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For this Question: Given an array A of size N containing 0's, 1's and 2's. The task is to segregate the 0's, 1's and 2's in the array as all the 0's should appear in the first part of the array, 1's should appear in middle part of the array and finally all the 2's in the remaining part of the array.
Note: Do not use inbuilt sort function. Try to solve in O(N) per test caseThe first line contains an integer T denoting the total number of test cases. Then T testcases follow.
Each testcases contains two lines of input. The first line denotes the size of the array N.
The second lines contains the elements of the array A separated by spaces.
Constraints:
1 <= T <= 100
1 <= N <= 100000
0 <= Ai <= 2
Sum of N for each test case does not exceed 10^5For each testcase, in a newline, print the sorted array.Input :
2
5
0 2 1 2 0
3
0 1 0
Output:
0 0 1 2 2
0 0 1
Explanation:
Testcase 1: After segragating the 0s, 1s and 2s, we have 0 0 1 2 2 which shown in the output.
Testcase 2: For the given array input, output will be 0 0 1., I have written this Solution Code: #include "bits/stdc++.h"
#pragma GCC optimize "03"
using namespace std;
#define int long long int
#define ld long double
#define pi pair<int, int>
#define pb push_back
#define fi first
#define se second
#define IOS ios::sync_with_stdio(false); cin.tie(0); cout.tie(0)
#ifndef LOCAL
#define endl '\n'
#endif
const int N = 2e5 + 5;
const int mod = 1e9 + 7;
const int inf = 1e9 + 9;
signed main() {
IOS;
int t; cin >> t;
while(t--){
int n; cin >> n;
int a[3] = {0};
for(int i = 1; i <= n; i++){
int p; cin >> p;
a[p]++;
}
a[1] += a[0];
a[2] += a[1];
for(int i = 1; i <= n; i++){
if(i <= a[0]) cout << "0 ";
else if(i <= a[1]) cout << "1 ";
else cout << "2 ";
}
cout << endl;
}
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A of size N containing 0's, 1's and 2's. The task is to segregate the 0's, 1's and 2's in the array as all the 0's should appear in the first part of the array, 1's should appear in middle part of the array and finally all the 2's in the remaining part of the array.
Note: Do not use inbuilt sort function. Try to solve in O(N) per test caseThe first line contains an integer T denoting the total number of test cases. Then T testcases follow.
Each testcases contains two lines of input. The first line denotes the size of the array N.
The second lines contains the elements of the array A separated by spaces.
Constraints:
1 <= T <= 100
1 <= N <= 100000
0 <= Ai <= 2
Sum of N for each test case does not exceed 10^5For each testcase, in a newline, print the sorted array.Input :
2
5
0 2 1 2 0
3
0 1 0
Output:
0 0 1 2 2
0 0 1
Explanation:
Testcase 1: After segragating the 0s, 1s and 2s, we have 0 0 1 2 2 which shown in the output.
Testcase 2: For the given array input, output will be 0 0 1., I have written this Solution Code: function zeroOneTwoSort(arr, n) {
// write code here
// do not console.log the answer
// return sorted array
return arr.sort((a, b) => a - b)
}, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a temperature F in Farenheit, your task is to convert it into Celsius using the following equation:-
T(°c) = (T(°f) - 32)*5/9You don't have to worry about the input, you just have to complete the function <b>fahrenheitToCelsius</b>
Constraints:-
-10^3 <= F <= 10^3
<b>Note:-</b> It is guaranteed that F - 32 will be a multiple of 9.Print an integer containing converted temperature in Fahrenheit.Sample Input 1:
77
Sample Output 1:
25
Sample Input 2:-
-40
Sample Output 2:-
-40
<b>Explanation 1</b>:
T(°c) = (T(°f) - 32)*5/9
T(°c) = (77-32)*5/9
T(°c) =25, I have written this Solution Code: void farhenheitToCelsius(int n){
n-=32;
n/=9;
n*=5;
cout<<n;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a temperature F in Farenheit, your task is to convert it into Celsius using the following equation:-
T(°c) = (T(°f) - 32)*5/9You don't have to worry about the input, you just have to complete the function <b>fahrenheitToCelsius</b>
Constraints:-
-10^3 <= F <= 10^3
<b>Note:-</b> It is guaranteed that F - 32 will be a multiple of 9.Print an integer containing converted temperature in Fahrenheit.Sample Input 1:
77
Sample Output 1:
25
Sample Input 2:-
-40
Sample Output 2:-
-40
<b>Explanation 1</b>:
T(°c) = (T(°f) - 32)*5/9
T(°c) = (77-32)*5/9
T(°c) =25, I have written this Solution Code: Fahrenheit= int(input())
Celsius = int(((Fahrenheit-32)*5)/9 )
print(Celsius), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a temperature F in Farenheit, your task is to convert it into Celsius using the following equation:-
T(°c) = (T(°f) - 32)*5/9You don't have to worry about the input, you just have to complete the function <b>fahrenheitToCelsius</b>
Constraints:-
-10^3 <= F <= 10^3
<b>Note:-</b> It is guaranteed that F - 32 will be a multiple of 9.Print an integer containing converted temperature in Fahrenheit.Sample Input 1:
77
Sample Output 1:
25
Sample Input 2:-
-40
Sample Output 2:-
-40
<b>Explanation 1</b>:
T(°c) = (T(°f) - 32)*5/9
T(°c) = (77-32)*5/9
T(°c) =25, I have written this Solution Code: static void farhrenheitToCelsius(int farhrenheit)
{
int celsius = ((farhrenheit-32)*5)/9;
System.out.println(celsius);
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: John is confused about the number of ways to propose to Olivia. So, he asked for your help. Now, you need to determine the number of ways for John to propose to Olivia.
For that, You are given an integer N and you need to count the number of pairs (x<sub>1</sub>, x<sub>2</sub>) such that x<sub>1</sub><sup>2</sup> + x<sub>2</sub><sup>2</sup> = N and x<sub>1</sub>, x<sub>2</sub> both are positive integer.The first line contains a single integer T, the number of test cases.
T lines follow. Each line describes a single test case and contains a single integer N.
<b>Constraints:</b>
1 <= T <= 100
2 <= N <= 10<sup>5</sup>For each test case, print a single integer count of such pairs.Sample Input 1:
2
13
4
Sample Output 2:
2
0, I have written this Solution Code: //HEADER FILES AND NAMESPACES
#include<bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#pragma GCC target("popcnt")
using namespace std;
using namespace __gnu_pbds;
template <typename T>
using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
template <typename T>
using ordered_multiset = tree<T, null_type, less_equal<T>, rb_tree_tag, tree_order_statistics_node_update>;
using cd = complex<double>;
const double PI = acos(-1);
// DEFINE STATEMENTS
const long long infty = 1e18;
#define num1 1000000007
#define num2 998244353
#define REP(i,a,n) for(ll i=a;i<n;i++)
#define REPd(i,a,n) for(ll i=a; i>=n; i--)
#define pb push_back
#define pob pop_back
#define fr first
#define sc second
#define fix(f,n) std::fixed<<std::setprecision(n)<<f
#define all(x) x.begin(), x.end()
#define M_PI 3.14159265358979323846
#define epsilon (double)(0.000000001)
#define popcount __builtin_popcountll
#define fileio(x) freopen("input.txt", "r", stdin); freopen(x, "w", stdout);
#define out(x) cout << ((x) ? "Yes\n" : "No\n")
#define sz(x) x.size()
typedef long long ll;
typedef long long unsigned int llu;
typedef vector<long long> vll;
typedef pair<long long, long long> pll;
typedef vector<pair<long long, long long>> vpll;
typedef vector<int> vii;
// DEBUG FUNCTIONS
#ifndef ONLINE_JUDGE
template<typename T>
void __p(T a) {
cout<<a;
}
template<typename T, typename F>
void __p(pair<T, F> a) {
cout<<"{";
__p(a.first);
cout<<",";
__p(a.second);
cout<<"}";
}
template<typename T>
void __p(std::vector<T> a) {
cout<<"{";
for(auto it=a.begin(); it<a.end(); it++)
__p(*it),cout<<",}"[it+1==a.end()];
}
template<typename T>
void __p(std::set<T> a) {
cout<<"{";
for(auto it=a.begin(); it!=a.end();){
__p(*it);
cout<<",}"[++it==a.end()];
}
}
template<typename T>
void __p(std::multiset<T> a) {
cout<<"{";
for(auto it=a.begin(); it!=a.end();){
__p(*it);
cout<<",}"[++it==a.end()];
}
}
template<typename T, typename F>
void __p(std::map<T,F> a) {
cout<<"{\n";
for(auto it=a.begin(); it!=a.end();++it)
{
__p(it->first);
cout << ": ";
__p(it->second);
cout<<"\n";
}
cout << "}\n";
}
template<typename T, typename ...Arg>
void __p(T a1, Arg ...a) {
__p(a1);
__p(a...);
}
template<typename Arg1>
void __f(const char *name, Arg1 &&arg1) {
cout<<name<<" : ";
__p(arg1);
cout<<endl;
}
template<typename Arg1, typename ... Args>
void __f(const char *names, Arg1 &&arg1, Args &&... args) {
int bracket=0,i=0;
for(;; i++)
if(names[i]==','&&bracket==0)
break;
else if(names[i]=='(')
bracket++;
else if(names[i]==')')
bracket--;
const char *comma=names+i;
cout.write(names,comma-names)<<" : ";
__p(arg1);
cout<<" | ";
__f(comma+1,args...);
}
#define trace(...) cout<<"Line:"<<__LINE__<<" ", __f(#__VA_ARGS__, __VA_ARGS__)
#else
#define trace(...)
#define error(...)
#endif
void solve(){
ll n; cin >> n;
ll ans = 0;
for(ll i = 1;i*i<n;i++){
ll x = sqrt(n - i*i);
if(x*x + i*i == n){
ans++;
}
}
cout << ans << "\n";
}
int main(){
ios_base::sync_with_stdio(false);
cin.tie(NULL);
// cout.tie(NULL);
#ifdef LOCALFLAG
freopen("input.txt", "r", stdin);
freopen("output.txt", "w", stdout);
#endif
ll t = 1;
cin >> t;
while(t--){
solve();
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array with repeated elements, the task is to find the maximum distance between two occurrences of an element.
Note:- It is guaranteed that atleast one number is repeated.The first line of the input contains an integer N denoting the number of elements in the array, the next line contains N space separated integers.
Constraints:
1 <= N <= 100000
1 <= Arr[i] <= 10^9For each test case in new line print the Maximum distance between two occurrences of an elementSample Input
6
1 1 2 2 2 1
Sample Output
5
Explanation:-
The index of two occurrences are:- (1, 2), (1, 6), (2, 6), (3, 4), (3, 5), (4, 5)
it can be seen that the maximum distance is between (1, 6) i. e. 5
Sample Input
5
1 1 1 1 1
Sample Output
4, I have written this Solution Code: import numpy as np
from collections import defaultdict
d=defaultdict(list)
n=input()
a=np.array([input().strip().split()]).flatten()
for i in range(len(a)):
d[a[i]].append(i)
for i in d:
d[i].sort()
ans=0
for i in d:
ans=max(ans,d[i][-1]-d[i][0])
print(ans), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array with repeated elements, the task is to find the maximum distance between two occurrences of an element.
Note:- It is guaranteed that atleast one number is repeated.The first line of the input contains an integer N denoting the number of elements in the array, the next line contains N space separated integers.
Constraints:
1 <= N <= 100000
1 <= Arr[i] <= 10^9For each test case in new line print the Maximum distance between two occurrences of an elementSample Input
6
1 1 2 2 2 1
Sample Output
5
Explanation:-
The index of two occurrences are:- (1, 2), (1, 6), (2, 6), (3, 4), (3, 5), (4, 5)
it can be seen that the maximum distance is between (1, 6) i. e. 5
Sample Input
5
1 1 1 1 1
Sample Output
4, 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 arrSize = sc.nextInt();
long arr[] = new long[arrSize];
for(int i = 0; i < arrSize; i++)
arr[i] = sc.nextLong();
System.out.println(maxDist(arr, arrSize));
}
static int maxDist(long arr[], int arrSize)
{
int ans = 0;
HashMap<Long, Integer> hMap = new HashMap<>();
for(int i = 0; i < arrSize; i++)
{
if(hMap.containsKey(arr[i]) == true)
ans = Math.max(ans, i-hMap.get(arr[i]));
else
hMap.put(arr[i], i);
}
return ans;
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array with repeated elements, the task is to find the maximum distance between two occurrences of an element.
Note:- It is guaranteed that atleast one number is repeated.The first line of the input contains an integer N denoting the number of elements in the array, the next line contains N space separated integers.
Constraints:
1 <= N <= 100000
1 <= Arr[i] <= 10^9For each test case in new line print the Maximum distance between two occurrences of an elementSample Input
6
1 1 2 2 2 1
Sample Output
5
Explanation:-
The index of two occurrences are:- (1, 2), (1, 6), (2, 6), (3, 4), (3, 5), (4, 5)
it can be seen that the maximum distance is between (1, 6) i. e. 5
Sample Input
5
1 1 1 1 1
Sample Output
4, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main()
{
int n;
cin>>n;
long a[n];
unordered_map<long,int> m;
int ans=0;
for(int i=0;i<n;i++){
cin>>a[i];
if(m.find(a[i])!=m.end()){ans=max(ans,i-m[a[i]]);}
else{
m[a[i]]=i;
}
}
cout<<ans;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara's Phone has N apps and each app takes K unit of memory. Now Sara wants to release M units of memory. Your task is to tell the minimum apps Sara needs to delete or say it is not possible.<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>Phone()</b> that takes integers N, K, and M as arguments.
Constraints:-
1 <= N <= 1000
1 <= K <= 100
0 <= M <= 10000Return minimum apps to delete and if it is not possible return -1.Sample Input:-
10 3 10
Sample Output:-
4
Sample Input:-
10 3 40
Sample Output:-
-1, I have written this Solution Code:
int Phone(int n, int k, int m){
if(n*k<m){
return -1;
}
int x = m/k;
if(m%k!=0){x++;}
return x;
}, In this Programming Language: C, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara's Phone has N apps and each app takes K unit of memory. Now Sara wants to release M units of memory. Your task is to tell the minimum apps Sara needs to delete or say it is not possible.<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>Phone()</b> that takes integers N, K, and M as arguments.
Constraints:-
1 <= N <= 1000
1 <= K <= 100
0 <= M <= 10000Return minimum apps to delete and if it is not possible return -1.Sample Input:-
10 3 10
Sample Output:-
4
Sample Input:-
10 3 40
Sample Output:-
-1, I have written this Solution Code:
int Phone(int n, int k, int m){
if(n*k<m){
return -1;
}
int x = m/k;
if(m%k!=0){x++;}
return x;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara's Phone has N apps and each app takes K unit of memory. Now Sara wants to release M units of memory. Your task is to tell the minimum apps Sara needs to delete or say it is not possible.<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>Phone()</b> that takes integers N, K, and M as arguments.
Constraints:-
1 <= N <= 1000
1 <= K <= 100
0 <= M <= 10000Return minimum apps to delete and if it is not possible return -1.Sample Input:-
10 3 10
Sample Output:-
4
Sample Input:-
10 3 40
Sample Output:-
-1, I have written this Solution Code: static int Phone(int n, int k, int m){
if(n*k<m){
return -1;
}
int x = m/k;
if(m%k!=0){x++;}
return x;
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara's Phone has N apps and each app takes K unit of memory. Now Sara wants to release M units of memory. Your task is to tell the minimum apps Sara needs to delete or say it is not possible.<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>Phone()</b> that takes integers N, K, and M as arguments.
Constraints:-
1 <= N <= 1000
1 <= K <= 100
0 <= M <= 10000Return minimum apps to delete and if it is not possible return -1.Sample Input:-
10 3 10
Sample Output:-
4
Sample Input:-
10 3 40
Sample Output:-
-1, I have written this Solution Code: def Phone(N,K,M):
if N*K < M :
return -1
x = M//K
if M%K!=0:
x=x+1
return x, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer n, For each i (1<=i<=n) if i is even print "<b>even</b>" else print "<b>odd</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 functions <b>For_Loop()</b> that take the integer n as a parameter.
</b>Constraints:</b>
1 ≤ n ≤ 100Print even or odd for each i, separated by white spaces.Sample Input:
5
Sample Output:
odd even odd even odd
Sample Input:
2
Sample Output:
odd even, I have written this Solution Code: public static void For_Loop(int n){
for(int i=1;i<=n;i++){
if(i%2==1){System.out.print("odd ");}
else{
System.out.print("even ");
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer n, For each i (1<=i<=n) if i is even print "<b>even</b>" else print "<b>odd</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 functions <b>For_Loop()</b> that take the integer n as a parameter.
</b>Constraints:</b>
1 ≤ n ≤ 100Print even or odd for each i, separated by white spaces.Sample Input:
5
Sample Output:
odd even odd even odd
Sample Input:
2
Sample Output:
odd even, I have written this Solution Code: n = int(input())
for i in range(1, n+1):
if(i%2)==0:
print("even ",end="")
else:
print("odd ",end=""), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number N for each i (1 < = i < = N), you have to print the number except :-
For each multiple of 3, print "Newton" instead of the number.
For each multiple of 5, print "School" instead of the number.
For numbers that are multiples of both 3 and 5, print "NewtonSchool" instead of the number.The first line of the input contains N.
<b>Constraints</b>
1 < = N < = 1000
Print N space separated number or Newton School according to the condition.Sample Input:-
3
Sample Output:-
1 2 Newton
Sample Input:-
5
Sample Output:-
1 2 Newton 4 School, I have written this Solution Code: n=int(input())
for i in range(1,n+1):
if i%3==0 and i%5==0:
print("NewtonSchool",end=" ")
elif i%3==0:
print("Newton",end=" ")
elif i%5==0:
print("School",end=" ")
else:
print(i,end=" "), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number N for each i (1 < = i < = N), you have to print the number except :-
For each multiple of 3, print "Newton" instead of the number.
For each multiple of 5, print "School" instead of the number.
For numbers that are multiples of both 3 and 5, print "NewtonSchool" instead of the number.The first line of the input contains N.
<b>Constraints</b>
1 < = N < = 1000
Print N space separated number or Newton School according to the condition.Sample Input:-
3
Sample Output:-
1 2 Newton
Sample Input:-
5
Sample Output:-
1 2 Newton 4 School, I have written this Solution Code: import java.util.*;
import java.lang.*;
import java.io.*;
class Main
{
static void NewtonSchool(int n){
for(int i=1;i<=n;i++){
if(i%3==0 && i%5==0){System.out.print("NewtonSchool ");}
else if(i%5==0){System.out.print("School ");}
else if(i%3==0){System.out.print("Newton ");}
else{System.out.print(i+" ");}
}
}
public static void main (String[] args) throws java.lang.Exception
{
Scanner sc = new Scanner(System.in);
int x= sc.nextInt();
NewtonSchool(x);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: When learning a new language, we first learn to output some messages. Here, we will start with the famous <b>"Hello World"</b> message. Now, here you are given a function to complete. <i>Don't worry about the ins and outs of functions, <b>just add the printing command to print "Hello World", </b></i>.your task is to just print "Hello World", without the quotes.Hello WorldHello World must be printed., I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) {
System.out.print("Hello World");
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: When learning a new language, we first learn to output some messages. Here, we will start with the famous <b>"Hello World"</b> message. Now, here you are given a function to complete. <i>Don't worry about the ins and outs of functions, <b>just add the printing command to print "Hello World", </b></i>.your task is to just print "Hello World", without the quotes.Hello WorldHello World must be printed., I have written this Solution Code: def print_fun():
print ("Hello World")
def main():
print_fun()
if __name__ == '__main__':
main(), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given four positive integers A, B, C, and D, can a rectangle have these integers as its sides?The input consists of a single line containing four space-separated integers A, B, C and D.
<b> Constraints: </b>
1 ≤ A, B, C, D ≤ 1000Output "Yes" if a rectangle is possible; otherwise, "No" (without quotes).Sample Input 1:
3 4 4 3
Sample Output 1:
Yes
Sample Explanation 1:
A rectangle is possible with 3, 4, 3, and 4 as sides in clockwise order.
Sample Input 2:
3 4 3 5
Sample Output 2:
No
Sample Explanation 2:
No rectangle can be made with these side lengths., I have written this Solution Code: #include<iostream>
using namespace std;
int main(){
int a,b,c,d;
cin >> a >> b >> c >> d;
if((a==c) &&(b==d)){
cout << "Yes\n";
}else if((a==b) && (c==d)){
cout << "Yes\n";
}else if((a==d) && (b==c)){
cout << "Yes\n";
}else{
cout << "No\n";
}
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given an array of integers. Consider absolute difference between all the pairs of the the elements. You need to find Kth smallest absolute difference. If the size of the array is N then value of K will be less than N and more than or equal to 1.The first line of input contains number of test cases T.
The first line of each test case contains a two integers N and K denoting the number of elements in the array A and difference you need to output. The second line of each test case contains N space separated integers denoting the elements of the array A
Constraints:
1<= T <= 10
2 <= N <= 100000
1 <= K < N < 100000
0 <= A[i] <= 100000For each test case, output Kth smallest absolute difference.Input :
1
6 2
1 3 4 1 3 8
Output :
0
Explanation :
Test case 1: First smallest difference is 0, between the pair (1, 1) and second smallest absolute difference difference is also 0 between the pairs (3, 3)., I have written this Solution Code: import java.util.*;
import java.io.*;
import java.lang.*;
class Main{
public static void main(String[] args)throws IOException {
BufferedReader read = new BufferedReader(new InputStreamReader(System.in));
int t = Integer.parseInt(read.readLine().trim());
while (t-- > 0) {
String str[] = read.readLine().trim().split(" ");
int n = Integer.parseInt(str[0]);
int k = Integer.parseInt(str[1]);
int arr[] = new int[n];
str = read.readLine().trim().split(" ");
for (int i = 0; i < n; i++)
arr[i] = Integer.parseInt(str[i]);
System.out.println(Math.abs(small(arr, k)));
}
}
public static int small(int arr[], int k) {
Arrays.sort(arr);
int l = 0, r = arr[arr.length - 1] - arr[0];
while (r > l) {
int mid = l + (r - l) / 2;
if (count(arr, mid) < k) {
l = mid + 1;
} else {
r = mid;
}
}
return r;
}
public static int count(int arr[], int mid) {
int ans = 0, j = 0;
for (int i = 1; i < arr.length; ++i) {
while (j < i && arr[i] - arr[j] > mid) {
++j;
}
ans += i - j;
}
return ans;
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given three integers A, B, and C, your task is to print the sum of these three integers.The input contains 3 integers separated by spaces A, B, and C.
Constraints:-
1 <= A, B, C <= 100Print the sum of A, B and C.Sample Input
1 2 3
Sample Output:-
6
Sample Input:-
5 4 2
Sample Output:-
11, I have written this Solution Code: static void simpleSum(int a, int b, int c){
System.out.println(a+b+c);
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given three integers A, B, and C, your task is to print the sum of these three integers.The input contains 3 integers separated by spaces A, B, and C.
Constraints:-
1 <= A, B, C <= 100Print the sum of A, B and C.Sample Input
1 2 3
Sample Output:-
6
Sample Input:-
5 4 2
Sample Output:-
11, I have written this Solution Code: void simpleSum(int a, int b, int c){
cout<<a+b+c;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given three integers A, B, and C, your task is to print the sum of these three integers.The input contains 3 integers separated by spaces A, B, and C.
Constraints:-
1 <= A, B, C <= 100Print the sum of A, B and C.Sample Input
1 2 3
Sample Output:-
6
Sample Input:-
5 4 2
Sample Output:-
11, I have written this Solution Code: x = input()
a, b, c = x.split()
a = int(a)
b = int(b)
c = int(c)
print(a+b+c), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an N*N matrix. Print the elements of the matrix in anticlockwise order (see the sample for better understanding).First line contains N.
N lines follow each containing N space seperated integers.
Constraints:-
2 <= N <= 500
1 <= Mat[i][j] <= 1000Output N*N integers in a single line separated by spaces, which are the elements of the matrix in anti-clockwise order.Sample Input
4
1 2 3 4
5 6 7 8
9 10 11 12
13 14 15 16
Sample output
1 5 9 13 14 15 16 12 8 4 3 2 6 10 11 7
Explanation:
We start from 1 , go down till 13 and then go right till 16 then go up till 4 , then we go left till 2 then down and so on in anti-clockwise fashion .
Sample Input
3
1 2 3
4 5 6
7 8 9
Sample output
1 4 7 8 9 6 3 2 5
, I have written this Solution Code: import java.util.*;
import java.lang.*;
import java.io.*;
class Main{
public static void main(String args[])throws IOException{
BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
int n = Integer.parseInt(in.readLine());
int x, y, start, end, num=n, firstcond, secondcond, thirdcond, fourthcond, flag;
if(n%2!=0)
num++;
num = num/2;
int[][] a = new int[n][n];
for(x=0; x<n; x++){
String nextLine[] = in.readLine().split(" ");
for(y=0; y<n; y++){
a[x][y] = Integer.parseInt(nextLine[y]);
}
}
start = 0;
end = n-1;
while(num>=1){
flag=0;
firstcond = secondcond = thirdcond = fourthcond = 1;
for(x=start, y=start;
(firstcond==1 || secondcond==1 || thirdcond==1 || fourthcond==1) && (x>=start && y>=start && x<=end && y<=end);
){
System.out.print(a[x][y] + " ");
if(firstcond==1){
x++;
if(x==end+1){
firstcond=0;
x--;
y++;
}
}else if(secondcond==1){
y++;
if(y==end+1){
secondcond=0;
y--;
x--;
}
}else if(thirdcond==1){
x--;
if(x==start-1){
if(flag==0)
break;
thirdcond=0;
x++;
y--;
}
flag=1;
}else{
y--;
if(y==start){
fourthcond=0;
x++;
y++;
}
}
}
start++;
end--;
num--;
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an N*N matrix. Print the elements of the matrix in anticlockwise order (see the sample for better understanding).First line contains N.
N lines follow each containing N space seperated integers.
Constraints:-
2 <= N <= 500
1 <= Mat[i][j] <= 1000Output N*N integers in a single line separated by spaces, which are the elements of the matrix in anti-clockwise order.Sample Input
4
1 2 3 4
5 6 7 8
9 10 11 12
13 14 15 16
Sample output
1 5 9 13 14 15 16 12 8 4 3 2 6 10 11 7
Explanation:
We start from 1 , go down till 13 and then go right till 16 then go up till 4 , then we go left till 2 then down and so on in anti-clockwise fashion .
Sample Input
3
1 2 3
4 5 6
7 8 9
Sample output
1 4 7 8 9 6 3 2 5
, I have written this Solution Code: n = int(input())
arr = []
for i in range(n):
j = input().split()
arr.append([int(xx) for xx in j])
def counterClockspiralPrint(m, n, arr) :
k = 0; l = 0
cnt = 0
total = m * n
while (k < m and l < n) :
if (cnt == total) :
break
for i in range(k, m) :
print(arr[i][l], end = " ")
cnt += 1
l += 1
if (cnt == total) :
break
for i in range (l, n) :
print( arr[m - 1][i], end = " ")
cnt += 1
m -= 1
if (cnt == total) :
break
if (k < m) :
for i in range(m - 1, k - 1, -1) :
print(arr[i][n - 1], end = " ")
cnt += 1
n -= 1
if (cnt == total) :
break
if (l < n) :
for i in range(n - 1, l - 1, -1) :
print( arr[k][i], end = " ")
cnt += 1
k += 1
counterClockspiralPrint(n, n, arr), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an N*N matrix. Print the elements of the matrix in anticlockwise order (see the sample for better understanding).First line contains N.
N lines follow each containing N space seperated integers.
Constraints:-
2 <= N <= 500
1 <= Mat[i][j] <= 1000Output N*N integers in a single line separated by spaces, which are the elements of the matrix in anti-clockwise order.Sample Input
4
1 2 3 4
5 6 7 8
9 10 11 12
13 14 15 16
Sample output
1 5 9 13 14 15 16 12 8 4 3 2 6 10 11 7
Explanation:
We start from 1 , go down till 13 and then go right till 16 then go up till 4 , then we go left till 2 then down and so on in anti-clockwise fashion .
Sample Input
3
1 2 3
4 5 6
7 8 9
Sample output
1 4 7 8 9 6 3 2 5
, I have written this Solution Code:
#include <bits/stdc++.h>
using namespace std;
#define N 1000
void counterClockspiralPrint( int m,
int n,
int arr[][N])
{
int i, k = 0, l = 0;
// k - starting row index
// m - ending row index
// l - starting column index
// n - ending column index
// i - iterator
// initialize the count
int cnt = 0;
// total number of
// elements in matrix
int total = m * n;
while (k < m && l < n)
{
if (cnt == total)
break;
// Print the first column
// from the remaining columns
for (i = k; i < m; ++i)
{
cout << arr[i][l] << " ";
cnt++;
}
l++;
if (cnt == total)
break;
// Print the last row from
// the remaining rows
for (i = l; i < n; ++i)
{
cout << arr[m - 1][i] << " ";
cnt++;
}
m--;
if (cnt == total)
break;
// Print the last column
// from the remaining columns
if (k < m)
{
for (i = m - 1; i >= k; --i)
{
cout << arr[i][n - 1] << " ";
cnt++;
}
n--;
}
if (cnt == total)
break;
// Print the first row
// from the remaining rows
if (l < n)
{
for (i = n - 1; i >= l; --i)
{
cout << arr[k][i] << " ";
cnt++;
}
k++;
}
}
}
int main()
{
int n;
cin>>n;
int arr[n][N];
for(int i=0;i<n;i++){
for(int j=0;j<n;j++){
cin>>arr[i][j];
}}
counterClockspiralPrint(n,n, arr);
return 0;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an N*N matrix. Print the elements of the matrix in anticlockwise order (see the sample for better understanding).First line contains N.
N lines follow each containing N space seperated integers.
Constraints:-
2 <= N <= 500
1 <= Mat[i][j] <= 1000Output N*N integers in a single line separated by spaces, which are the elements of the matrix in anti-clockwise order.Sample Input
4
1 2 3 4
5 6 7 8
9 10 11 12
13 14 15 16
Sample output
1 5 9 13 14 15 16 12 8 4 3 2 6 10 11 7
Explanation:
We start from 1 , go down till 13 and then go right till 16 then go up till 4 , then we go left till 2 then down and so on in anti-clockwise fashion .
Sample Input
3
1 2 3
4 5 6
7 8 9
Sample output
1 4 7 8 9 6 3 2 5
, I have written this Solution Code: function printAntiClockWise(mat)
{
let i, k = 0, l = 0;
let m = N;
let n = N;
let cnt = 0, total = m*n;
while(k < m && l < n)
{
if (cnt == total)
break;
// Print the first column
// from the remaining columns
for (i = k; i < m; ++i)
{
process.stdout.write(mat[i][l] + " ");
cnt++;
}
l++;
if (cnt == total)
break;
// Print the last row from
// the remaining rows
for (i = l; i < n; ++i)
{
process.stdout.write(mat[m - 1][i] + " ");
cnt++;
}
m--;
if (cnt == total)
break;
// Print the last column
// from the remaining columns
if (k < m)
{
for (i = m - 1; i >= k; --i)
{
process.stdout.write(mat[i][n - 1] + " ");
cnt++;
}
n--;
}
if (cnt == total)
break;
// Print the first row
// from the remaining rows
if (l < n)
{
for (i = n - 1; i >= l; --i)
{
process.stdout.write(mat[k][i] + " ");
cnt++;
}
k++;
}
}
}, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number N for each i (1 < = i < = N), you have to print the number except :-
For each multiple of 3, print "Newton" instead of the number.
For each multiple of 5, print "School" instead of the number.
For numbers that are multiples of both 3 and 5, print "NewtonSchool" instead of the number.The first line of the input contains N.
<b>Constraints</b>
1 < = N < = 1000
Print N space separated number or Newton School according to the condition.Sample Input:-
3
Sample Output:-
1 2 Newton
Sample Input:-
5
Sample Output:-
1 2 Newton 4 School, I have written this Solution Code: n=int(input())
for i in range(1,n+1):
if i%3==0 and i%5==0:
print("NewtonSchool",end=" ")
elif i%3==0:
print("Newton",end=" ")
elif i%5==0:
print("School",end=" ")
else:
print(i,end=" "), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number N for each i (1 < = i < = N), you have to print the number except :-
For each multiple of 3, print "Newton" instead of the number.
For each multiple of 5, print "School" instead of the number.
For numbers that are multiples of both 3 and 5, print "NewtonSchool" instead of the number.The first line of the input contains N.
<b>Constraints</b>
1 < = N < = 1000
Print N space separated number or Newton School according to the condition.Sample Input:-
3
Sample Output:-
1 2 Newton
Sample Input:-
5
Sample Output:-
1 2 Newton 4 School, I have written this Solution Code: import java.util.*;
import java.lang.*;
import java.io.*;
class Main
{
static void NewtonSchool(int n){
for(int i=1;i<=n;i++){
if(i%3==0 && i%5==0){System.out.print("NewtonSchool ");}
else if(i%5==0){System.out.print("School ");}
else if(i%3==0){System.out.print("Newton ");}
else{System.out.print(i+" ");}
}
}
public static void main (String[] args) throws java.lang.Exception
{
Scanner sc = new Scanner(System.in);
int x= sc.nextInt();
NewtonSchool(x);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Nobita wants to score well in his upcoming test, but he is not able to solve the simple division problems, seeing Nobita's determination Doraemon gives him a gadget that can do division problems easily but somehow Nobita deleted the internal program which calculates the division.
As an excellent coder, Nobita came to you for help. Help Nobita to write a code for his gadget.
You will be given two integers <b>D</b> and <b>Q</b>, you have to print the value of <b>D/Q</b> rounded down .The input contains two space- separated integers depicting the values of D and Q.
Constraints:-
0 <= D, Q <= 100Print the values of D/Q if the value can be calculated else print -1 if it is undefined.
Note:- Remember division by 0 is an undefined value that will give runtime error in your program.Sample Input:-
9 3
Sample Output:-
3
Sample Input:-
8 5
Sample Output:-
1
Explanation:-
8/5 = 1.6 = 1(floor), I have written this Solution Code: import java.io.*;
import java.util.*;
import java.lang.Math.*;
class Main {
public static void main (String[] args) throws IOException{
BufferedReader bf = new BufferedReader(new InputStreamReader(System.in));
String[] st = bf.readLine().split(" ");
if(Integer.parseInt(st[1])==0)
System.out.print(-1);
else {
int f = (Integer.parseInt(st[0])/Integer.parseInt(st[1]));
System.out.print(f);
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Nobita wants to score well in his upcoming test, but he is not able to solve the simple division problems, seeing Nobita's determination Doraemon gives him a gadget that can do division problems easily but somehow Nobita deleted the internal program which calculates the division.
As an excellent coder, Nobita came to you for help. Help Nobita to write a code for his gadget.
You will be given two integers <b>D</b> and <b>Q</b>, you have to print the value of <b>D/Q</b> rounded down .The input contains two space- separated integers depicting the values of D and Q.
Constraints:-
0 <= D, Q <= 100Print the values of D/Q if the value can be calculated else print -1 if it is undefined.
Note:- Remember division by 0 is an undefined value that will give runtime error in your program.Sample Input:-
9 3
Sample Output:-
3
Sample Input:-
8 5
Sample Output:-
1
Explanation:-
8/5 = 1.6 = 1(floor), I have written this Solution Code: D,Q = input().split()
D = int(D)
Q = int(Q)
if(0<=D and Q<=100 and Q >0):
print(int(D/Q))
else:
print('-1'), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Nobita wants to score well in his upcoming test, but he is not able to solve the simple division problems, seeing Nobita's determination Doraemon gives him a gadget that can do division problems easily but somehow Nobita deleted the internal program which calculates the division.
As an excellent coder, Nobita came to you for help. Help Nobita to write a code for his gadget.
You will be given two integers <b>D</b> and <b>Q</b>, you have to print the value of <b>D/Q</b> rounded down .The input contains two space- separated integers depicting the values of D and Q.
Constraints:-
0 <= D, Q <= 100Print the values of D/Q if the value can be calculated else print -1 if it is undefined.
Note:- Remember division by 0 is an undefined value that will give runtime error in your program.Sample Input:-
9 3
Sample Output:-
3
Sample Input:-
8 5
Sample Output:-
1
Explanation:-
8/5 = 1.6 = 1(floor), I have written this Solution Code: #include <iostream>
using namespace std;
int main(){
int n,m;
cin>>n>>m;
if(m==0){cout<<-1;return 0;}
cout<<n/m;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Some Data types are given below:-
Integer
Long
float
Double
char
Your task is to take input in the given format and print them in the same order.You don't have to worry about input, you just have to complete the function <b>printDataTypes()</b>Print each element in a new line in the same order as the input.
Note:- <b>Print float round off to two decimal places and double to 4 decimal places.</b>Sample Input:-
2
2312351235
1.21
543.1321
c
Sample Output:-
2
2312351235
1.21
543.1321
c, I have written this Solution Code: static void printDataTypes(int a, long b, float c, double d, char e)
{
System.out.println(a);
System.out.println(b);
System.out.printf("%.2f",c);
System.out.println();
System.out.printf("%.4f",d);
System.out.println();
System.out.println(e);
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Some Data types are given below:-
Integer
Long
float
Double
char
Your task is to take input in the given format and print them in the same order.You don't have to worry about input, you just have to complete the function <b>printDataTypes()</b>Print each element in a new line in the same order as the input.
Note:- <b>Print float round off to two decimal places and double to 4 decimal places.</b>Sample Input:-
2
2312351235
1.21
543.1321
c
Sample Output:-
2
2312351235
1.21
543.1321
c, I have written this Solution Code: void printDataTypes(int a, long long b, float c, double d, char e){
cout<<a<<endl;
cout<<b<<endl;
cout <<fixed<< std::setprecision(2) << c << '\n';
cout <<fixed<< std::setprecision(4) << d << '\n';
cout<<e<<endl;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Some Data types are given below:-
Integer
Long
float
Double
char
Your task is to take input in the given format and print them in the same order.You don't have to worry about input, you just have to complete the function <b>printDataTypes()</b>Print each element in a new line in the same order as the input.
Note:- <b>Print float round off to two decimal places and double to 4 decimal places.</b>Sample Input:-
2
2312351235
1.21
543.1321
c
Sample Output:-
2
2312351235
1.21
543.1321
c, I have written this Solution Code: a=int(input())
b=int(input())
x=float(input())
g = "{:.2f}".format(x)
d=float(input())
e = "{:.4f}".format(d)
u=input()
print(a)
print(b)
print(g)
print(e)
print(u), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a Doubly linked list consisting of <b>N</b> nodes and given a number <b>K</b>. The task is to delete the Kth node from the end of the linked list.<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>deleteElement()</b> that takes head node and the position K as parameter.
Constraints:
1 <=K<=N<= 1000
1 <=value<= 1000Return the head of the modified Doubly linked listInput:
5 3
1 2 3 4 5
Output:
1 2 4 5
Explanation:
After deleting 3rd node from the end of the linked list, 3 will be deleted and the list will become 1, 2, 4, 5., I have written this Solution Code: public static Node deleteElement(Node head,int k) {
int cnt=0;
Node temp=head;
while(temp!=null){
cnt++;
temp=temp.next;
}
k=cnt-k;
if(k==0){head=head.next;
head.prev=null;
return head;}
temp=head;
int i=0;
while(i!=k-1){
temp=temp.next;
i++;
}
temp.next=temp.next.next;
if(k!=cnt-1){temp.next.prev=temp;}
return head;
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A[] of size N, containing positive integers. You need to sort the elements of array using Insertion sort algorithm.First line of the input denotes number of test cases T. First line of the testcase is the size of array N and second line consists of array elements separated by space.
Constraints:
1 <= T <= 100
1 <= N <= 10^3
1 <= A[i] <= 10^3For each testcase print the sorted array in a new line.Input:
2
5
4 1 3 9 7
10
10 9 8 7 6 5 4 3 2 1
Output:
1 3 4 7 9
1 2 3 4 5 6 7 8 9 10
Explanation:
Testcase 1: The array after perfoming insertion sort: 1 3 4 7 9.
Testcase 2: The array after performing insertion sort: 1 2 3 4 5 6 7 8 9 10., I have written this Solution Code: // arr is unsorted array
// n is the number of elements in the array
function insertionSort(arr, n) {
// write code here
// do not console.log the answer
// return sorted array
return arr.sort((a, b) => a - b)
}
, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A[] of size N, containing positive integers. You need to sort the elements of array using Insertion sort algorithm.First line of the input denotes number of test cases T. First line of the testcase is the size of array N and second line consists of array elements separated by space.
Constraints:
1 <= T <= 100
1 <= N <= 10^3
1 <= A[i] <= 10^3For each testcase print the sorted array in a new line.Input:
2
5
4 1 3 9 7
10
10 9 8 7 6 5 4 3 2 1
Output:
1 3 4 7 9
1 2 3 4 5 6 7 8 9 10
Explanation:
Testcase 1: The array after perfoming insertion sort: 1 3 4 7 9.
Testcase 2: The array after performing insertion sort: 1 2 3 4 5 6 7 8 9 10., I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void insertionSort(int[] arr){
for(int i = 0; i < arr.length-1; i++){
for(int j = i+1; j < arr.length; j++){
if(arr[i] > arr[j]){
int temp = arr[j];
arr[j] = arr[i];
arr[i] = temp;
}
}
}
}
public static void main (String[] args) {
Scanner scan = new Scanner(System.in);
int T = scan.nextInt();
while(T > 0){
int n = scan.nextInt();
int arr[] = new int[n];
for(int i = 0; i<n; i++){
arr[i] = scan.nextInt();
}
insertionSort(arr);
for(int i = 0; i<n; i++){
System.out.print(arr[i] + " ");
}
System.out.println();
T--;
System.gc();
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A[] of size N, containing positive integers. You need to sort the elements of array using Insertion sort algorithm.First line of the input denotes number of test cases T. First line of the testcase is the size of array N and second line consists of array elements separated by space.
Constraints:
1 <= T <= 100
1 <= N <= 10^3
1 <= A[i] <= 10^3For each testcase print the sorted array in a new line.Input:
2
5
4 1 3 9 7
10
10 9 8 7 6 5 4 3 2 1
Output:
1 3 4 7 9
1 2 3 4 5 6 7 8 9 10
Explanation:
Testcase 1: The array after perfoming insertion sort: 1 3 4 7 9.
Testcase 2: The array after performing insertion sort: 1 2 3 4 5 6 7 8 9 10., I have written this Solution Code: #include "bits/stdc++.h"
#pragma GCC optimize "03"
using namespace std;
#define int long long int
#define ld long double
#define pi pair<int, int>
#define pb push_back
#define fi first
#define se second
#define IOS ios::sync_with_stdio(false); cin.tie(0); cout.tie(0)
#ifndef LOCAL
#define endl '\n'
#endif
const int N = 2e5 + 5;
const int mod = 1e9 + 7;
const int inf = 1e9 + 9;
int a[N];
signed main() {
IOS;
int t; cin >> t;
while(t--){
int n; cin >> n;
for(int i = 1; i <= n; i++)
cin >> a[i];
sort(a + 1, a + n + 1);
for(int i = 1; i <= n; i++)
cout << a[i] << " ";
cout << endl;
}
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A[] of size N, containing positive integers. You need to sort the elements of array using Insertion sort algorithm.First line of the input denotes number of test cases T. First line of the testcase is the size of array N and second line consists of array elements separated by space.
Constraints:
1 <= T <= 100
1 <= N <= 10^3
1 <= A[i] <= 10^3For each testcase print the sorted array in a new line.Input:
2
5
4 1 3 9 7
10
10 9 8 7 6 5 4 3 2 1
Output:
1 3 4 7 9
1 2 3 4 5 6 7 8 9 10
Explanation:
Testcase 1: The array after perfoming insertion sort: 1 3 4 7 9.
Testcase 2: The array after performing insertion sort: 1 2 3 4 5 6 7 8 9 10., I have written this Solution Code: def InsertionSort(arr):
arr.sort()
return arr, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara is giving a True False exam consisting of 10 questions. In which she knows that exactly X of the given questions are True and the rest are false (thanks to her friend) but she does not know the order so she randomly marks X questions true and rest False.
Given the value of X, your task is to tell Sara what will be the minimum marks she can get<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>minimumMarks()</b> that takes integer X as argument.
Constraints:-
0 <= X <= 10Return the minimum marks she can get.Sample Input:-
X = 3
Sample Output:-
4
Sample Input:-
10
Sample Output:-
10, I have written this Solution Code:
int minimumMarks(int X){
if(X>5){
return 10-2*(10-X);
}
return 10-2*X;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara is giving a True False exam consisting of 10 questions. In which she knows that exactly X of the given questions are True and the rest are false (thanks to her friend) but she does not know the order so she randomly marks X questions true and rest False.
Given the value of X, your task is to tell Sara what will be the minimum marks she can get<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>minimumMarks()</b> that takes integer X as argument.
Constraints:-
0 <= X <= 10Return the minimum marks she can get.Sample Input:-
X = 3
Sample Output:-
4
Sample Input:-
10
Sample Output:-
10, I have written this Solution Code: def minimumMarks(X):
if X>5:
return 10-2*(10-X)
return 10-2*(X)
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara is giving a True False exam consisting of 10 questions. In which she knows that exactly X of the given questions are True and the rest are false (thanks to her friend) but she does not know the order so she randomly marks X questions true and rest False.
Given the value of X, your task is to tell Sara what will be the minimum marks she can get<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>minimumMarks()</b> that takes integer X as argument.
Constraints:-
0 <= X <= 10Return the minimum marks she can get.Sample Input:-
X = 3
Sample Output:-
4
Sample Input:-
10
Sample Output:-
10, I have written this Solution Code: static int minimumMarks(int X){
if(X>5){
return 10-2*(10-X);
}
return 10-2*X;
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara is giving a True False exam consisting of 10 questions. In which she knows that exactly X of the given questions are True and the rest are false (thanks to her friend) but she does not know the order so she randomly marks X questions true and rest False.
Given the value of X, your task is to tell Sara what will be the minimum marks she can get<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>minimumMarks()</b> that takes integer X as argument.
Constraints:-
0 <= X <= 10Return the minimum marks she can get.Sample Input:-
X = 3
Sample Output:-
4
Sample Input:-
10
Sample Output:-
10, I have written this Solution Code:
int minimumMarks(int X){
if(X>5){
return 10-2*(10-X);
}
return 10-2*X;
}
, In this Programming Language: C, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of integers of size N, your task is to find the maximum parity index of this array.
<b>Parity Index is the maximum difference between two indices i and j (1 <= i < j <= N) of an array A such that A<sub>i</sub> < A<sub>j</sub>.</b>The first line contains a single integer N, next line contains N space-separated integers depicting the values of the array.
<b>Constraints:-</b>
1 < = N < = 10<sup>5</sup>
1 < = Arr[i] < = 10<sup>5</sup>Print the maximum value of <b>j- i</b> under the given condition, if no pair satisfies the condition print -1.Sample Input 1:-
5
1 2 3 4 5
Sample Output 1:-
4
Sample Input 2:-
5
5 4 3 2 1
Sample Output 2:-
-1
<b>Explanation 1:</b>
The maximum difference of j<sub>th</sub> - i<sub>th</sub> index is 4:(4<sub>th</sub> - 0<sub>th</sub>), also arr[4] > arr[0]
, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
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') {
if (cnt != 0) {
break;
}
else {
continue;
}
}
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();
}
}
public static void main (String[] args)throws IOException {
Reader sc = new Reader();
int N = sc.nextInt();
int[] arr = new int[N];
for(int i=0;i<N;i++){
arr[i] = sc.nextInt();
}
int max=0;
if(arr[0]<arr[N-1])
System.out.print(N-1);
else{
for(int i=0;i<N-1;i++){
int j = N-1;
while(j>i){
if(arr[i]<arr[j]){
if(max<j-i){
max = j-i;
} break;
}
j--;
}
if(i==j)
break;
if(j==N-1)
break;
}
if(max==0)
System.out.print("-1");
else
System.out.print(max);
}
}
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of integers of size N, your task is to find the maximum parity index of this array.
<b>Parity Index is the maximum difference between two indices i and j (1 <= i < j <= N) of an array A such that A<sub>i</sub> < A<sub>j</sub>.</b>The first line contains a single integer N, next line contains N space-separated integers depicting the values of the array.
<b>Constraints:-</b>
1 < = N < = 10<sup>5</sup>
1 < = Arr[i] < = 10<sup>5</sup>Print the maximum value of <b>j- i</b> under the given condition, if no pair satisfies the condition print -1.Sample Input 1:-
5
1 2 3 4 5
Sample Output 1:-
4
Sample Input 2:-
5
5 4 3 2 1
Sample Output 2:-
-1
<b>Explanation 1:</b>
The maximum difference of j<sub>th</sub> - i<sub>th</sub> index is 4:(4<sub>th</sub> - 0<sub>th</sub>), also arr[4] > arr[0]
, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define int long long
/* For a given array arr[],
returns the maximum j – i such that
arr[j] > arr[i] */
int maxIndexDiff(int arr[], int n)
{
int maxDiff;
int i, j;
int *LMin = new int[(sizeof(int) * n)];
int *RMax = new int[(sizeof(int) * n)];
/* Construct LMin[] such that
LMin[i] stores the minimum value
from (arr[0], arr[1], ... arr[i]) */
LMin[0] = arr[0];
for (i = 1; i < n; ++i)
LMin[i] = min(arr[i], LMin[i - 1]);
/* Construct RMax[] such that
RMax[j] stores the maximum value from
(arr[j], arr[j+1], ..arr[n-1]) */
RMax[n - 1] = arr[n - 1];
for (j = n - 2; j >= 0; --j)
RMax[j] = max(arr[j], RMax[j + 1]);
/* Traverse both arrays from left to right
to find optimum j - i. This process is similar to
merge() of MergeSort */
i = 0, j = 0, maxDiff = -1;
while (j < n && i < n)
{
if (LMin[i] < RMax[j])
{
maxDiff = max(maxDiff, j - i);
j = j + 1;
}
else
i = i + 1;
}
return maxDiff;
}
// Driver Code
signed main()
{
int n;
cin>>n;
int a[n];
for(int i=0;i<n;i++){
cin>>a[i];
}
int maxDiff = maxIndexDiff(a, n);
cout << maxDiff;
return 0;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of integers of size N, your task is to find the maximum parity index of this array.
<b>Parity Index is the maximum difference between two indices i and j (1 <= i < j <= N) of an array A such that A<sub>i</sub> < A<sub>j</sub>.</b>The first line contains a single integer N, next line contains N space-separated integers depicting the values of the array.
<b>Constraints:-</b>
1 < = N < = 10<sup>5</sup>
1 < = Arr[i] < = 10<sup>5</sup>Print the maximum value of <b>j- i</b> under the given condition, if no pair satisfies the condition print -1.Sample Input 1:-
5
1 2 3 4 5
Sample Output 1:-
4
Sample Input 2:-
5
5 4 3 2 1
Sample Output 2:-
-1
<b>Explanation 1:</b>
The maximum difference of j<sub>th</sub> - i<sub>th</sub> index is 4:(4<sub>th</sub> - 0<sub>th</sub>), also arr[4] > arr[0]
, I have written this Solution Code: n=int(input())
arr=list(map(int,input().split()))
rightMax = [0] * n
rightMax[n - 1] = arr[n - 1]
for i in range(n - 2, -1, -1):
rightMax[i] = max(rightMax[i + 1], arr[i])
maxDist = -2**31
i = 0
j = 0
while (i < n and j < n):
if (rightMax[j] >= arr[i]):
maxDist = max(maxDist, j - i)
j += 1
else:
i += 1
if maxDist==0:
maxDist=-1
print(maxDist), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: During the elections, Bob is in charge of conducting voting in his village, but the EVM system malfunctioned, and there was a long line of voters waiting outside to vote. The following is how the Advanced EVM Machine works.
Each time when a voter scans his VoterId Card and votes for the party of his choice, the Voter's id and Party Name are registered in the background, and if the same voter votes again, the EVM does not capture his vote, then the vote is skipped and the vote given the first time is used.
Now that you are Bob's best mate, you can't bear to see him in such a strained situation when outside voters are being very aggressive and screaming at him. Can you easily write a piece of code to save your friend's life while Bob is busy calming down the outside situation?The number N (1 ≤ N ≤ 1e5) appears on the first line. The queries to the machine are included in the next n lines. Each request consists of two strings and is written on a non-empty line. The first string is a Voter Card Id, and the second string is the Party Name, all of which are at most 32 characters long both upper case and lower case possible.
<b>Constraints</b>
1 ≤ N ≤ 100000
1 ≤ Voter Id length ≤ 40
1 ≤ PartyName length ≤ 32Output a single line indicating which party has won and by how many votes. In case of a draw between parties print all the parties with the same winning vote, in lexographically increasing order on basis of Party Name.Sample Input
4
12678345 BJP
57891082 congress
12678345 AAP
65489 TMC
Sample Output
BJP 1
Congress 1
TMC 1
<b>Explanation :</b>
As Winning Parties here as BJP, Congress, TMC with 1 vote(s) each, but AAP vote is not considered because the same VoterId - 12678345 has done a vote again., I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws IOException{
HashSet<String> set = new HashSet<String> ();
HashMap<String, Long> map = new HashMap<>();
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int n = Integer.parseInt(br.readLine());
long max = 1L;
for(int i=0;i<n;i++){
String[] num = br.readLine().split(" ");
String voterID = num[0];
String partyName = num[1];
if(!set.contains(voterID)){
set.add(voterID);
if(!map.containsKey(partyName)){
map.put(partyName, 1L);
}
else{
map.put(partyName, map.get(partyName)+1);
max = Math.max(max, map.get(partyName));
}
}
}
ArrayList<String> winningParty = new ArrayList<>();
for (Map.Entry<String,Long> entry : map.entrySet()){
if(entry.getValue()==max) winningParty.add(entry.getKey());
}
Collections.sort(winningParty);
for(int i=0;i<winningParty.size();i++){
System.out.println(winningParty.get(i)+" "+max);
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: During the elections, Bob is in charge of conducting voting in his village, but the EVM system malfunctioned, and there was a long line of voters waiting outside to vote. The following is how the Advanced EVM Machine works.
Each time when a voter scans his VoterId Card and votes for the party of his choice, the Voter's id and Party Name are registered in the background, and if the same voter votes again, the EVM does not capture his vote, then the vote is skipped and the vote given the first time is used.
Now that you are Bob's best mate, you can't bear to see him in such a strained situation when outside voters are being very aggressive and screaming at him. Can you easily write a piece of code to save your friend's life while Bob is busy calming down the outside situation?The number N (1 ≤ N ≤ 1e5) appears on the first line. The queries to the machine are included in the next n lines. Each request consists of two strings and is written on a non-empty line. The first string is a Voter Card Id, and the second string is the Party Name, all of which are at most 32 characters long both upper case and lower case possible.
<b>Constraints</b>
1 ≤ N ≤ 100000
1 ≤ Voter Id length ≤ 40
1 ≤ PartyName length ≤ 32Output a single line indicating which party has won and by how many votes. In case of a draw between parties print all the parties with the same winning vote, in lexographically increasing order on basis of Party Name.Sample Input
4
12678345 BJP
57891082 congress
12678345 AAP
65489 TMC
Sample Output
BJP 1
Congress 1
TMC 1
<b>Explanation :</b>
As Winning Parties here as BJP, Congress, TMC with 1 vote(s) each, but AAP vote is not considered because the same VoterId - 12678345 has done a vote again., I have written this Solution Code: /**
* author: tourist1256
* created: 2021-4-2 13:58:11
**/
#include <algorithm>
#include <array>
#include <bitset>
#include <cassert>
#include <chrono>
#include <cmath>
#include <cstring>
#include <functional>
#include <iomanip>
#include <iostream>
#include <map>
#include <numeric>
#include <queue>
#include <random>
#include <set>
#include <vector>
using namespace std;
double pi = acos(-1);
#define tezi \
ios_base::sync_with_stdio(false); \
cin.tie(0); \
cout.tie(0);
#define F_OR(i, a, b, s) \
for (ll i = (a); (s) > 0 ? i < (b) : i >= (b); i += (s))
#define F_OR1(e) F_OR(i, 0, e, 1)
#define F_OR2(i, e) F_OR(i, 0, e, 1)
#define F_OR3(i, b, e) F_OR(i, b, e, 1)
#define F_OR4(i, b, e, s) F_OR(i, b, e, s)
#define GET5(a, b, c, d, e, ...) e
#define F_ORC(...) GET5(__VA_ARGS__, F_OR4, F_OR3, F_OR2, F_OR1)
#define FOR(...) \
F_ORC(__VA_ARGS__) \
(__VA_ARGS__)
#define EACH(x, a) for (auto &x : a)
#define itr(it, a) for (auto it = a.begin(); it != a.end(); it++)
#define fill(a, b) memset(a, b, sizeof(a))
#define LOCAL
#define time cout << (0.1 * clock()) / CLOCKS_PER_SEC << endl;
#define countBits(x) __builtin_popcount(ll(x))
#define countZeroesAtBegin(x) __builtin_clz(ll(x))
#define countZeroesAtEnd(x) __builtin_ctz(ll(x))
#define last(x) x[x.end() - x.begin() - 1]
#define pb push_back
#define bg begin
#define ff first
#define ss second
#define pi 3.1415926535897932384626
#define infll 0x3f3f3f3f3f3f3f3f
using ll = long long;
using pl = pair<ll, ll>;
using vll = vector<ll>;
using vpl = vector<pl>;
using mat = vector<vll>;
const ll MAX = 1e5;
const ll mod = 1e9 + 7;
#define all(x) x.begin(), x.end()
#define clr(x) memset(x, 0, sizeof(x))
#define sortall(x) sort(all(x))
static int row[4] = {-1, 0, 0, 1};
static int col[4] = {0, -1, 1, 0};
template <typename T>
ostream &operator<<(ostream &os, const vector<T> &v) {
EACH(ele, v) { cout << ele << " "; }
return os;
}
template <typename T, typename U>
ostream &operator<<(ostream &os, const pair<T, U> &p) {
return os << '(' << p.ff << "," << p.ss << ')';
}
template <typename T>
istream &operator>>(istream &in, vector<T> &v) {
EACH(ele, v) { cin >> ele; }
return in;
}
mt19937 rng(chrono::high_resolution_clock::now().time_since_epoch().count());
void _runtime_terror_() {
ll N, winningVotes = -infll;
cin >> N;
string voterId, partyName;
map<string, ll> database;
map<string, bool> voters;
for (int i = 0; i < N; i++) {
cin >> voterId >> partyName;
if (voters.find(voterId) == voters.end()) {
voters[voterId] = true;
database[partyName] += 1;
winningVotes = max(database[partyName], winningVotes);
}
}
vector<pair<string, ll>> ans;
for (auto &it : database) {
if (winningVotes == it.ss) {
ans.pb({it.ff, it.ss});
}
}
for (auto &it : ans) {
cout << it.ff << " " << it.ss << "\n";
}
}
int main() {
#ifndef ONLINE_JUDGE
freopen("input.txt", "r", stdin);
freopen("output.txt", "w", stdout);
freopen("debug.txt", "w", stderr);
#endif
tezi;
int tc = 1;
// cin >> tc;
while (tc--) {
_runtime_terror_();
}
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given four positive integers A, B, C, and D, can a rectangle have these integers as its sides?The input consists of a single line containing four space-separated integers A, B, C and D.
<b> Constraints: </b>
1 ≤ A, B, C, D ≤ 1000Output "Yes" if a rectangle is possible; otherwise, "No" (without quotes).Sample Input 1:
3 4 4 3
Sample Output 1:
Yes
Sample Explanation 1:
A rectangle is possible with 3, 4, 3, and 4 as sides in clockwise order.
Sample Input 2:
3 4 3 5
Sample Output 2:
No
Sample Explanation 2:
No rectangle can be made with these side lengths., I have written this Solution Code: #include<iostream>
using namespace std;
int main(){
int a,b,c,d;
cin >> a >> b >> c >> d;
if((a==c) &&(b==d)){
cout << "Yes\n";
}else if((a==b) && (c==d)){
cout << "Yes\n";
}else if((a==d) && (b==c)){
cout << "Yes\n";
}else{
cout << "No\n";
}
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: <HTML>
  <HEAD>
   <TITLE>IS THIS PROBLEM IN CEPHEUS 2022?</TITLE>
  </HEAD>
  <BODY>
   <H1> YOU WILL FIND SOMETHING USEFUL FROM THIS PROBLEM STATEMENT</H1>
   <H2> LAUGHS IN SITH LORD </H2>
  </BODY>
</HTML>The only line of input contains N, a random number between 1 and 2022.
The above number is irrelevant to the problem statement.<cepheus>Crack the output!</cepheus>Sample Input:
2022
Sample Output:
YES
, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) {
System.out.print("YES");
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: <HTML>
  <HEAD>
   <TITLE>IS THIS PROBLEM IN CEPHEUS 2022?</TITLE>
  </HEAD>
  <BODY>
   <H1> YOU WILL FIND SOMETHING USEFUL FROM THIS PROBLEM STATEMENT</H1>
   <H2> LAUGHS IN SITH LORD </H2>
  </BODY>
</HTML>The only line of input contains N, a random number between 1 and 2022.
The above number is irrelevant to the problem statement.<cepheus>Crack the output!</cepheus>Sample Input:
2022
Sample Output:
YES
, I have written this Solution Code: #include<bits/stdc++.h>
using namespace std;
signed main(){
int n; cin>>n;
cout<<"YES";
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: <HTML>
  <HEAD>
   <TITLE>IS THIS PROBLEM IN CEPHEUS 2022?</TITLE>
  </HEAD>
  <BODY>
   <H1> YOU WILL FIND SOMETHING USEFUL FROM THIS PROBLEM STATEMENT</H1>
   <H2> LAUGHS IN SITH LORD </H2>
  </BODY>
</HTML>The only line of input contains N, a random number between 1 and 2022.
The above number is irrelevant to the problem statement.<cepheus>Crack the output!</cepheus>Sample Input:
2022
Sample Output:
YES
, I have written this Solution Code: print ("YES");, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number N for each i (1 < = i < = N), you have to print the number except :-
For each multiple of 3, print "Newton" instead of the number.
For each multiple of 5, print "School" instead of the number.
For numbers that are multiples of both 3 and 5, print "NewtonSchool" instead of the number.The first line of the input contains N.
<b>Constraints</b>
1 < = N < = 1000
Print N space separated number or Newton School according to the condition.Sample Input:-
3
Sample Output:-
1 2 Newton
Sample Input:-
5
Sample Output:-
1 2 Newton 4 School, I have written this Solution Code: n=int(input())
for i in range(1,n+1):
if i%3==0 and i%5==0:
print("NewtonSchool",end=" ")
elif i%3==0:
print("Newton",end=" ")
elif i%5==0:
print("School",end=" ")
else:
print(i,end=" "), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number N for each i (1 < = i < = N), you have to print the number except :-
For each multiple of 3, print "Newton" instead of the number.
For each multiple of 5, print "School" instead of the number.
For numbers that are multiples of both 3 and 5, print "NewtonSchool" instead of the number.The first line of the input contains N.
<b>Constraints</b>
1 < = N < = 1000
Print N space separated number or Newton School according to the condition.Sample Input:-
3
Sample Output:-
1 2 Newton
Sample Input:-
5
Sample Output:-
1 2 Newton 4 School, I have written this Solution Code: import java.util.*;
import java.lang.*;
import java.io.*;
class Main
{
static void NewtonSchool(int n){
for(int i=1;i<=n;i++){
if(i%3==0 && i%5==0){System.out.print("NewtonSchool ");}
else if(i%5==0){System.out.print("School ");}
else if(i%3==0){System.out.print("Newton ");}
else{System.out.print(i+" ");}
}
}
public static void main (String[] args) throws java.lang.Exception
{
Scanner sc = new Scanner(System.in);
int x= sc.nextInt();
NewtonSchool(x);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number <b>N</b>, find the count of pairs(x, y) such that the absolute value of sum of x<sup>3</sup> and y<sup>3</sup> is equal to N, i. e |x<sup>3</sup> + y<sup>3</sup>|=N<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>Pairs()</b> that takes the integer N as parameter
<b>Constraints</b>
1 <= N <= 10<sup>8</sup>Return the <b>count of pairs</b> which satisfies the given condition.Sample Input:-
8
Sample Output:-
2
Explanation:-
The required pairs are (0, -2), (0, 2)
Sample Input:-
3
Sample Output:-
0, I have written this Solution Code: def Pairs(N):
cnt=0
for i in range(-1000, 1001):
for j in range (i,1001):
a=i*i*i
b=j*j*j
if abs(a+b)==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: Given a number <b>N</b>, find the count of pairs(x, y) such that the absolute value of sum of x<sup>3</sup> and y<sup>3</sup> is equal to N, i. e |x<sup>3</sup> + y<sup>3</sup>|=N<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>Pairs()</b> that takes the integer N as parameter
<b>Constraints</b>
1 <= N <= 10<sup>8</sup>Return the <b>count of pairs</b> which satisfies the given condition.Sample Input:-
8
Sample Output:-
2
Explanation:-
The required pairs are (0, -2), (0, 2)
Sample Input:-
3
Sample Output:-
0, I have written this Solution Code: int Pairs(long N){
int cnt=0;
long int sum=0,a,b,i,j;
for(i=-6000 ;i<=6000;i++){
for(j=i; j<=6000;j++){
a=i*i*i;
b=j*j*j;
if((a+b)>0 && (a+b)==N){
cnt++;}
else if((a+b)<0 && (a+b) ==(-N)){cnt++;}
}}
return cnt;
}
, In this Programming Language: C, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number <b>N</b>, find the count of pairs(x, y) such that the absolute value of sum of x<sup>3</sup> and y<sup>3</sup> is equal to N, i. e |x<sup>3</sup> + y<sup>3</sup>|=N<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>Pairs()</b> that takes the integer N as parameter
<b>Constraints</b>
1 <= N <= 10<sup>8</sup>Return the <b>count of pairs</b> which satisfies the given condition.Sample Input:-
8
Sample Output:-
2
Explanation:-
The required pairs are (0, -2), (0, 2)
Sample Input:-
3
Sample Output:-
0, I have written this Solution Code: public static int Pairs(int N){
int cnt=0;
long sum=0,a,b,i;
for(i=-6000 ;i<=6000;i++){
for(long j=i; j<=6000;j++){
a=i*i*i;
b=j*j*j;
if(Math.abs(a+b)==N){
cnt++;}
}
}
return cnt;
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number <b>N</b>, find the count of pairs(x, y) such that the absolute value of sum of x<sup>3</sup> and y<sup>3</sup> is equal to N, i. e |x<sup>3</sup> + y<sup>3</sup>|=N<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>Pairs()</b> that takes the integer N as parameter
<b>Constraints</b>
1 <= N <= 10<sup>8</sup>Return the <b>count of pairs</b> which satisfies the given condition.Sample Input:-
8
Sample Output:-
2
Explanation:-
The required pairs are (0, -2), (0, 2)
Sample Input:-
3
Sample Output:-
0, I have written this Solution Code:
int Pairs(long N){
int cnt=0;
long int sum=0,a,b,i,j;
for(i=-6000 ;i<=6000;i++){
for(j=i; j<=6000;j++){
a=i*i*i;
b=j*j*j;
if(abs(a+b)==N){
cnt++;}
}
}
return cnt;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of N elements where each element is either 1 or 0. You have to divide the array into maximum number of subarrays such that each element of the array is in exactly one subarray such that each subarray has equal number of 1's and 0's.First line of input contains N.
Second line of input contains N space separated elements of the array.
Constraints:
1 <= N <= 100000
0 <= elements of the array <= 1Print the single integer which is the maximum number of subarrays the array can be divided into. If it is not possible then print -1.Sample input 1
4
1 0 1 0
Sample output 1
2
Sample input 2
4
1 1 0 0
Sample output 2
1, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws Exception {
BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
int N = Integer.parseInt(reader.readLine());
StringTokenizer tokens = new StringTokenizer(reader.readLine());
int partitions = 0;
int c0 = 0, c1 = 0;
for(int i = 0; i < N; i++) {
int x = Integer.parseInt(tokens.nextToken());
if(x == 0) {
c0++;
} else {
c1++;
}
if(c0 > 0 || c1 > 0) {
if(c0 == c1) {
partitions++;
c0 = c1 = 0;
}
}
}
if(c1 > 0 || c0 > 0) {
partitions = -1;
}
System.out.println(partitions);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of N elements where each element is either 1 or 0. You have to divide the array into maximum number of subarrays such that each element of the array is in exactly one subarray such that each subarray has equal number of 1's and 0's.First line of input contains N.
Second line of input contains N space separated elements of the array.
Constraints:
1 <= N <= 100000
0 <= elements of the array <= 1Print the single integer which is the maximum number of subarrays the array can be divided into. If it is not possible then print -1.Sample input 1
4
1 0 1 0
Sample output 1
2
Sample input 2
4
1 1 0 0
Sample output 2
1, I have written this Solution Code: n = int(input())
a = input().split()
o,z = 0,0
cnt = 0
for i in a:
if i == '1':
o += 1
else:
z +=1
if o == z:
cnt += 1
if o == z:
print(cnt)
else:
print(-1), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of N elements where each element is either 1 or 0. You have to divide the array into maximum number of subarrays such that each element of the array is in exactly one subarray such that each subarray has equal number of 1's and 0's.First line of input contains N.
Second line of input contains N space separated elements of the array.
Constraints:
1 <= N <= 100000
0 <= elements of the array <= 1Print the single integer which is the maximum number of subarrays the array can be divided into. If it is not possible then print -1.Sample input 1
4
1 0 1 0
Sample output 1
2
Sample input 2
4
1 1 0 0
Sample output 2
1, I have written this Solution Code: #include<bits/stdc++.h>
using namespace std;
#define ll long long
#define VV vector
#define pb push_back
#define bitc __builtin_popcountll
#define m_p make_pair
#define infi 1e18+1
#define eps 0.000000000001
#define fastio ios_base::sync_with_stdio(false);cin.tie(NULL);
string char_to_str(char c){string tem(1,c);return tem;}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
template<class T>//usage rand<long long>()
T rand() {
return uniform_int_distribution<T>()(rng);
}
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
typedef tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> oset;
// string to integer stoi()
// string to long long stoll()
// string.substr(position,length);
// integer to string to_string();
//////////////
auto clk=clock();
#define all(x) x.begin(),x.end()
#define S second
#define F first
#define sz(x) ((long long)x.size())
#define int long long
#define f80 __float128
#define pii pair<int,int>
/////////////
signed main()
{
fastio;
#ifdef ANIKET_GOYAL
freopen("inputf.in","r",stdin);
freopen("outputf.in","w",stdout);
#endif
int n;
cin>>n;
int c=0;
int ans=0;
for(int i=0;i<n;++i){
int x;
cin>>x;
if(x==0)
++c;
else
--c;
if(c==0)
++ans;
}
if(c==0){
cout<<ans;
}else
{
cout<<"-1";
}
#ifdef ANIKET_GOYAL
// cout<<endl<<endl<<endl<<endl<<"Time elapsed: "<<(double)(clock()-clk)/CLOCKS_PER_SEC<<endl;
#endif
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array <b>Arr[]</b> of size <b>N</b> as input, your task is to count the number of triplets Arr[i], Arr[j] and Arr[k] such that:-
i < j < k and the difference between every 2 elements of triplets is less than or equal to P i. e |Arr[i] - Arr[j]| <= P, |Arr[i] - Arr[k]| <= P and |Arr[j] - Arr[k]| <= PThe first line of input contains two space- separated integers N and P.
next line contains N space separated integers depicting the values of the Arr[].
Constraints:-
3 <= N <= 10<sup>5</sup>
1 <= Arr[i], P <= 10<sup>9</sup>
0 <= i <= N-1Return the count of triplets that satisfies the above conditions.Sample Input:-
5 4
1 3 2 5 9
Sample Output:-
4
Explanation:-
(1, 3, 2), (1, 3, 5), (1, 2, 5), (2, 3, 5) are the required triplets
Sample Input:-
5 3
1 8 4 2 9
Sample Output:-
1, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
String str1 = br.readLine();
String str2[] = str1.split(" ");
int n = Integer.parseInt(str2[0]);
int k = Integer.parseInt(str2[1]);
String str3 = br.readLine();
String str4[] = str3.split(" ");
long[] arr = new long[n];
for(int i = 0; i < n; ++i) {
arr[i] = Long.parseLong(str4[i]);
}
Arrays.sort(arr);
int i=0,j=2;
long ans=0;
while(j!=n){
if(i==j-1){j++;continue;}
if((arr[j]-arr[i])>k){i++;}
else{
int x = j-i;
ans+=(x*(x-1))/2;
j++;
}
}
System.out.print(ans);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array <b>Arr[]</b> of size <b>N</b> as input, your task is to count the number of triplets Arr[i], Arr[j] and Arr[k] such that:-
i < j < k and the difference between every 2 elements of triplets is less than or equal to P i. e |Arr[i] - Arr[j]| <= P, |Arr[i] - Arr[k]| <= P and |Arr[j] - Arr[k]| <= PThe first line of input contains two space- separated integers N and P.
next line contains N space separated integers depicting the values of the Arr[].
Constraints:-
3 <= N <= 10<sup>5</sup>
1 <= Arr[i], P <= 10<sup>9</sup>
0 <= i <= N-1Return the count of triplets that satisfies the above conditions.Sample Input:-
5 4
1 3 2 5 9
Sample Output:-
4
Explanation:-
(1, 3, 2), (1, 3, 5), (1, 2, 5), (2, 3, 5) are the required triplets
Sample Input:-
5 3
1 8 4 2 9
Sample Output:-
1, I have written this Solution Code: n, p = input().split(' ')
n = int(n)
p = int(p)
arr = input().split(' ')[:n]
for i in range(n):
arr[i] = int(arr[i])
arr.sort()
i = 0
k = 2
count = 0
while k!=n:
if i == k-1:
k = k + 1
continue
if (arr[k] - arr[i]) <= p:
count = count + (int)(((k-i)*(k-i-1))/2)
k = k + 1
else:
i = i + 1
print(count)
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array <b>Arr[]</b> of size <b>N</b> as input, your task is to count the number of triplets Arr[i], Arr[j] and Arr[k] such that:-
i < j < k and the difference between every 2 elements of triplets is less than or equal to P i. e |Arr[i] - Arr[j]| <= P, |Arr[i] - Arr[k]| <= P and |Arr[j] - Arr[k]| <= PThe first line of input contains two space- separated integers N and P.
next line contains N space separated integers depicting the values of the Arr[].
Constraints:-
3 <= N <= 10<sup>5</sup>
1 <= Arr[i], P <= 10<sup>9</sup>
0 <= i <= N-1Return the count of triplets that satisfies the above conditions.Sample Input:-
5 4
1 3 2 5 9
Sample Output:-
4
Explanation:-
(1, 3, 2), (1, 3, 5), (1, 2, 5), (2, 3, 5) are the required triplets
Sample Input:-
5 3
1 8 4 2 9
Sample Output:-
1, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define int long long
signed main(){
int n,p;
cin>>n>>p;
int a[n];
for(int i=0;i<n;i++){
cin>>a[i];
}
sort(a,a+n);
int i=0,j=2;
int ans=0;
while(j!=n){
if(i==j-1){j++;continue;}
if((a[j]-a[i])>p){i++;}
else{
int x = j-i;
ans+=(x*(x-1))/2;
j++;
}
}
cout<<ans;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Write a program to check whether a number is positive, negative or zero using switch case.The first line of the input contains the number
<b>Constraints</b>
-10<sup>9</sup> ≤ n ≤ 10<sup>9</sup>Print the single line wether it's "Positive", "Negative" or "Zero"Sample Input :
13
Sample Output :
Positive
Sample Input :
-13
Sample Output :
Negative, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
String s=br.readLine().toString();
int n=Integer.parseInt(s);
int ch=0;
if(n>0){
ch=1;
}
else if(n<0) ch=-1;
switch(ch){
case 1: System.out.println("Positive");break;
case 0: System.out.println("Zero");break;
case -1: System.out.println("Negative");break;
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Write a program to check whether a number is positive, negative or zero using switch case.The first line of the input contains the number
<b>Constraints</b>
-10<sup>9</sup> ≤ n ≤ 10<sup>9</sup>Print the single line wether it's "Positive", "Negative" or "Zero"Sample Input :
13
Sample Output :
Positive
Sample Input :
-13
Sample Output :
Negative, I have written this Solution Code: n = input()
if '-' in list(n):
print('Negative')
elif int(n) == 0 :
print('Zero')
else:
print('Positive'), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Write a program to check whether a number is positive, negative or zero using switch case.The first line of the input contains the number
<b>Constraints</b>
-10<sup>9</sup> ≤ n ≤ 10<sup>9</sup>Print the single line wether it's "Positive", "Negative" or "Zero"Sample Input :
13
Sample Output :
Positive
Sample Input :
-13
Sample Output :
Negative, I have written this Solution Code: #include <stdio.h>
int main()
{
int num;
scanf("%d", &num);
switch (num > 0)
{
// Num is positive
case 1:
printf("Positive");
break;
// Num is either negative or zero
case 0:
switch (num < 0)
{
case 1:
printf("Negative");
break;
case 0:
printf("Zero");
break;
}
break;
}
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a string <i>s</i> consisting of lowercase English letters. Find the number of different palindromes you can make by changing <b>exactly</b> one charecter from the string to some other lowercase English letter.The first and single line contains string <i>s</i> (1 ≤ |<i>s</i>| ≤ 10).Print the number of different palindromes you can make by changing <b>exactly</b> one charecter from the string to some other lowercase English letter. Sample Input 1
abbb
Sample Output 1
2
Explanation:
The possible palindromes are:
1. abba
2. bbbb
========================================================================
Sample Input 2
abba
Sample Output 2
0
, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws Exception {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
char str[] = br.readLine().toCharArray();
int ans = 0;
char arr[] = {'a','b','c','d','e','f','g','h','i','j','k','l','m','n','o','p','q','r','s','t','u','v','w','x','y','z'};
Set<String> set = new HashSet<>();
for(int i=0;i<str.length;i++){
char p = str[i];
for(char ch:arr){
if(ch==p) continue;
str[i] = ch;
if(isPallindrome(str)){
if(set.contains(String.valueOf(str))==false){
set.add(String.valueOf(str));
ans++;
}
}
str[i] = p;
}
}
System.out.println(ans);
}
static boolean isPallindrome(char[] str){
int i = 0;
int j = str.length-1;
while(i<j){
if(str[i]!=str[j]) return false;
i++;
j--;
}
return true;
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a string <i>s</i> consisting of lowercase English letters. Find the number of different palindromes you can make by changing <b>exactly</b> one charecter from the string to some other lowercase English letter.The first and single line contains string <i>s</i> (1 ≤ |<i>s</i>| ≤ 10).Print the number of different palindromes you can make by changing <b>exactly</b> one charecter from the string to some other lowercase English letter. Sample Input 1
abbb
Sample Output 1
2
Explanation:
The possible palindromes are:
1. abba
2. bbbb
========================================================================
Sample Input 2
abba
Sample Output 2
0
, I have written this Solution Code: n=input()
n=list(n)
ln=len(n)
cnt=0
for i in range(ln//2):
if not(n[i]==n[ln-i-1]):
cnt+=1
if(cnt==1):
print(2)
elif(cnt==0 and ln%2==1):
print(25)
else:
print(0), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a string <i>s</i> consisting of lowercase English letters. Find the number of different palindromes you can make by changing <b>exactly</b> one charecter from the string to some other lowercase English letter.The first and single line contains string <i>s</i> (1 ≤ |<i>s</i>| ≤ 10).Print the number of different palindromes you can make by changing <b>exactly</b> one charecter from the string to some other lowercase English letter. Sample Input 1
abbb
Sample Output 1
2
Explanation:
The possible palindromes are:
1. abba
2. bbbb
========================================================================
Sample Input 2
abba
Sample Output 2
0
, I have written this Solution Code: #include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
using namespace std;
using namespace __gnu_pbds;
template <typename T> using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
#define endl '\n'
#define pb push_back
#define ub upper_bound
#define lb lower_bound
#define fi first
#define se second
#define int long long
typedef long long ll;
typedef long double ld;
#define pii pair<int,int>
#define sz(x) ((ll)x.size())
#define fr(a,b,c) for(int a=b; a<=c; a++)
#define frev(a,b,c) for(int a=c; a>=b; a--)
#define rep(a,b,c) for(int a=b; a<c; a++)
#define trav(a,x) for(auto &a:x)
#define all(con) con.begin(),con.end()
#define done(x) {cout << x << endl;return;}
#define mini(x,y) x = min(x,y)
#define maxi(x,y) x = max(x,y)
const ll infl = 0x3f3f3f3f3f3f3f3fLL;
const int infi = 0x3f3f3f3f;
mt19937_64 mt(chrono::steady_clock::now().time_since_epoch().count());
//const int mod = 998244353;
const int mod = 1e9 + 7;
typedef vector<int> vi;
typedef vector<string> vs;
typedef vector<vector<int>> vvi;
typedef vector<pair<int, int>> vpii;
typedef map<int, int> mii;
typedef set<int> si;
typedef set<pair<int,int>> spii;
typedef queue<int> qi;
uniform_int_distribution<int> rng(0, 1e9);
// DEBUG FUNCTIONS START
void __print(int x) {cerr << x;}
void __print(double x) {cerr << x;}
void __print(long double x) {cerr << x;}
void __print(char x) {cerr << '\'' << x << '\'';}
void __print(const char *x) {cerr << '\"' << x << '\"';}
void __print(const string &x) {cerr << '\"' << x << '\"';}
void __print(bool x) {cerr << (x ? "true" : "false");}
template<typename T, typename V> void __print(const pair<T, V> &x) {cerr << '{'; __print(x.first); cerr << ','; __print(x.second); cerr << '}';}
template<typename T> void __print(const T &x) {int f = 0; cerr << '{'; for (auto &i: x) cerr << (f++ ? "," : ""), __print(i); cerr << "}";}
void deb() {cerr << "\n";}
template <typename T, typename... V> void deb(T t, V... v) {__print(t); if (sizeof...(v)) cerr << ", "; deb(v...);}
// DEBUG FUNCTIONS END
const int N = 2e5 + 5;
void solve(){
string s;
cin >> s;
int n = sz(s);
int x = 0;
rep(i, 0, n / 2){
x += s[i] != s[n - 1 - i];
}
if(x == 1){
cout << 2 << endl;
}
else if(x > 1){
cout << 0 << endl;
}
else{
if(n & 1){
cout << 25 << endl;
}
else{
cout << 0 << endl;
}
}
}
signed main(){
ios_base::sync_with_stdio(0), cin.tie(0);
cout << fixed << setprecision(15);
int t = 1;
//cin >> t;
while (t--)
solve();
return 0;
}
int powm(int a, int b){
int res = 1;
while (b) {
if (b & 1)
res = res * a % mod;
a = a * a % mod;
b >>= 1;
}
return res;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A[] of integers, sort the array according to frequency of elements. That is elements that have higher frequency come first. If frequencies of two elements are same, then smaller number comes first.The first line of input contains an integer T denoting the number of test cases. The description of T test cases follows. The first line of each test case contains a single integer N denoting the size of an array. The second line contains N space-separated integers A<sub>1</sub>, A<sub>2</sub>, ..., A<sub>N</sub> denoting the elements of the array.
Constraints:
1 ≤ T ≤ 100
1 ≤ N ≤10<sup>5</sup>
1 ≤A[i] ≤ 10<sup>4</sup>
The Sum of N over all test cases does not exceed 2*10<sup>5</sup>
For each testcase, in a new line, print each sorted array in a separate line. For each array its numbers should be separated by space.Sample Input:
2
5
5 5 4 6 4
5
9 9 9 2 5
Sample Output:
4 4 5 5 6
9 9 9 2 5
<b>Explanation:</b>
Test Case 1: The highest frequency here is 2. Both 5 and 4 have that frequency. Now since the frequencies are the same then smaller element comes first. So 4 4 comes first then comes 5 5. Finally comes 6.
The output is 4 4 5 5 6.
Test Case2: The highest frequency here is 3. The element 9 has the highest frequency. So 9 9 9 comes first. Now both 2 and 5 have same frequency. So we print smaller element first.
The output is 9 9 9 2 5., I have written this Solution Code: #include "bits/stdc++.h"
#pragma GCC optimize "03"
using namespace std;
#define int long long int
#define ld long double
#define pi pair<int, int>
#define pb push_back
#define fi first
#define se second
#define IOS ios::sync_with_stdio(false); cin.tie(0); cout.tie(0)
#ifndef LOCAL
#define endl '\n'
#endif
const int N = 1e5+ 5;
const int mod = 1e9 + 7;
const int inf = 1e9 + 9;
int a[N], f[N];
bool cmp(int a, int b){
if(f[a] == f[b])
return a < b;
return f[a] > f[b];
}
signed main() {
IOS;
int t; cin >> t;
while(t--){
memset(f, 0, sizeof f);
int n; cin >> n;
for(int i = 1; i <= n; i++){
cin >> a[i];
f[a[i]]++;
}
sort(a + 1, a + n + 1, cmp);
for(int i = 1; i <= n; i++)
cout << a[i] << " ";
cout << endl;
}
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A[] of integers, sort the array according to frequency of elements. That is elements that have higher frequency come first. If frequencies of two elements are same, then smaller number comes first.The first line of input contains an integer T denoting the number of test cases. The description of T test cases follows. The first line of each test case contains a single integer N denoting the size of an array. The second line contains N space-separated integers A<sub>1</sub>, A<sub>2</sub>, ..., A<sub>N</sub> denoting the elements of the array.
Constraints:
1 ≤ T ≤ 100
1 ≤ N ≤10<sup>5</sup>
1 ≤A[i] ≤ 10<sup>4</sup>
The Sum of N over all test cases does not exceed 2*10<sup>5</sup>
For each testcase, in a new line, print each sorted array in a separate line. For each array its numbers should be separated by space.Sample Input:
2
5
5 5 4 6 4
5
9 9 9 2 5
Sample Output:
4 4 5 5 6
9 9 9 2 5
<b>Explanation:</b>
Test Case 1: The highest frequency here is 2. Both 5 and 4 have that frequency. Now since the frequencies are the same then smaller element comes first. So 4 4 comes first then comes 5 5. Finally comes 6.
The output is 4 4 5 5 6.
Test Case2: The highest frequency here is 3. The element 9 has the highest frequency. So 9 9 9 comes first. Now both 2 and 5 have same frequency. So we print smaller element first.
The output is 9 9 9 2 5., I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) {
Scanner in = new Scanner (System.in);
int t = in.nextInt();
while (t-- > 0) {
int n = in.nextInt();
int arr[] = new int[n];
for (int i=0; i<n; i++) {
arr[i] = in.nextInt();
}
int farr[] = new int[10003];
for (int i=0; i<n; i++) {
farr[arr[i]] ++;
}
int max = 0;
int index = 0;
while (true) {
for (int i=0; i<10003; i++) {
if (max < farr[i]) {
max = farr[i];
index = i;
}
}
for (int i=0; i<max; i++) {
System.out.print(index + " ");
}
if (max == 0) {
break;
}
farr[index] = 0;
index =0;
max = 0;
}
System.out.println();
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A[] of integers, sort the array according to frequency of elements. That is elements that have higher frequency come first. If frequencies of two elements are same, then smaller number comes first.The first line of input contains an integer T denoting the number of test cases. The description of T test cases follows. The first line of each test case contains a single integer N denoting the size of an array. The second line contains N space-separated integers A<sub>1</sub>, A<sub>2</sub>, ..., A<sub>N</sub> denoting the elements of the array.
Constraints:
1 ≤ T ≤ 100
1 ≤ N ≤10<sup>5</sup>
1 ≤A[i] ≤ 10<sup>4</sup>
The Sum of N over all test cases does not exceed 2*10<sup>5</sup>
For each testcase, in a new line, print each sorted array in a separate line. For each array its numbers should be separated by space.Sample Input:
2
5
5 5 4 6 4
5
9 9 9 2 5
Sample Output:
4 4 5 5 6
9 9 9 2 5
<b>Explanation:</b>
Test Case 1: The highest frequency here is 2. Both 5 and 4 have that frequency. Now since the frequencies are the same then smaller element comes first. So 4 4 comes first then comes 5 5. Finally comes 6.
The output is 4 4 5 5 6.
Test Case2: The highest frequency here is 3. The element 9 has the highest frequency. So 9 9 9 comes first. Now both 2 and 5 have same frequency. So we print smaller element first.
The output is 9 9 9 2 5., I have written this Solution Code: from collections import defaultdict
test = int(input())
while(test != 0):
test -= 1
n = int(input())
arr = list(map(int, input().split()))
freq = defaultdict(list)
temp = [0]*(max(arr) + 1)
for i in arr:
temp[i] += 1
for i in range(1,len(temp)):
if(temp[i] == 0):
continue
freq[temp[i]].append(i)
for i in sorted(freq.keys(), reverse=True):
x = freq[i]
x = sorted(x*i)
print(*x, end=" ")
print(), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Shinchan and Kazama are standing in a horizontal line, Shinchan is standing at point A and Kazama is standing at point B. Kazama is very intelligent and recently he learned how to calculate the speed if the distance and time are given and now he wants to check if the formula he learned is correct or not So he starts running at a speed of S unit/s towards Shinhan and noted the time he reaches to Shinhan. Since Kazama is disturbed by Shinchan, can you calculate the time for him?The input contains three integers A, B, and S separated by spaces.
Constraints:-
1 <= A, B, V <= 1000
Note:- It is guaranteed that the calculated distance will always be divisible by V.Print the Time taken in seconds by Kazama to reach Shinchan.
Note:- Remember Distance can not be negativeSample Input:-
5 2 3
Sample Output:-
1
Explanation:-
Distance = 5-2 = 3, Speed = 3
Time = Distance/Speed
Sample Input:-
9 1 2
Sample Output:-
4, I have written this Solution Code: a, b, v = map(int, input().strip().split(" "))
c = abs(a-b)
t = c//v
print(t), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Shinchan and Kazama are standing in a horizontal line, Shinchan is standing at point A and Kazama is standing at point B. Kazama is very intelligent and recently he learned how to calculate the speed if the distance and time are given and now he wants to check if the formula he learned is correct or not So he starts running at a speed of S unit/s towards Shinhan and noted the time he reaches to Shinhan. Since Kazama is disturbed by Shinchan, can you calculate the time for him?The input contains three integers A, B, and S separated by spaces.
Constraints:-
1 <= A, B, V <= 1000
Note:- It is guaranteed that the calculated distance will always be divisible by V.Print the Time taken in seconds by Kazama to reach Shinchan.
Note:- Remember Distance can not be negativeSample Input:-
5 2 3
Sample Output:-
1
Explanation:-
Distance = 5-2 = 3, Speed = 3
Time = Distance/Speed
Sample Input:-
9 1 2
Sample Output:-
4, I have written this Solution Code: /* package codechef; // don't place package name! */
import java.util.*;
import java.lang.*;
import java.io.*;
/* Name of the class has to be "Main" only if the class is public. */
class Main
{
public static void main (String[] args) throws java.lang.Exception
{
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
int m = sc.nextInt();
int k = sc.nextInt();
System.out.print(Time(n,m,k));
}
static int Time(int A, int B, int S){
if(B>A){
return (B-A)/S;
}
return (A-B)/S;
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Some Data types are given below:-
Integer
Long
float
Double
char
Your task is to take input in the given format and print them in the same order.You don't have to worry about input, you just have to complete the function <b>printDataTypes()</b>Print each element in a new line in the same order as the input.
Note:- <b>Print float round off to two decimal places and double to 4 decimal places.</b>Sample Input:-
2
2312351235
1.21
543.1321
c
Sample Output:-
2
2312351235
1.21
543.1321
c, I have written this Solution Code: static void printDataTypes(int a, long b, float c, double d, char e)
{
System.out.println(a);
System.out.println(b);
System.out.printf("%.2f",c);
System.out.println();
System.out.printf("%.4f",d);
System.out.println();
System.out.println(e);
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Some Data types are given below:-
Integer
Long
float
Double
char
Your task is to take input in the given format and print them in the same order.You don't have to worry about input, you just have to complete the function <b>printDataTypes()</b>Print each element in a new line in the same order as the input.
Note:- <b>Print float round off to two decimal places and double to 4 decimal places.</b>Sample Input:-
2
2312351235
1.21
543.1321
c
Sample Output:-
2
2312351235
1.21
543.1321
c, I have written this Solution Code: void printDataTypes(int a, long long b, float c, double d, char e){
cout<<a<<endl;
cout<<b<<endl;
cout <<fixed<< std::setprecision(2) << c << '\n';
cout <<fixed<< std::setprecision(4) << d << '\n';
cout<<e<<endl;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Some Data types are given below:-
Integer
Long
float
Double
char
Your task is to take input in the given format and print them in the same order.You don't have to worry about input, you just have to complete the function <b>printDataTypes()</b>Print each element in a new line in the same order as the input.
Note:- <b>Print float round off to two decimal places and double to 4 decimal places.</b>Sample Input:-
2
2312351235
1.21
543.1321
c
Sample Output:-
2
2312351235
1.21
543.1321
c, I have written this Solution Code: a=int(input())
b=int(input())
x=float(input())
g = "{:.2f}".format(x)
d=float(input())
e = "{:.4f}".format(d)
u=input()
print(a)
print(b)
print(g)
print(e)
print(u), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a natural number N, your task is to print all the digits of the number in English words. The words have to separate by space and in lowercase English letters.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>Print_Digit()</b> that takes integer N as a parameter.
<b>Constraints:-</b>
1 ≤ N ≤ 10<sup>7</sup>Print the digits of the number as shown in the example.
<b>Note:-</b>
Print all digits in lowercase English lettersSample Input:-
1024
Sample Output:-
one zero two four
Sample Input:-
2
Sample Output:-
two, I have written this Solution Code: def Print_Digit(n):
dc = {1: "one", 2: "two", 3: "three", 4: "four",
5: "five", 6: "six", 7: "seven", 8: "eight", 9: "nine", 0: "zero"}
final_list = []
while (n > 0):
final_list.append(dc[int(n%10)])
n = int(n / 10)
for val in final_list[::-1]:
print(val, end=' '), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a natural number N, your task is to print all the digits of the number in English words. The words have to separate by space and in lowercase English letters.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>Print_Digit()</b> that takes integer N as a parameter.
<b>Constraints:-</b>
1 ≤ N ≤ 10<sup>7</sup>Print the digits of the number as shown in the example.
<b>Note:-</b>
Print all digits in lowercase English lettersSample Input:-
1024
Sample Output:-
one zero two four
Sample Input:-
2
Sample Output:-
two, I have written this Solution Code: class Solution {
public static void Print_Digits(int N){
if(N==0){return;}
Print_Digits(N/10);
int x=N%10;
if(x==1){System.out.print("one ");}
else if(x==2){System.out.print("two ");}
else if(x==3){System.out.print("three ");}
else if(x==4){System.out.print("four ");}
else if(x==5){System.out.print("five ");}
else if(x==6){System.out.print("six ");}
else if(x==7){System.out.print("seven ");}
else if(x==8){System.out.print("eight ");}
else if(x==9){System.out.print("nine ");}
else if(x==0){System.out.print("zero ");}
}
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given 2 non-negative integers m and n, find gcd(m, n)
GCD of 2 integers m and n is defined as the greatest integer g such that g is a divisor of both m and n. Both m and n fit in a 32 bit signed integer.
NOTE: DO NOT USE LIBRARY FUNCTIONSInput contains two space separated integers, m and n
Constraints:-
1 < = m, n < = 10^18Output the single integer denoting the gcd of the above integers.Sample Input:
6 9
Sample Output:
3
Sample Input:-
5 6
Sample Output:-
1, I have written this Solution Code: #include "bits/stdc++.h"
#pragma GCC optimize "03"
using namespace std;
#define int long long int
#define ld long double
#define pi pair<int, int>
#define pb push_back
#define fi first
#define se second
#define IOS ios::sync_with_stdio(false); cin.tie(0); cout.tie(0)
#ifndef LOCAL
#define endl '\n'
#endif
const int N = 2e5 + 5;
const int mod = 1e9 + 7;
const int inf = 1e9 + 9;
signed main() {
IOS;
int n, m;
cin >> n >> m;
cout << __gcd(n, m);
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given 2 non-negative integers m and n, find gcd(m, n)
GCD of 2 integers m and n is defined as the greatest integer g such that g is a divisor of both m and n. Both m and n fit in a 32 bit signed integer.
NOTE: DO NOT USE LIBRARY FUNCTIONSInput contains two space separated integers, m and n
Constraints:-
1 < = m, n < = 10^18Output the single integer denoting the gcd of the above integers.Sample Input:
6 9
Sample Output:
3
Sample Input:-
5 6
Sample Output:-
1, I have written this Solution Code: def hcf(a, b):
if(b == 0):
return a
else:
return hcf(b, a % b)
li= list(map(int,input().strip().split()))
print(hcf(li[0], li[1])), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given 2 non-negative integers m and n, find gcd(m, n)
GCD of 2 integers m and n is defined as the greatest integer g such that g is a divisor of both m and n. Both m and n fit in a 32 bit signed integer.
NOTE: DO NOT USE LIBRARY FUNCTIONSInput contains two space separated integers, m and n
Constraints:-
1 < = m, n < = 10^18Output the single integer denoting the gcd of the above integers.Sample Input:
6 9
Sample Output:
3
Sample Input:-
5 6
Sample Output:-
1, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws IOException{
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
String[] sp = br.readLine().trim().split(" ");
long m = Long.parseLong(sp[0]);
long n = Long.parseLong(sp[1]);
System.out.println(GCDAns(m,n));
}
private static long GCDAns(long m,long n){
if(m==0)return n;
if(n==0)return m;
return GCDAns(n%m,m);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: We have A cards, each of which has an integer 1 written on it. Similarly, we also have B cards with 0s and C cards with −1s.
We will pick up K among these cards. What is the maximum possible sum of the numbers written on the cards chosen?The input consists of 4 space separated integers as follows :
A B C K
<b>Constraints</b>
All values in input are integers.
0≤A, B, C
1≤K≤A+B+C≤2×10^9Print the maximum possible sum of the numbers written on the cards chosen.<b>Sample Input 1</b>
2 1 1 3
<b>Sample Output 1</b>
2
<b>Sample Input 2</b>
1 2 3 4
<b>Sample Output 2</b>
0
<b>Sample Input 3</b>
2000000000 0 0 2000000000
<b>Sample Output 3</b>
2000000000, I have written this Solution Code: #include<bits/stdc++.h>
using namespace std;
using ll = long long;
using P = pair<ll, ll>;
#define rep(i, n) for(ll i = 0; i < (ll)n; i++)
int main() {
int a, b, c, k;
cin >> a >> b >> c >> k;
if(k <= a) cout << k << endl;
else if(k <= a + b) cout << a << endl;
else cout << a - (k - a - b) << endl;
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: There are n cities in the universe and our beloved Spider-Man is in city 1. He doesn't like to travel by vehicles, so he shot webs forming edges between some pairs of cities. Eventually, there were m edges and each had some cost associated with it.
Spider-Man now defines the cost of a path p from cities p<sub>1</sub> to p<sub>k</sub> as w<sub>1</sub> + 2w<sub>2</sub> + 3w<sub>3</sub> . . . + (k-1)*w<sub>k-1</sub>, where w<sub>i</sub> is the cost of an edge from p<sub>i</sub> to p<sub>i+1</sub>.
Thus, the minimum distance between cities i and j is the smallest cost of a path starting from i and ending at j.
Find the minimum distance from city 1 to all the cities i (1 ≤ i ≤ n). If there exists no way to go from city 1 to city i, print -1.
<b>Note: </b>
All the edges are bidirectional. There may be multiple edges and self-loops in the input.The first line contains two space separated integers n and m - the number of nodes and edges respectively.
The next m lines contain three-space separated integers x, y, w - representing an edge between x and y with cost w.
<b>Constraints:</b>
1 ≤ n ≤ 3000
0 ≤ m ≤ 10000
1 ≤ x, y ≤ n
1 ≤ w ≤ 10<sup>9</sup>Output n lines. In the i<sup>th</sup> line, output the minimum distance from city 1 to the i<sup>th</sup> city. If there exists no such path, output -1.Sample Input
6 5
2 4 3
2 3 4
2 1 2
2 5 6
1 5 2
Sample Output
0
2
10
8
2
-1
Explanation:
Shortest path from 1 to 3 is (1->2->3) with total weight= 1*2+2*4=10
Shortest path from 1 to 5 is (1->5) with total weight= 1*2=2
There doesn't exist any path from 1 to 6 so print -1
, I have written this Solution Code:
import sys
from collections import defaultdict
from heapq import heappush, heappop
n, m = map(int, input().split())
d = defaultdict(list)
dist = [sys.maxsize]*n
dist[0] = 0
for _ in range(m):
start, dest, wt = map(int, input().split())
d[start-1].append((dest-1, wt))
d[dest-1].append((start-1, wt))
heap = [(0, 0, 0)]
while heap:
count, cost, u= heappop(heap)
for vertex, weight in d[u]:
if dist[vertex] > cost + weight*(count+1):
dist[vertex] = cost + weight*(count+1)
heappush(heap, (count+1, dist[vertex], vertex))
for d in dist:
if d == sys.maxsize:
print(-1)
else:
print(d), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: There are n cities in the universe and our beloved Spider-Man is in city 1. He doesn't like to travel by vehicles, so he shot webs forming edges between some pairs of cities. Eventually, there were m edges and each had some cost associated with it.
Spider-Man now defines the cost of a path p from cities p<sub>1</sub> to p<sub>k</sub> as w<sub>1</sub> + 2w<sub>2</sub> + 3w<sub>3</sub> . . . + (k-1)*w<sub>k-1</sub>, where w<sub>i</sub> is the cost of an edge from p<sub>i</sub> to p<sub>i+1</sub>.
Thus, the minimum distance between cities i and j is the smallest cost of a path starting from i and ending at j.
Find the minimum distance from city 1 to all the cities i (1 ≤ i ≤ n). If there exists no way to go from city 1 to city i, print -1.
<b>Note: </b>
All the edges are bidirectional. There may be multiple edges and self-loops in the input.The first line contains two space separated integers n and m - the number of nodes and edges respectively.
The next m lines contain three-space separated integers x, y, w - representing an edge between x and y with cost w.
<b>Constraints:</b>
1 ≤ n ≤ 3000
0 ≤ m ≤ 10000
1 ≤ x, y ≤ n
1 ≤ w ≤ 10<sup>9</sup>Output n lines. In the i<sup>th</sup> line, output the minimum distance from city 1 to the i<sup>th</sup> city. If there exists no such path, output -1.Sample Input
6 5
2 4 3
2 3 4
2 1 2
2 5 6
1 5 2
Sample Output
0
2
10
8
2
-1
Explanation:
Shortest path from 1 to 3 is (1->2->3) with total weight= 1*2+2*4=10
Shortest path from 1 to 5 is (1->5) with total weight= 1*2=2
There doesn't exist any path from 1 to 6 so print -1
, I have written this Solution Code: #include<bits/stdc++.h>
using namespace std;
#define int long long
const int INF =1e18;
vector<tuple<int, int, int>> adj;
void solve()
{
int n, m;
cin>>n>>m;
// assert(1<=n && n<=3000);
// assert(0<=m && m<=10000);
adj.resize(n);
for(int i = 0;i<m;i++)
{
int x, y, w;
cin>>x>>y>>w;
x--;
y--;
// assert(0<=x && x<n);
// assert(0<=y && y<n);
// assert(1<=w && w<=1e9);
adj.push_back({x, y, w});
adj.push_back({y, x, w});
}
vector<int> dp_old(n, INF);
vector<int> dp_new(n, INF);
dp_old[0] = 0;
for(int i = 1;i<=n;i++)
{
fill(dp_new.begin(), dp_new.end(), INF);
for(auto [x, y,w]:adj)
{
dp_new[y]= min({dp_new[y], dp_old[x] + i * w});
}
for(int j = 0;j<n;j++)
dp_new[j] = min(dp_new[j], dp_old[j]);
swap(dp_new, dp_old);
}
for(int i = 0;i<n;i++)
{
if(dp_old[i] == INF)
dp_old[i] = -1;
cout<<dp_old[i]<<"\n";
}
}
signed main()
{
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cerr.tie(NULL);
#ifndef ONLINE_JUDGE
if (fopen("INPUT.txt", "r"))
{
freopen("INPUT.txt", "r", stdin);
freopen("OUTPUT.txt", "w", stdout);
}
#endif
solve();
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: There are n cities in the universe and our beloved Spider-Man is in city 1. He doesn't like to travel by vehicles, so he shot webs forming edges between some pairs of cities. Eventually, there were m edges and each had some cost associated with it.
Spider-Man now defines the cost of a path p from cities p<sub>1</sub> to p<sub>k</sub> as w<sub>1</sub> + 2w<sub>2</sub> + 3w<sub>3</sub> . . . + (k-1)*w<sub>k-1</sub>, where w<sub>i</sub> is the cost of an edge from p<sub>i</sub> to p<sub>i+1</sub>.
Thus, the minimum distance between cities i and j is the smallest cost of a path starting from i and ending at j.
Find the minimum distance from city 1 to all the cities i (1 ≤ i ≤ n). If there exists no way to go from city 1 to city i, print -1.
<b>Note: </b>
All the edges are bidirectional. There may be multiple edges and self-loops in the input.The first line contains two space separated integers n and m - the number of nodes and edges respectively.
The next m lines contain three-space separated integers x, y, w - representing an edge between x and y with cost w.
<b>Constraints:</b>
1 ≤ n ≤ 3000
0 ≤ m ≤ 10000
1 ≤ x, y ≤ n
1 ≤ w ≤ 10<sup>9</sup>Output n lines. In the i<sup>th</sup> line, output the minimum distance from city 1 to the i<sup>th</sup> city. If there exists no such path, output -1.Sample Input
6 5
2 4 3
2 3 4
2 1 2
2 5 6
1 5 2
Sample Output
0
2
10
8
2
-1
Explanation:
Shortest path from 1 to 3 is (1->2->3) with total weight= 1*2+2*4=10
Shortest path from 1 to 5 is (1->5) with total weight= 1*2=2
There doesn't exist any path from 1 to 6 so print -1
, I have written this Solution Code: import java.io.*;import java.util.*;import java.math.*;import static java.lang.Math.*;import static java.
util.Map.*;import static java.util.Arrays.*;import static java.util.Collections.*;
import static java.lang.System.*;
public class Main
{
public void tq()throws Exception
{
st=new StringTokenizer(bq.readLine());
int tq=1;
sb=new StringBuilder(2000000);
o:
while(tq-->0)
{
int n=i();
int m=i();
LinkedList<int[]> l[]=new LinkedList[n];
for(int x=0;x<n;x++)l[x]=new LinkedList<>();
for(int x=0;x<m;x++)
{
int a=i()-1;
int b=i()-1;
int c=i();
l[a].add(new int[]{b,c});
l[b].add(new int[]{a,c});
}
long d[]=new long[n];
for(int x=0;x<n;x++)d[x]=maxl;
d[0]=0l;
PriorityQueue<long[]> p=new PriorityQueue<>(5000,(a,b)->a[2]-b[2]<1l?-1:1);
p.add(new long[]{0l,0,0});
while(p.size()>0)
{
long r[]=p.poll();
long di=r[0];
int no=(int)r[1];
long mu=r[2];
for(int e[]:l[no])
{
int node=e[0];
int w=e[1];
long de=di+w*(mu+1);
if(d[node]>de)
{
d[node]=de;
p.add(new long[]{de,node,mu+1});
}
}
}
for(long x:d)
{
sl(x==maxl?-1:x);
}
}
p(sb);
}
int di[][]={{-1,0},{1,0},{0,-1},{0,1}};
int de[][]={{-1,0},{1,0},{0,-1},{0,1},{-1,-1},{1,1},{-1,1},{1,-1}};
long mod=1000000007l;int max=Integer.MAX_VALUE,min=Integer.MIN_VALUE;long maxl=Long.MAX_VALUE, minl=Long.
MIN_VALUE;BufferedReader bq=new BufferedReader(new InputStreamReader(in));StringTokenizer st;
StringBuilder sb;public static void main(String[] a)throws Exception{new Main().tq();}int[] so(int ar[])
{Integer r[]=new Integer[ar.length];for(int x=0;x<ar.length;x++)r[x]=ar[x];sort(r);for(int x=0;x<ar.length;
x++)ar[x]=r[x];return ar;}long[] so(long ar[]){Long r[]=new Long[ar.length];for(int x=0;x<ar.length;x++)
r[x]=ar[x];sort(r);for(int x=0;x<ar.length;x++)ar[x]=r[x];return ar;}
char[] so(char ar[]) {Character
r[]=new Character[ar.length];for(int x=0;x<ar.length;x++)r[x]=ar[x];sort(r);for(int x=0;x<ar.length;x++)
ar[x]=r[x];return ar;}void s(String s){sb.append(s);}void s(int s){sb.append(s);}void s(long s){sb.
append(s);}void s(char s){sb.append(s);}void s(double s){sb.append(s);}void ss(){sb.append(' ');}void sl
(String s){sb.append(s);sb.append("\n");}void sl(int s){sb.append(s);sb.append("\n");}void sl(long s){sb
.append(s);sb.append("\n");}void sl(char s) {sb.append(s);sb.append("\n");}void sl(double s){sb.append(s)
;sb.append("\n");}void sl(){sb.append("\n");}int l(int v){return 31-Integer.numberOfLeadingZeros(v);}
long l(long v){return 63-Long.numberOfLeadingZeros(v);}int sq(int a){return (int)sqrt(a);}long sq(long a)
{return (long)sqrt(a);}long gcd(long a,long b){while(b>0l){long c=a%b;a=b;b=c;}return a;}int gcd(int a,int b)
{while(b>0){int c=a%b;a=b;b=c;}return a;}boolean pa(String s,int i,int j){while(i<j)if(s.charAt(i++)!=
s.charAt(j--))return false;return true;}boolean[] si(int n) {boolean bo[]=new boolean[n+1];bo[0]=true;bo[1]
=true;for(int x=4;x<=n;x+=2)bo[x]=true;for(int x=3;x*x<=n;x+=2){if(!bo[x]){int vv=(x<<1);for(int y=x*x;y<=n;
y+=vv)bo[y]=true;}}return bo;}long mul(long a,long b,long m) {long r=1l;a%=m;while(b>0){if((b&1)==1)
r=(r*a)%m;b>>=1;a=(a*a)%m;}return r;}int i()throws IOException{if(!st.hasMoreTokens())st=new
StringTokenizer(bq.readLine());return Integer.parseInt(st.nextToken());}long l()throws IOException
{if(!st.hasMoreTokens())st=new StringTokenizer(bq.readLine());return Long.parseLong(st.nextToken());}String
s()throws IOException {if (!st.hasMoreTokens())st=new StringTokenizer(bq.readLine());return st.nextToken();}
double d()throws IOException{if(!st.hasMoreTokens())st=new StringTokenizer(bq.readLine());return Double.
parseDouble(st.nextToken());}void p(Object p){out.print(p);}void p(String p){out.print(p);}void p(int p)
{out.print(p);}void p(double p){out.print(p);}void p(long p){out.print(p);}void p(char p){out.print(p);}void
p(boolean p){out.print(p);}void pl(Object p){out.println(p);}void pl(String p){out.println(p);}void pl(int p)
{out.println(p);}void pl(char p){out.println(p);}void pl(double p){out.println(p);}void pl(long p){out.
println(p);}void pl(boolean p)
{out.println(p);}void pl(){out.println();}void s(int a[]){for(int e:a)
{sb.append(e);sb.append(' ');}sb.append("\n");}
void s(long a[])
{for(long e:a){sb.append(e);sb.append(' ')
;}sb.append("\n");}void s(int ar[][]){for(int a[]:ar){for(int e:a){sb.append(e);sb.append(' ');}sb.append
("\n");}}
void s(char a[])
{for(char e:a){sb.append(e);sb.append(' ');}sb.append("\n");}void s(char ar[][])
{for(char a[]:ar){for(char e:a){sb.append(e);sb.append(' ');}sb.append("\n");}}int[] ari(int n)throws
IOException {int ar[]=new int[n];if(!st.hasMoreTokens())st=new StringTokenizer(bq.readLine());for(int x=0;
x<n;x++)ar[x]=Integer.parseInt(st.nextToken());return ar;}int[][] ari(int n,int m)throws
IOException {int ar[][]=new int[n][m];for(int x=0;x<n;x++){if (!st.hasMoreTokens())st=new StringTokenizer
(bq.readLine());for(int y=0;y<m;y++)ar[x][y]=Integer.parseInt(st.nextToken());}return ar;}long[] arl
(int n)throws IOException {long ar[]=new long[n];if(!st.hasMoreTokens()) st=new StringTokenizer(bq.readLine())
;for(int x=0;x<n;x++)ar[x]=Long.parseLong(st.nextToken());return ar;}long[][] arl(int n,int m)throws
IOException {long ar[][]=new long[n][m];for(int x=0;x<n;x++) {if(!st.hasMoreTokens()) st=new
StringTokenizer(bq.readLine());for(int y=0;y<m;y++)ar[x][y]=Long.parseLong(st.nextToken());}return ar;}
String[] ars(int n)throws IOException {String ar[] =new String[n];if(!st.hasMoreTokens())st=new
StringTokenizer(bq.readLine());for(int x=0;x<n;x++)ar[x]=st.nextToken();return ar;}double[] ard
(int n)throws IOException {double ar[] =new double[n];if(!st.hasMoreTokens())st=new StringTokenizer
(bq.readLine());for(int x=0;x<n;x++)ar[x]=Double.parseDouble(st.nextToken());return ar;}double[][] ard
(int n,int m)throws IOException{double ar[][]=new double[n][m];for(int x=0;x<n;x++) {if(!st.hasMoreTokens())
st=new StringTokenizer(bq.readLine());for(int y=0;y<m;y++) ar[x][y]=Double.parseDouble(st.nextToken());}
return ar;}char[] arc(int n)throws IOException{char ar[]=new char[n];if(!st.hasMoreTokens())st=new
StringTokenizer(bq.readLine());for(int x=0;x<n;x++)ar[x]=st.nextToken().charAt(0);return ar;}char[][]
arc(int n,int m)throws IOException {char ar[][]=new char[n][m];for(int x=0;x<n;x++){String s=bq.readLine();
for(int y=0;y<m;y++)ar[x][y]=s.charAt(y);}return ar;}void p(int ar[])
{StringBuilder sb=new StringBuilder
(2*ar.length);for(int a:ar){sb.append(a);sb.append(' ');}out.println(sb);}void p(int ar[][])
{StringBuilder sb=new StringBuilder(2*ar.length*ar[0].length);for(int a[]:ar){for(int aa:a){sb.append(aa);
sb.append(' ');}sb.append("\n");}p(sb);}void p(long ar[]){StringBuilder sb=new StringBuilder
(2*ar.length);for(long a:ar){ sb.append(a);sb.append(' ');}out.println(sb);}
void p(long ar[][])
{StringBuilder sb=new StringBuilder(2*ar.length*ar[0].length);for(long a[]:ar){for(long aa:a){sb.append(aa);
sb.append(' ');}sb.append("\n");}p(sb);}void p(String ar[]){int c=0;for(String s:ar)c+=s.length()+1;
StringBuilder sb=new StringBuilder(c);for(String a:ar){sb.append(a);sb.append(' ');}out.println(sb);}
void p(double ar[])
{StringBuilder sb=new StringBuilder(2*ar.length);for(double a:ar){sb.append(a);
sb.append(' ');}out.println(sb);}void p
(double ar[][]){StringBuilder sb=new StringBuilder(2*
ar.length*ar[0].length);for(double a[]:ar){for(double aa:a){sb.append(aa);sb.append(' ');}sb.append("\n")
;}p(sb);}void p(char ar[])
{StringBuilder sb=new StringBuilder(2*ar.length);for(char aa:ar){sb.append(aa);
sb.append(' ');}out.println(sb);}void p(char ar[][]){StringBuilder sb=new StringBuilder(2*ar.length*ar[0]
.length);for(char a[]:ar){for(char aa:a){sb.append(aa);sb.append(' ');}sb.append("\n");}p(sb);}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A of size N, you need to find its maximum, 2<sup>nd</sup> maximum and 3<sup>rd</sup> maximum element.
Try solving it in O(N) per test caseThe first line of the input contains the number of test cases T.
For each test case, the first line of the input contains an integer N denoting the number of elements in the array A. The following line contains N (space-separated) elements of A.
<b>Constraints:</b>
1 <= T <= 100
3 <= N <= 10<sup>6</sup>
1 <= A[i] <= 10<sup>9</sup>
<b>Note</b>:-It is guaranteed that the sum of N over all text cases does not exceed 10<sup>6</sup>For each test case, output the first, second and third maximum elements in the array.Sample Input:
3
5
1 4 2 4 5
6
1 3 5 7 9 8
7
11 22 33 44 55 66 77
Sample Output:
5 4 4
9 8 7
77 66 55
<b>Explanation:</b>
Testcase 1:
[1 4 2 4 5]
First max = 5
Second max = 4
Third max = 4, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) {
Scanner sc =new Scanner(System.in);
int T= sc.nextInt();
for(int i=0;i<T;i++){
int arrsize=sc.nextInt();
int max=0,secmax=0,thirdmax=0,j;
for(int k=0;k<arrsize;k++){
j=sc.nextInt();
if(j>max){
thirdmax=secmax;
secmax=max;
max=j;
}
else if(j>secmax){
thirdmax=secmax;
secmax=j;
}
else if(j>thirdmax){
thirdmax=j;
}
if(k%10000==0){
System.gc();
}
}
System.out.println(max+" "+secmax+" "+thirdmax+" ");
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A of size N, you need to find its maximum, 2<sup>nd</sup> maximum and 3<sup>rd</sup> maximum element.
Try solving it in O(N) per test caseThe first line of the input contains the number of test cases T.
For each test case, the first line of the input contains an integer N denoting the number of elements in the array A. The following line contains N (space-separated) elements of A.
<b>Constraints:</b>
1 <= T <= 100
3 <= N <= 10<sup>6</sup>
1 <= A[i] <= 10<sup>9</sup>
<b>Note</b>:-It is guaranteed that the sum of N over all text cases does not exceed 10<sup>6</sup>For each test case, output the first, second and third maximum elements in the array.Sample Input:
3
5
1 4 2 4 5
6
1 3 5 7 9 8
7
11 22 33 44 55 66 77
Sample Output:
5 4 4
9 8 7
77 66 55
<b>Explanation:</b>
Testcase 1:
[1 4 2 4 5]
First max = 5
Second max = 4
Third max = 4, I have written this Solution Code: t=int(input())
while t>0:
t-=1
n=int(input())
l=list(map(int,input().strip().split()))
li=[0,0,0]
for i in l:
x=i
for j in range(0,3):
y=min(x,li[j])
li[j]=max(x,li[j])
x=y
print(li[0],end=" ")
print(li[1],end=" ")
print(li[2]), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A of size N, you need to find its maximum, 2<sup>nd</sup> maximum and 3<sup>rd</sup> maximum element.
Try solving it in O(N) per test caseThe first line of the input contains the number of test cases T.
For each test case, the first line of the input contains an integer N denoting the number of elements in the array A. The following line contains N (space-separated) elements of A.
<b>Constraints:</b>
1 <= T <= 100
3 <= N <= 10<sup>6</sup>
1 <= A[i] <= 10<sup>9</sup>
<b>Note</b>:-It is guaranteed that the sum of N over all text cases does not exceed 10<sup>6</sup>For each test case, output the first, second and third maximum elements in the array.Sample Input:
3
5
1 4 2 4 5
6
1 3 5 7 9 8
7
11 22 33 44 55 66 77
Sample Output:
5 4 4
9 8 7
77 66 55
<b>Explanation:</b>
Testcase 1:
[1 4 2 4 5]
First max = 5
Second max = 4
Third max = 4, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main(){
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
int t;
cin>>t;
while(t--){
long long n;
cin>>n;
vector<long> a(n);
long ans[3]={0};
long x,y;
for(int i=0;i<n;i++){
cin>>a[i];
x=a[i];
for(int j=0;j<3;j++){
y=min(x,ans[j]);
ans[j]=max(x,ans[j]);
// cout<<ans[j]<<" ";
x=y;
}
}
if(ans[1]<ans[0]){
swap(ans[1],ans[0]);
}
if(ans[2]<ans[1]){
swap(ans[1],ans[2]);
}
if(ans[1]<ans[0]){
swap(ans[1],ans[0]);
}
cout<<ans[2]<<" "<<ans[1]<<" "<<ans[0]<<endl;
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A of size N, you need to find its maximum, 2<sup>nd</sup> maximum and 3<sup>rd</sup> maximum element.
Try solving it in O(N) per test caseThe first line of the input contains the number of test cases T.
For each test case, the first line of the input contains an integer N denoting the number of elements in the array A. The following line contains N (space-separated) elements of A.
<b>Constraints:</b>
1 <= T <= 100
3 <= N <= 10<sup>6</sup>
1 <= A[i] <= 10<sup>9</sup>
<b>Note</b>:-It is guaranteed that the sum of N over all text cases does not exceed 10<sup>6</sup>For each test case, output the first, second and third maximum elements in the array.Sample Input:
3
5
1 4 2 4 5
6
1 3 5 7 9 8
7
11 22 33 44 55 66 77
Sample Output:
5 4 4
9 8 7
77 66 55
<b>Explanation:</b>
Testcase 1:
[1 4 2 4 5]
First max = 5
Second max = 4
Third max = 4, I have written this Solution Code: function maxNumbers(arr,n) {
// write code here
// do not console.log the answer
// return the answer as an array of 3 numbers
return arr.sort((a,b)=>b-a).slice(0,3)
};
, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Write a curried function called <code>reverseWords</code> that takes two string arguments. The first argument will be a sentence and the second argument will be a word. The function should append the word at the end of the sentence after leaving a space. The <code>reverseWords</code> function would return a function called <code>reverseFunc</code>.
The <code>reverseFunc</code> function, when called, should split the new sentence formed into individual words using <code>split()</code> function, reverse the order of the words using <code>reverse()</code> function, and join them back together with a space between each word using <code>join()</code> function. Then the new string formed should be returned by the <code>reverseFunc</code> functionThe <code>reverseWords</code> function takes two string arguments. The first argument will be a sentence consisting of multiple words and the second argument will be a single word.
The <code>reverseFunc</code> function takes no arguments. It used the new sentence formed by appending the second argument to the first argument of the <code>reverseWords</code> function.The <code>reverseWords</code> function should return a function called <code>reverseFunc</code>.
The <code>reverseFunc</code> function should take the new string formed by appending the second argument to the first one of the <code>reverseWords</code> function, split it into individual words, reverse their order, join them back with a space between each word, and return the string formed.const arg1 = "hello world";
const arg2 = "everyone";
const reverseFunc = reverseWords(arg1, arg2);
console.log(reverseFunc()); //prints "everyone world hello"
, I have written this Solution Code: function reverseWords(str, word){
str = str + " " + word;
function reverseFunc() {
const words = str.split(' ');
const reversedWords = words.reverse();
const reversedString = reversedWords.join(' ');
return reversedString;
}
return reverseFunc;
}
, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Newton has started studying for the upcoming test. The textbook has N pages in total. Newton wants to read at most X pages a day for Y days.
Find out whether it is possible for Newton to complete the whole book.The first line of input contains three space- separated integers N, X, and Y — the number of pages, the number of pages Newton can read in a day, and the number of days.
<b>Constraints :</b>
1 ≤ N ≤ 100
1 ≤ X, Y ≤ 10Output on a new line, YES, if Newton can complete the whole book in a given time, and NO otherwise.Sample Input:-
5 2 3
Sample Output:-
YES
Sample Input:-
10 3 3
Sample Output:-
NO, I have written this Solution Code:
import java.util.*;
import java.lang.*;
import java.io.*;
class Solution {
public void Solve(int n,int x,int y) {
if(x*y>=n)
System.out.println("YES");
else
System.out.println("NO");
}
}
class Main
{
public static void main (String[] args) throws java.lang.Exception
{
//StringBuilder sss = new StringBuilder("");
Scanner sc=new Scanner(System.in);
int n=sc.nextInt();
assert(n >= 1 && n <= 100);
int x=sc.nextInt();
assert(x >= 1 && x <= 10);
int o=sc.nextInt();
assert(o >= 1 && o <= 10);
Solution obj=new Solution();
obj.Solve(n,x,o);
}
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are N integers A[1], A[2],. , A[N]. You need to find whether their product is divisible by 4.The first line of input contains a single integer N.
The next line contains N singly spaced integers, A[1], A[2],. , A[N].
Constraints
1 <= N <= 10
1 <= A[i] <= 1000000000Output "YES" if the product is divisible by 4, else output "NO". (without quotes)Sample Input
5
3 5 2 5 1
Sample Output
NO
Explanation: Product = 150 and 150 is not divisible by 4.
Sample Input
4
1 3 8 2
Sample Output
YES
Explanation: Product = 48 and 48 is divisible by 4., I have written this Solution Code: n = int(input())
arr = []
x = 4
prod = 1
a = input().split()
for i in range(0,n):
prod = prod*int(a[i])
if prod % 4 == 0:
print("YES")
else:
print("NO"), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are N integers A[1], A[2],. , A[N]. You need to find whether their product is divisible by 4.The first line of input contains a single integer N.
The next line contains N singly spaced integers, A[1], A[2],. , A[N].
Constraints
1 <= N <= 10
1 <= A[i] <= 1000000000Output "YES" if the product is divisible by 4, else output "NO". (without quotes)Sample Input
5
3 5 2 5 1
Sample Output
NO
Explanation: Product = 150 and 150 is not divisible by 4.
Sample Input
4
1 3 8 2
Sample Output
YES
Explanation: Product = 48 and 48 is divisible by 4., I have written this Solution Code:
import java.util.*;
import java.lang.*;
import java.io.*;
class Main
{
public static void main (String[] args) throws java.lang.Exception
{
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
int a[] = new int[n];
for(int i=0;i<n;i++){
a[i]=sc.nextInt();
}
int two=0,four=0;
for(int i=0;i<n;i++){
if(a[i]%2==0){
two++;
}
if(a[i]%4==0){
four++;
}
}
if(four>=1 || two>=2){
System.out.print("YES");
}
else{
System.out.print("NO");
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are N integers A[1], A[2],. , A[N]. You need to find whether their product is divisible by 4.The first line of input contains a single integer N.
The next line contains N singly spaced integers, A[1], A[2],. , A[N].
Constraints
1 <= N <= 10
1 <= A[i] <= 1000000000Output "YES" if the product is divisible by 4, else output "NO". (without quotes)Sample Input
5
3 5 2 5 1
Sample Output
NO
Explanation: Product = 150 and 150 is not divisible by 4.
Sample Input
4
1 3 8 2
Sample Output
YES
Explanation: Product = 48 and 48 is divisible by 4., I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define sd(x) scanf("%d", &x)
#define sz(v) (int) v.size()
#define pr(v) For(i, 0, sz(v)) {cout<<v[i]<<" ";} cout<<endl;
#define slld(x) scanf("%lld", &x)
#define all(x) x.begin(), x.end()
#define For(i, st, en) for(ll i=st; i<en; i++)
#define tr(x) for(auto it=x.begin(); it!=x.end(); it++)
#define fast std::ios::sync_with_stdio(false);cin.tie(NULL);
#define pb push_back
#define ll long long
#define int long long
#define ld long double
#define mp make_pair
#define F first
#define S second
typedef pair<int, int> pii;
typedef vector<int> vi;
#define pi 3.141592653589793238
#define MOD 1000000007
#define INF 1000000000000000007LL
const int N = 200005;
// it's swapnil07 ;)
#ifdef SWAPNIL07
#define trace(...) __f(#__VA_ARGS__, __VA_ARGS__)
template <typename Arg1>
void __f(const char* name, Arg1&& arg1){
cout << name << " : " << arg1 << endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args){
const char* comma = strchr(names + 1, ',');cout.write(names, comma - names) << " : " << arg1<<" | ";__f(comma+1, args...);
}
int begtime = clock();
#define end_routine() cout << "\n\nTime elapsed: " << (clock() - begtime)*1000/CLOCKS_PER_SEC << " ms\n\n";
#else
#define endl '\n'
#define trace(...)
#define end_routine()
#endif
signed main()
{
fast
#ifdef SWAPNIL07
freopen("input.txt","r",stdin);
freopen("output.txt","w",stdout);
#endif
int n; cin>>n;
int ct = 0;
For(i, 0, n){
int a; cin>>a;
while(a%2==0){
ct++;
a/=2;
}
}
if(ct>=2){
cout<<"YES";
}
else{
cout<<"NO";
}
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: A number X (X>=0) is called strange if the sum of its digits is divisible by 9. Given an integer N, your task is to find the Nth strange number.The input contains a single line containing the value of N.
Constraints:-
1 <= N <= 1000Print the Nth strange number.Sample Input:-
3
Sample Output:-
18
Explanation:-
0, 9, and 18 are the first three strange numbers.
Sample Input:-
2
Sample Output:-
9, I have written this Solution Code: import java.util.*;
import java.io.*;
import java.lang.*;
public class Main
{
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int x=sc.nextInt();
int ans = 9 * (x-1);
System.out.print(ans);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: A number X (X>=0) is called strange if the sum of its digits is divisible by 9. Given an integer N, your task is to find the Nth strange number.The input contains a single line containing the value of N.
Constraints:-
1 <= N <= 1000Print the Nth strange number.Sample Input:-
3
Sample Output:-
18
Explanation:-
0, 9, and 18 are the first three strange numbers.
Sample Input:-
2
Sample Output:-
9, 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 1000000000000007
#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 cnt[max1];
signed main(){
int n;
cin>>n;
out((n-1)*9);
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
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