector_product(vector<vector<int>> a, vector<int> b, int in_mod)
{
int n =a.size();
vector<int> output(n);
for(int i=0;i<n;i++)
{
for(int j=0;j<n;j++)
{
output[i]+=a[i][j]*b[j];
output[i]%=in_mod;
}
}
return output;
}
auto power(auto a, auto b, const int in_mod)
{
auto prod = mult_identity(a);
auto mult = a % in_mod;
while (b != 0)
{
if (b % 2)
{
prod = multiply(prod, mult, in_mod);
}
if(b/2)
mult = multiply(mult, mult, in_mod);
b /= 2;
}
return prod;
}
auto mod_inv(auto q, const int in_mod)
{
return power(q, in_mod - 2, in_mod);
}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
#define stp cout << fixed << setprecision(20);
int digit_sum(int x){
int sm = 0;
while(x){
sm += x%10;
x/= 10;
}
return sm;
}
void solv(){
int n;
cin>>n;
int cnt = 0;
for(int sm = 1;sm<= 200;sm++){
if(n %sm == 0){
if( digit_sum(n/sm) == sm){
cnt++;
}
}
}
cout<<cnt<<endl;
}
void solve()
{
int t = 1;
cin>>t;
for(int T=1;T<=t;T++)
{
// cout<<"Case #"<<T<<": ";
solv();
}
}
signed main()
{
// freopen("input.txt", "r", stdin);
// freopen("output.txt", "w", stdout);
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cerr.tie(NULL);
auto clk = clock();
// -------------------------------------Code starts here---------------------------------------------------------------------
signed t = 1;
// cin >> t;
for (signed test = 1; test <= t; test++)
{
// cout<<"Case #"<<test<<": ";
solve();
}
// -------------------------------------Code ends here------------------------------------------------------------------
clk = clock() - clk;
#ifndef ONLINE_JUDGE
cerr << fixed << setprecision(6) << "\nTime: " << ((float)clk) / CLOCKS_PER_SEC << "\n";
#endif
return 0;
}
/*
000100
1000100
1 0 -1 -2 -1 -2 -3
*/
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Andy loves threes and fours and only likes numbers which are formed by threes and fours. Andy is curious about the Nth positive integer that Andy loves.First and only line of input contains a single integer N.
Constraints :
1 <= N <= 1000000000000000Output should contain a single number, the Nth positive integer that Andy loves.Sample Input 1
1
Sample Output 1
3
Sample Input 2
3
Sample Output 2
33
Explanation:-
series is like 3, 4, 33, 34, 44, 333......, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) {
Scanner scan=new Scanner(System.in);
long n=scan.nextLong();
nthTerm(n);
}
public static void nthTerm(long n){
if(n==1 || n==2){
System.out.print(n-1+3);
return;
}
n--;
nthTerm(n/2);
System.out.print(n%2+3);
return;
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Andy loves threes and fours and only likes numbers which are formed by threes and fours. Andy is curious about the Nth positive integer that Andy loves.First and only line of input contains a single integer N.
Constraints :
1 <= N <= 1000000000000000Output should contain a single number, the Nth positive integer that Andy loves.Sample Input 1
1
Sample Output 1
3
Sample Input 2
3
Sample Output 2
33
Explanation:-
series is like 3, 4, 33, 34, 44, 333......, I have written this Solution Code: import math
N = int(input())
num_of_digits = int(math.log2(N+1))
nearest_number = pow(2,num_of_digits)-1
mask = bin(N-nearest_number)
string = ""
dic = {"0":"3","1":"4"}
for i in range(2,len(mask)):
string += dic[mask[i]]
print("3"*(num_of_digits-(len(mask)-2)),end="")
print(string), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Andy loves threes and fours and only likes numbers which are formed by threes and fours. Andy is curious about the Nth positive integer that Andy loves.First and only line of input contains a single integer N.
Constraints :
1 <= N <= 1000000000000000Output should contain a single number, the Nth positive integer that Andy loves.Sample Input 1
1
Sample Output 1
3
Sample Input 2
3
Sample Output 2
33
Explanation:-
series is like 3, 4, 33, 34, 44, 333......, I have written this Solution Code: #include <bits/stdc++.h>
// #define ll long long
using namespace std;
int main(){
long long n;
cin>>n;
long long p=2;
int q=1;
long long t=2;
while(p<n){
n=n-p;
p=p*t;
q++;
}
string s="";
int b[q];
for(int i=0;i<q;i++){
b[i]=3;
}
int x=log(n)+1;
int r=q-1;
n--;
while(n>0){
if(n&1){b[r]=4;}
n=n/2;
r--;
}
for(int i=0;i<q;i++){
cout<<b[i];
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number N. Write a program to the find number of diagonals possible in N sided convex polygon.<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>numberOfDiagonal()</b> that takes the integer N parameter.
Constraints:
1 <= T <= 100
1 <= N <= 10^3Return number of diagonals possible in N sided convex polygon.Sample Input:
3
3
5
6
Sample Output:
0
5
9
Explanation:
For test case 2: The number of diagonals of 5 sided polygon: 5., I have written this Solution Code: static int numberOfDiagonal(int N){
if(N<=3){return 0;}
return (N*(N-3))/2;
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number N. Write a program to the find number of diagonals possible in N sided convex polygon.<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>numberOfDiagonal()</b> that takes the integer N parameter.
Constraints:
1 <= T <= 100
1 <= N <= 10^3Return number of diagonals possible in N sided convex polygon.Sample Input:
3
3
5
6
Sample Output:
0
5
9
Explanation:
For test case 2: The number of diagonals of 5 sided polygon: 5., I have written this Solution Code:
int numberOfDiagonals(int n){
if(n<=3){return 0;}
return (n*(n-3))/2;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number N. Write a program to the find number of diagonals possible in N sided convex polygon.<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>numberOfDiagonal()</b> that takes the integer N parameter.
Constraints:
1 <= T <= 100
1 <= N <= 10^3Return number of diagonals possible in N sided convex polygon.Sample Input:
3
3
5
6
Sample Output:
0
5
9
Explanation:
For test case 2: The number of diagonals of 5 sided polygon: 5., I have written this Solution Code: def numberOfDiagonals(n):
if n <=3:
return 0
return (n*(n-3))//2
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number N. Write a program to the find number of diagonals possible in N sided convex polygon.<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>numberOfDiagonal()</b> that takes the integer N parameter.
Constraints:
1 <= T <= 100
1 <= N <= 10^3Return number of diagonals possible in N sided convex polygon.Sample Input:
3
3
5
6
Sample Output:
0
5
9
Explanation:
For test case 2: The number of diagonals of 5 sided polygon: 5., I have written this Solution Code:
int numberOfDiagonals(int n){
if(n<=3){return 0;}
return (n*(n-3))/2;
}
, In this Programming Language: C, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Assume you are an awesome parent and want to give your children some cookies. But, you should give each child at most one cookie.
Each child i has a greed factor greed[i], which is the minimum size of a cookie that the child will be content with; and each cookie j has a size sz[j]. If sz[j] >= greed[i], we can assign the cookie j to the child i, and the child i will be content. Your goal is to maximize the number of your content children and output the maximum number.There are two number n(number of childrens) and m(number of cookies) are given in first line of input.
in second line, n space seperated integers, greed of children are given in a row..
in third line, m space seperated integers, sizes of cookies are given in a row..
1 <= greed. length <= 3 * 104
0 <= sz. length <= 3 * 104
1 <= greed[i], sz[j] <= 231 - 1Your goal is to maximize the number of your content children and output the maximum number and print what to report in case of exceptions.Sample Input:
3 2
1 2 3
1 1
Sample Output:
1
Explanation: You have 3 children and 2 cookies. The greed factors of 3 children are 1, 2, 3.
And even though you have 2 cookies, since their size is both 1, you could only make the child whose greed factor is 1 content.
You need to output 1., I have written this Solution Code: [n,m] = [x for x in input().split()]
ch = [int(x) for x in input().split()]
co = [int(x) for x in input().split()]
count = 0
ch = sorted(ch)
for c in ch:
try:
avco = min([x for x in co if x>=c])
co.remove(avco)
count = count+1
except:
continue
print(count), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Assume you are an awesome parent and want to give your children some cookies. But, you should give each child at most one cookie.
Each child i has a greed factor greed[i], which is the minimum size of a cookie that the child will be content with; and each cookie j has a size sz[j]. If sz[j] >= greed[i], we can assign the cookie j to the child i, and the child i will be content. Your goal is to maximize the number of your content children and output the maximum number.There are two number n(number of childrens) and m(number of cookies) are given in first line of input.
in second line, n space seperated integers, greed of children are given in a row..
in third line, m space seperated integers, sizes of cookies are given in a row..
1 <= greed. length <= 3 * 104
0 <= sz. length <= 3 * 104
1 <= greed[i], sz[j] <= 231 - 1Your goal is to maximize the number of your content children and output the maximum number and print what to report in case of exceptions.Sample Input:
3 2
1 2 3
1 1
Sample Output:
1
Explanation: You have 3 children and 2 cookies. The greed factors of 3 children are 1, 2, 3.
And even though you have 2 cookies, since their size is both 1, you could only make the child whose greed factor is 1 content.
You need to output 1., I have written this Solution Code: import java.lang.reflect.Array;
import java.util.*;
class Main {
public static void main(String args[]) {
Scanner sc = new Scanner(System.in);
int n,m;
int[] v;
try{
n=sc.nextInt();
m=sc.nextInt();
v=new int[n];
for(int i=0;i<n;i++){
int a=sc.nextInt();
v[i]=a;
}
PriorityQueue<Integer> pq=new PriorityQueue<>();
for(int i=0;i<m;i++){
int a=sc.nextInt();
pq.add(a);
}
int ans=0;
for(int i=0;i<n;i++){
int val=v[i];
while(pq.size()>0&&pq.peek()<val){
pq.poll();
}
Iterator it = pq.iterator();
if(it.hasNext()){
ans++;
pq.poll();
}
}
System.out.println(ans);
}
catch(Exception e){
System.out.println(e);
}
}
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: We were trying to add a `COMMENT` table to store the user comments on various posts. But we noobs were unable to do it, can you check our query and help us in solving the error here:
Query: CREATE TABLE COMMENTS (
USERNAME VARCHAR(24),
COMMENT_TEXT TEXT,
POST_ID INT;
);
Error: near line 1: near ";": syntax error
Error: near line 5: near ")": syntax error
<schema>[{'name': 'COMMENTS', 'columns': [{'name': 'USERNAME', 'type': 'VARCHAR(24)'}, {'name': 'COMMENT_TEXT', 'type': 'TEXT'}, {'name': 'POST_ID', 'type': 'INT'}]}]</schema>nannanCREATE TABLE comments (
username varchar(24),
comment_text text,
post_id int;
);, I have written this Solution Code:
CREATE TABLE COMMENTS (
USERNAME VARCHAR(24),
COMMENT_TEXT TEXT,
POST_ID INT
);, In this Programming Language: SQL, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a sorted array arr[] of N integers and a number K is given. The task is to check if the element K is present in the array or not.
<b>Note:</b> Use binary search to solve the problemThe first line of input contains a number of test cases T. For each test case, the first line of input contains a number of elements in the array, and the number K is separated by space. The next line contains N elements.
<b>Constraints:</b>
1 ≤ T ≤ 10
1 ≤ N ≤ 10<sup>5</sup>
1 ≤ K ≤ 10<sup>9</sup>
1 ≤ arr[i] ≤ 10<sup>9</sup>
<b>Sum of N over all test cases doesn't exceed 10<sup>6</sup></b>If the element is present in the array print "1" else print "-1".Sample Input:
2
5 6
1 2 3 4 6
5 2
1 3 4 5 6
Sample Output:
1
-1, I have written this Solution Code: static int isPresent(long arr[], int n, long k)
{
int left = 0;
int right = n-1;
int res = -1;
while(left<=right){
int mid = (left+right)/2;
if(arr[mid] == k){
res = 1;
break;
}else if(arr[mid] < k){
left = mid + 1;
}else{
right = mid - 1;
}
}
return res;
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a sorted array arr[] of N integers and a number K is given. The task is to check if the element K is present in the array or not.
<b>Note:</b> Use binary search to solve the problemThe first line of input contains a number of test cases T. For each test case, the first line of input contains a number of elements in the array, and the number K is separated by space. The next line contains N elements.
<b>Constraints:</b>
1 ≤ T ≤ 10
1 ≤ N ≤ 10<sup>5</sup>
1 ≤ K ≤ 10<sup>9</sup>
1 ≤ arr[i] ≤ 10<sup>9</sup>
<b>Sum of N over all test cases doesn't exceed 10<sup>6</sup></b>If the element is present in the array print "1" else print "-1".Sample Input:
2
5 6
1 2 3 4 6
5 2
1 3 4 5 6
Sample Output:
1
-1, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main(){
int t;
cin>>t;
while(t--){
int n;
cin>>n;
unordered_map<long long,int> m;
long k;
cin>>k;
long long a;
for(int i=0;i<n;i++){
cin>>a;
m[a]++;
}
if(m.find(k)!=m.end()){
cout<<1<<endl;
}
else{
cout<<-1<<endl;
}
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a sorted array arr[] of N integers and a number K is given. The task is to check if the element K is present in the array or not.
<b>Note:</b> Use binary search to solve the problemThe first line of input contains a number of test cases T. For each test case, the first line of input contains a number of elements in the array, and the number K is separated by space. The next line contains N elements.
<b>Constraints:</b>
1 ≤ T ≤ 10
1 ≤ N ≤ 10<sup>5</sup>
1 ≤ K ≤ 10<sup>9</sup>
1 ≤ arr[i] ≤ 10<sup>9</sup>
<b>Sum of N over all test cases doesn't exceed 10<sup>6</sup></b>If the element is present in the array print "1" else print "-1".Sample Input:
2
5 6
1 2 3 4 6
5 2
1 3 4 5 6
Sample Output:
1
-1, I have written this Solution Code: def binary_search(arr, low, high, x):
if high >= low:
mid = (high + low) // 2
if arr[mid] == x:
return 1
elif arr[mid] > x:
return binary_search(arr, low, mid - 1, x)
else:
return binary_search(arr, mid + 1, high, x)
else:
return -1
def position(n,arr,x):
return binary_search(arr,0,n-1,x)
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a sorted array arr[] of N integers and a number K is given. The task is to check if the element K is present in the array or not.
<b>Note:</b> Use binary search to solve the problemThe first line of input contains a number of test cases T. For each test case, the first line of input contains a number of elements in the array, and the number K is separated by space. The next line contains N elements.
<b>Constraints:</b>
1 ≤ T ≤ 10
1 ≤ N ≤ 10<sup>5</sup>
1 ≤ K ≤ 10<sup>9</sup>
1 ≤ arr[i] ≤ 10<sup>9</sup>
<b>Sum of N over all test cases doesn't exceed 10<sup>6</sup></b>If the element is present in the array print "1" else print "-1".Sample Input:
2
5 6
1 2 3 4 6
5 2
1 3 4 5 6
Sample Output:
1
-1, I have written this Solution Code: // arr is they array to search from
// x is target
function binSearch(arr, x) {
// write code here
// do not console.log
// return the 1 or -1
let l = 0;
let r = arr.length - 1;
let mid;
while (r >= l) {
mid = l + Math.floor((r - l) / 2);
// If the element is present at the middle
// itself
if (arr[mid] == x)
return 1;
// If element is smaller than mid, then
// it can only be present in left subarray
if (arr[mid] > x)
r = mid - 1;
// Else the element can only be present
// in right subarray
else
l = mid + 1;
}
// We reach here when element is not
// present in array
return -1;
}, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You think that integers divisible by K are nice. Given L and R (L<=R), find the number of nice integers from L to R (both inclusive).First Line of input contains three integers L R K
Constraints :
0 <= L <= R <= 1000000000000000000(10^18)
1 <= K <= 1000000000000000000(10^18)Output a single integer which is the number of nice integers from L to R (both inclusive).Sample input 1
1 10 2
Sample output 1
5
Sample intput 2
4 5 3
Sample output 2
0
Explanation:-
Required divisors = 2 4 6 8 10, I have written this Solution Code:
import java.io.*;
import java.util.*;
class Main{
public static void main(String[] args)throws IOException{
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
String str[]=br.readLine().split(" ");
long l=Long.parseLong(str[0]);
long r=Long.parseLong(str[1]);
long k=Long.parseLong(str[2]);
long count=(r/k)-((l-1)/k);
System.out.print(count);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You think that integers divisible by K are nice. Given L and R (L<=R), find the number of nice integers from L to R (both inclusive).First Line of input contains three integers L R K
Constraints :
0 <= L <= R <= 1000000000000000000(10^18)
1 <= K <= 1000000000000000000(10^18)Output a single integer which is the number of nice integers from L to R (both inclusive).Sample input 1
1 10 2
Sample output 1
5
Sample intput 2
4 5 3
Sample output 2
0
Explanation:-
Required divisors = 2 4 6 8 10, I have written this Solution Code: l,r,k=[int(x)for x in input().split()]
print(r//k - (l-1)//k), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You think that integers divisible by K are nice. Given L and R (L<=R), find the number of nice integers from L to R (both inclusive).First Line of input contains three integers L R K
Constraints :
0 <= L <= R <= 1000000000000000000(10^18)
1 <= K <= 1000000000000000000(10^18)Output a single integer which is the number of nice integers from L to R (both inclusive).Sample input 1
1 10 2
Sample output 1
5
Sample intput 2
4 5 3
Sample output 2
0
Explanation:-
Required divisors = 2 4 6 8 10, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main(){
unsigned long long x,l,r,k;
cin>>l>>r>>k;
x=l/k;
if(l%k==0){x--;}
cout<<r/k-x;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Arpit Gupta has brought a toy for his valentine. She is playing with that toy which runs for T seconds when winded.If winded when the toy is already running, from that moment it will run for T seconds (not additional T seconds) For example if T is 10 and toy has run for 5 seconds and winded at this moment then in total it will run for 15 seconds.
Arpit's Valentine winds the toy N times.She winds the toy at t[i] seconds after the first time she winds it.How long will the toy run in total?First Line of input contains two integers N and T
Second Line of input contains N integers, list of time Arpit's Valentine has wound the toy at.
Constraints
1 <= N <= 100000
1 <= T <= 1000000000
1 <= t[i] <= 1000000000
t[0] = 0Print a single integer the total time the toy has run.Sample input 1
2 4
0 3
Sample output 1
7
Sample input 2
2 10
0 5
Sample output 2
15
Explanation:
Testcase1:-
at first the toy is winded at 0 it will go till 4 but it again winded at 3 making it go for more 4 seconds
so the total is 7, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args)throws IOException {
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
String[] str=br.readLine().split(" ");
int n=Integer.parseInt(str[0]);
int t=Integer.parseInt(str[1]);
int[] arr=new int[n];
str=br.readLine().split(" ");
for(int i=0;i<n;i++){
arr[i]=Integer.parseInt(str[i]);
}
int sum=0;
for(int i=1;i<n;i++){
int dif=arr[i]-arr[i-1];
if(dif>t){
sum=sum+t;
}else{
sum=sum+dif;
}
}
sum+=t;
System.out.print(sum);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Arpit Gupta has brought a toy for his valentine. She is playing with that toy which runs for T seconds when winded.If winded when the toy is already running, from that moment it will run for T seconds (not additional T seconds) For example if T is 10 and toy has run for 5 seconds and winded at this moment then in total it will run for 15 seconds.
Arpit's Valentine winds the toy N times.She winds the toy at t[i] seconds after the first time she winds it.How long will the toy run in total?First Line of input contains two integers N and T
Second Line of input contains N integers, list of time Arpit's Valentine has wound the toy at.
Constraints
1 <= N <= 100000
1 <= T <= 1000000000
1 <= t[i] <= 1000000000
t[0] = 0Print a single integer the total time the toy has run.Sample input 1
2 4
0 3
Sample output 1
7
Sample input 2
2 10
0 5
Sample output 2
15
Explanation:
Testcase1:-
at first the toy is winded at 0 it will go till 4 but it again winded at 3 making it go for more 4 seconds
so the total is 7, I have written this Solution Code: n , t = [int(x) for x in input().split() ]
l= [int(x) for x in input().split() ]
c = 0
for i in range(len(l)-1):
if l[i+1] - l[i]<=t:
c+=l[i+1] - l[i]
else:
c+=t
c+=t
print(c), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Arpit Gupta has brought a toy for his valentine. She is playing with that toy which runs for T seconds when winded.If winded when the toy is already running, from that moment it will run for T seconds (not additional T seconds) For example if T is 10 and toy has run for 5 seconds and winded at this moment then in total it will run for 15 seconds.
Arpit's Valentine winds the toy N times.She winds the toy at t[i] seconds after the first time she winds it.How long will the toy run in total?First Line of input contains two integers N and T
Second Line of input contains N integers, list of time Arpit's Valentine has wound the toy at.
Constraints
1 <= N <= 100000
1 <= T <= 1000000000
1 <= t[i] <= 1000000000
t[0] = 0Print a single integer the total time the toy has run.Sample input 1
2 4
0 3
Sample output 1
7
Sample input 2
2 10
0 5
Sample output 2
15
Explanation:
Testcase1:-
at first the toy is winded at 0 it will go till 4 but it again winded at 3 making it go for more 4 seconds
so the total is 7, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main(){
long n,t;
cin>>n>>t;
long a[n];
for(int i=0;i<n;i++){
cin>>a[i];
}
long cur=t;
long ans=t;
for(int i=1;i<n;i++){
ans+=min(a[i]-a[i-1],t);
}
cout<<ans;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a matrix of N*N dimensions. Take the input and display the matrix as an output.First line contains N.
N lines follow each containing N space seperated integers.
Constraints:-
2 <= N <= 500
1 <= Mat[i][j] <= 1000Print the given matrix.Input:
2
3 4
7 6
Output:
3 4
7 6
Input:
3
1 2 3
4 5 6
7 8 9
Output:
1 2 3
4 5 6
7 8 9, 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 {
Scanner sc = new Scanner(System.in);
int N = sc.nextInt();
int mat[][] = new int[N][N];
for(int i = 0; i < N; i++)
{
for(int j = 0; j < N; j++){
mat[i][j] = sc.nextInt();
}
}
for(int i = 0; i < N; i++)
{
for(int j = 0; j < N; j++){
System.out.print(mat[i][j]+" ");
}
System.out.println();
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a matrix of N*N dimensions. Take the input and display the matrix as an output.First line contains N.
N lines follow each containing N space seperated integers.
Constraints:-
2 <= N <= 500
1 <= Mat[i][j] <= 1000Print the given matrix.Input:
2
3 4
7 6
Output:
3 4
7 6
Input:
3
1 2 3
4 5 6
7 8 9
Output:
1 2 3
4 5 6
7 8 9, I have written this Solution Code: n = int(input())
for _ in range(n):
print(input()), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a matrix of N*N dimensions. Take the input and display the matrix as an output.First line contains N.
N lines follow each containing N space seperated integers.
Constraints:-
2 <= N <= 500
1 <= Mat[i][j] <= 1000Print the given matrix.Input:
2
3 4
7 6
Output:
3 4
7 6
Input:
3
1 2 3
4 5 6
7 8 9
Output:
1 2 3
4 5 6
7 8 9, I have written this Solution Code:
#include <bits/stdc++.h>
using namespace std;
int main()
{
int n;
cin>>n;
int arr[n][n];
for(int i=0;i<n;i++){
for(int j=0;j<n;j++){
cin>>arr[i][j];
}}
for(int i=0;i<n;i++){
for(int j=0;j<n;j++){
cout<<arr[i][j]<<" ";
}
cout<<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 integers and an integer K, your task is to calculate the count of pairs whose sum is divisible by K.The first line of input contains two integers N and K, the next line contains N space-separated integers depicting values of an array.
Constraints:-
1 < = N < = 100000
1 < = Arr[i] <= 100000
1 <= K <= 100000Print the count of required pairs.Sample Input
5 4
1 2 3 4 5
Sample Output
2
Sample Input
5 3
1 2 3 4 5
Sample Output
4
Explanation:-
In Sample 2,
(1 5), (1 2), (2 4), and (4 5) are the required pairs, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static long subarraysDivByK(int[] A, int k)
{
long ans =0 ;
int rem;
int[] freq = new int[k];
for(int i=0;i<A.length;i++)
{
rem = A[i]%k;
ans += freq[(k - rem)% k] ;
freq[rem]++;
}
return ans;
}
public static void main (String[] args) throws IOException{
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
String[] str = br.readLine().split(" ");
String[] input = br.readLine().split(" ");
int n = Integer.parseInt(str[0]);
int k = Integer.parseInt(str[1]);
int [] a = new int [n];
for(int i=0; i<n; i++)
a[i] = Integer.parseInt(input[i]);
System.out.println(subarraysDivByK(a, k));
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of N integers and an integer K, your task is to calculate the count of pairs whose sum is divisible by K.The first line of input contains two integers N and K, the next line contains N space-separated integers depicting values of an array.
Constraints:-
1 < = N < = 100000
1 < = Arr[i] <= 100000
1 <= K <= 100000Print the count of required pairs.Sample Input
5 4
1 2 3 4 5
Sample Output
2
Sample Input
5 3
1 2 3 4 5
Sample Output
4
Explanation:-
In Sample 2,
(1 5), (1 2), (2 4), and (4 5) are the required pairs, I have written this Solution Code: def countKdivPairs(A, n, K):
freq = [0] * K
for i in range(n):
freq[A[i] % K]+= 1
sum = freq[0] * (freq[0] - 1) / 2;
i = 1
while(i <= K//2 and i != (K - i) ):
sum += freq[i] * freq[K-i]
i+= 1
if( K % 2 == 0 ):
sum += (freq[K//2] * (freq[K//2]-1)/2);
return int(sum)
a,b=input().split()
a=int(a)
b=int(b)
arr=input().split()
for i in range(0,a):
arr[i]=int(arr[i])
print (countKdivPairs(arr,a, b)), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of N integers and an integer K, your task is to calculate the count of pairs whose sum is divisible by K.The first line of input contains two integers N and K, the next line contains N space-separated integers depicting values of an array.
Constraints:-
1 < = N < = 100000
1 < = Arr[i] <= 100000
1 <= K <= 100000Print the count of required pairs.Sample Input
5 4
1 2 3 4 5
Sample Output
2
Sample Input
5 3
1 2 3 4 5
Sample Output
4
Explanation:-
In Sample 2,
(1 5), (1 2), (2 4), and (4 5) are the required pairs, I have written this Solution Code:
#include <bits/stdc++.h>
using namespace std;
#define MEM(a, b) memset(a, (b), sizeof(a))
#define FOREACH(it, l) for (auto it = l.begin(); it != l.end(); it++)
#define IN(A, B, C) assert( B <= A && A <= C)
#define MP make_pair
#define FOR(i,a) for(int i=0;i<a;i++)
#define FOR1(i,j,a) for(int i=j;i<a;i++)
#define EB emplace_back
#define INF (int)1e9
#define EPS 1e-9
#define PI 3.1415926535897932384626433832795
#define MOD 1000000007
#define read(type) readInt<type>()
#define max1 100001
#define out(x) cout<<x<<'\n'
#define out1(x) cout<<x<<" "
#define END cout<<'\n'
#define int long long
void fast(){
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
}
signed main(){
fast();
int n;
cin>>n;
int a;
int k;
cin>>k;
int fre[k];
FOR(i,k){
fre[i]=0;}
FOR(i,n){
cin>>a;
fre[a%k]++;
}
int ans=(fre[0]*(fre[0]-1))/2;
for(int i=1;i<=(k-1)/2;i++){
ans+=fre[i]*fre[k-i];
}
if(k%2==0){
ans+=(fre[k/2]*(fre[k/2]-1))/2;
}
out(ans);
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a string of length n consisting of lowercase English letters. Find the vowels in the string and replace each vowel with it's next vowel in the alphabetic order.
Next vowel of a vowel is another vowel that comes after it in alphabet. Next vowel of { 'a', 'e', 'i', 'o', 'u'} are { 'e', 'i', 'o', 'u', 'a' } respectively.First line contains n.
Next line contains a string.
<b>Constraints</b>
1 <= n <= 10<sup>5</sup>print the string after replacing each vowel with it's next vowel.Input:
4
deaf
Output:
dief
Explanation:
next vowel of 'e' = 'i'
next vowel of 'a' = 'e', I have written this Solution Code: import java.io.*;
import java.util.*;
public class Main {
public static void main(String[] args) {
InputStream inputStream = System.in;
OutputStream outputStream = System.out;
InputReader in = new InputReader(inputStream);
OutputWriter out = new OutputWriter(outputStream);
int n=Integer.parseInt(in.next());
String s=in.next();
char arr[] = s.toCharArray();
for(int i=0;i<n;i++){
if(arr[i] == 'a'){
arr[i] = 'e';
}
else if(arr[i] == 'e'){
arr[i] = 'i';
}
else if(arr[i] == 'i'){
arr[i] = 'o';
}
else if(arr[i] == 'o'){
arr[i] = 'u';
}
else if(arr[i] == 'u'){
arr[i] = 'a';
}
}
String ans=String.valueOf(arr);
out.print(ans);
out.close();
}
static class InputReader {
BufferedReader reader;
StringTokenizer tokenizer;
public InputReader(InputStream stream) {
reader = new BufferedReader(new InputStreamReader(stream), 32768);
tokenizer = null;
}
public String next() {
while (tokenizer == null || !tokenizer.hasMoreTokens()) {
try {
tokenizer = new StringTokenizer(reader.readLine());
} catch (IOException e) {
throw new RuntimeException(e);
}
}
return tokenizer.nextToken();
}
public int nextInt() {
return Integer.parseInt(next());
}
}
static class OutputWriter {
private final PrintWriter writer;
public OutputWriter(OutputStream outputStream) {
writer = new PrintWriter(new BufferedWriter(new OutputStreamWriter(outputStream)));
}
public OutputWriter(Writer writer) {
this.writer = new PrintWriter(writer);
}
public void print(Object... objects) {
for (int i = 0; i < objects.length; i++) {
if (i != 0) {
writer.print(' ');
}
writer.print(objects[i]);
}
}
public void println(Object... objects) {
print(objects);
writer.println();
}
public void close() {
writer.close();
}
public void println(int i) {
writer.println(i);
}
}
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given an array Arr of N integers. A subarray is good if the sum of elements of that subarray is greater than or equal to K. Print the length of good subarray of minimum length.First line of input contains N and K.
Second line of input contains N integers representing the elements of the array Arr.
Constraints
1 <= N <= 100000
1 <= Arr[i] <= 100000
1 <= K <= 1000000000000Print the length of good subarray of minimum length.Sample input
5 12
2 3 2 5 5
Sample output
3
Explanation :
Subarray from index 3 to 5 has sum 12 and is therefore good and its length(3) is minimum among all possible good subarray., 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 k;
cin>>k;
int a[n+1];
int to=0;
for(int i=1;i<=n;++i)
{
cin>>a[i];
}
int j=1;
int s=0;
int ans=n;
for(int i=1;i<=n;++i)
{
while(s<k&&j<=n)
{
s+=a[j];
++j;
}
if(s>=k)
{
ans=min(ans,j-i);
}
s-=a[i];
}
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: You are given an array Arr of N integers. A subarray is good if the sum of elements of that subarray is greater than or equal to K. Print the length of good subarray of minimum length.First line of input contains N and K.
Second line of input contains N integers representing the elements of the array Arr.
Constraints
1 <= N <= 100000
1 <= Arr[i] <= 100000
1 <= K <= 1000000000000Print the length of good subarray of minimum length.Sample input
5 12
2 3 2 5 5
Sample output
3
Explanation :
Subarray from index 3 to 5 has sum 12 and is therefore good and its length(3) is minimum among all possible good subarray., I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws Exception
{
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
String s[]=br.readLine().split(" ");
int n=Integer.parseInt(s[0]);
long k=Long.parseLong(s[1]);
int a[]=new int[n];
s=br.readLine().split(" ");
for(int i=0;i<n;i++)
{
a[i]=Integer.parseInt(s[i]);
}
int length=Integer.MAX_VALUE,i=0,j=0;
long currSum=0;
for(j=0;j<n;j++)
{
currSum+=a[j];
while(currSum>=k)
{
length=Math.min(length,j-i+1);
currSum-=a[i];
i++;
}
}
System.out.println(length);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
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