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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: Our five power rangers have powers P1, P2, P3, P4, and P5.
To ensure the proper distribution of power, the power of every power ranger must remain <b>less</b> than the sum of powers of other power rangers.
If the above condition is not met, there's an emergency. Can you let us know if there's an emergency?The first and the only line of input contains 5 integers P1, P2, P3, P4, and P5.
Constraints
0 <= P1, P2, P3, P4, P5 <= 100Output "SPD Emergency" (without quotes) if there's an emergency, else output "Stable".Sample Input
1 2 3 4 5
Sample Output
Stable
Explanation
The power of every power ranger is less than the sum of powers of other power rangers.
Sample Input
1 2 3 4 100
Sample Output
SPD Emergency
Explanation
The power of the 5th power ranger (100) is not less than the sum of powers of other power rangers (1+2+3+4=10)., I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws IOException {
int[] arr=new int[5];
BufferedReader rd=new BufferedReader(new InputStreamReader(System.in));
String[] s=rd.readLine().split(" ");
int sum=0;
for(int i=0;i<5;i++){
arr[i]=Integer.parseInt(s[i]);
sum+=arr[i];
}
int i=0,j=arr.length-1;
boolean isEmergency=false;
while(i<=j)
{
int temp=arr[i];
sum-=arr[i];
if(arr[i]>= sum)
{
isEmergency=true;
break;
}
sum+=temp;
i++;
}
if(isEmergency==false)
{
System.out.println("Stable");
}
else
{
System.out.println("SPD Emergency");
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Our five power rangers have powers P1, P2, P3, P4, and P5.
To ensure the proper distribution of power, the power of every power ranger must remain <b>less</b> than the sum of powers of other power rangers.
If the above condition is not met, there's an emergency. Can you let us know if there's an emergency?The first and the only line of input contains 5 integers P1, P2, P3, P4, and P5.
Constraints
0 <= P1, P2, P3, P4, P5 <= 100Output "SPD Emergency" (without quotes) if there's an emergency, else output "Stable".Sample Input
1 2 3 4 5
Sample Output
Stable
Explanation
The power of every power ranger is less than the sum of powers of other power rangers.
Sample Input
1 2 3 4 100
Sample Output
SPD Emergency
Explanation
The power of the 5th power ranger (100) is not less than the sum of powers of other power rangers (1+2+3+4=10)., I have written this Solution Code: arr = list(map(int,input().split()))
m = sum(arr)
f=[]
for i in range(len(arr)):
s = sum(arr[:i]+arr[i+1:])
if(arr[i]<s):
f.append(1)
else:
f.append(0)
if(all(f)):
print("Stable")
else:
print("SPD Emergency"), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Our five power rangers have powers P1, P2, P3, P4, and P5.
To ensure the proper distribution of power, the power of every power ranger must remain <b>less</b> than the sum of powers of other power rangers.
If the above condition is not met, there's an emergency. Can you let us know if there's an emergency?The first and the only line of input contains 5 integers P1, P2, P3, P4, and P5.
Constraints
0 <= P1, P2, P3, P4, P5 <= 100Output "SPD Emergency" (without quotes) if there's an emergency, else output "Stable".Sample Input
1 2 3 4 5
Sample Output
Stable
Explanation
The power of every power ranger is less than the sum of powers of other power rangers.
Sample Input
1 2 3 4 100
Sample Output
SPD Emergency
Explanation
The power of the 5th power ranger (100) is not less than the sum of powers of other power rangers (1+2+3+4=10)., I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define sd(x) scanf("%d", &x)
#define sz(v) (int) v.size()
#define pr(v) For(i, 0, sz(v)) {cout<<v[i]<<" ";} cout<<endl;
#define slld(x) scanf("%lld", &x)
#define all(x) x.begin(), x.end()
#define For(i, st, en) for(int i=st; i<en; i++)
#define tr(x) for(auto it=x.begin(); it!=x.end(); it++)
#define fast std::ios::sync_with_stdio(false);cin.tie(NULL);
#define pb push_back
#define ll long long
#define ld long double
#define int long long
#define double long double
#define mp make_pair
#define F first
#define S second
typedef pair<int, int> pii;
typedef vector<int> vi;
#define pi 3.141592653589793238
const int MOD = 1e9+7;
const int INF = 1LL<<60;
const int N = 2e5+5;
// it's swapnil07 ;)
#ifdef SWAPNIL07
#define trace(...) __f(#__VA_ARGS__, __VA_ARGS__)
template <typename Arg1>
void __f(const char* name, Arg1&& arg1){
cout << name << " : " << arg1 << endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args){
const char* comma = strchr(names + 1, ',');cout.write(names, comma - names) << " : " << arg1<<" | ";__f(comma+1, args...);
}
int begtime = clock();
#define end_routine() cout << "\n\nTime elapsed: " << (clock() - begtime)*1000/CLOCKS_PER_SEC << " ms\n\n";
#else
#define endl '\n'
#define trace(...)
#define end_routine()
#endif
void solve(){
vector<int> vect(5);
int tot = 0;
for(int i=0; i<5; i++){
cin>>vect[i];
tot += vect[i];
}
sort(all(vect));
tot -= vect[4];
if(vect[4] >= tot){
cout<<"SPD Emergency";
}
else{
cout<<"Stable";
}
}
signed main()
{
fast
#ifdef SWAPNIL07
freopen("input.txt","r",stdin);
freopen("output.txt","w",stdout);
#endif
int t=1;
// cin>>t;
while(t--){
solve();
cout<<"\n";
}
return 0;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: A pair is called lucky if its sum is even and positive. Given three numbers find if there exists a lucky pair or not.The only line contains three integers a<sub>1</sub>, a<sub>2</sub>, a<sub>3</sub>
<b>Constraints:</b>
-10<sup>9</sup> <= a<sub>1</sub>, a<sub>2</sub>, a<sub>3</sub> <= 10<sup>9</sup>Print "YES" without quotes if there exists a lucky pair otherwise print "NO" without quotes.Sample Input 1:
23 32 12
Sample Output 1:
YES
Sample Input 2:
1 -1 2
Sample Output 2:
NO, 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>;
// 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;
void solve(){
vector<ll>a(3);
for(ll i = 0;i<3;i++)cin >> a[i];
for(ll i = 0;i<3;i++){
for(ll j = i+1;j<3;j++){
if((a[i]+a[j])>0 && (a[i]+a[j])%2 == 0){
cout << "YES\n";
return;
}
}
}
cout << "NO\n";
}
int main(){
ios_base::sync_with_stdio(false);
cin.tie(NULL);
// cout.tie(NULL);
#ifdef LOCALFLAG
freopen("Input.txt", "r", stdin);
freopen("Output2.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: There are N buildings in a row with different heights H[i] (1 <= i <= N).
You are standing on the left side of the first building .From this position you can see the roof of a building <b>i</b> if no building to the left of the i<sup>th</sup> building has a height greater than or equal to the height of the i<sup>th</sup> building.
You are asked to find the number of buildings whose roofs you can see.The first line contains N denoting number of buildings.
The next line contains N space seperated integers denoting heights of the buildings from left to right.
Constraints
1 <= N <= 100000
1 <= H[i] <= 1000000000000000The output should contain one integer which is the number of buildings whose roofs you can see.Sample input:
5
1 2 2 4 3
Sample output:
3
Explanation:-
the building at index 3 will hide before building at index 2 and building at index 5 will hide before building at index 4
Sample input:
5
1 2 3 4 5
Sample output:
5
, I have written this Solution Code: n=int(input())
a=map(int,input().split())
b=[]
mx=-200000
cnt=0
for i in a:
if i>mx:
cnt+=1
mx=i
print(cnt), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: There are N buildings in a row with different heights H[i] (1 <= i <= N).
You are standing on the left side of the first building .From this position you can see the roof of a building <b>i</b> if no building to the left of the i<sup>th</sup> building has a height greater than or equal to the height of the i<sup>th</sup> building.
You are asked to find the number of buildings whose roofs you can see.The first line contains N denoting number of buildings.
The next line contains N space seperated integers denoting heights of the buildings from left to right.
Constraints
1 <= N <= 100000
1 <= H[i] <= 1000000000000000The output should contain one integer which is the number of buildings whose roofs you can see.Sample input:
5
1 2 2 4 3
Sample output:
3
Explanation:-
the building at index 3 will hide before building at index 2 and building at index 5 will hide before building at index 4
Sample input:
5
1 2 3 4 5
Sample output:
5
, I have written this Solution Code: function numberOfRoofs(arr)
{
let count=1;
let max = arr[0];
for(let i=1;i<arrSize;i++)
{
if(arr[i] > max)
{
count++;
max = arr[i];
}
}
return count;
}
, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: There are N buildings in a row with different heights H[i] (1 <= i <= N).
You are standing on the left side of the first building .From this position you can see the roof of a building <b>i</b> if no building to the left of the i<sup>th</sup> building has a height greater than or equal to the height of the i<sup>th</sup> building.
You are asked to find the number of buildings whose roofs you can see.The first line contains N denoting number of buildings.
The next line contains N space seperated integers denoting heights of the buildings from left to right.
Constraints
1 <= N <= 100000
1 <= H[i] <= 1000000000000000The output should contain one integer which is the number of buildings whose roofs you can see.Sample input:
5
1 2 2 4 3
Sample output:
3
Explanation:-
the building at index 3 will hide before building at index 2 and building at index 5 will hide before building at index 4
Sample input:
5
1 2 3 4 5
Sample output:
5
, I have written this Solution Code: import java.util.*;
import java.io.*;
class Main{
public static void main(String args[]){
Scanner s=new Scanner(System.in);
int n=s.nextInt();
int []a=new int[n];
for(int i=0;i<n;i++){
a[i]=s.nextInt();
}
int count=1;
int max = a[0];
for(int i=1;i<n;i++)
{
if(a[i] > max)
{
count++;
max = a[i];
}
}
System.out.println(count);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a linked list consisting of N nodes and an integer K, your 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 K as parameter.
Constraints:
1 <=K<=N<= 1000
1 <=Node.data<= 1000Return the head of the modified linked listInput 1:
5 3
1 2 3 4 5
Output 1:
1 2 4 5
Explanation:
After deleting 3rd node from the end of the linked list, 3 will be deleted and the list will be as 1, 2, 4, 5.
Input 2:-
5 5
8 1 8 3 6
Output 2:-
1 8 3 6 , 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;
}
temp=head;
k=cnt-k;
if(k==0){head=head.next; return head;}
temp=head;
cnt=0;
while(cnt!=k-1){
temp=temp.next;
cnt++;
}
temp.next=temp.next.next;
return head;
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: In a pascal triangle starting from top to bottom. Find the value of the element in a given row and column position i.e. ith row and at jth position.The first line of input contains an integer T denoting the number of test cases. Each test case will have two integers indexes of the ith row and jth column provided in the new line.
Constraints:
1 <= T <= 100
0 <= j <= i <= 500
Print the answer for each testcase in separated line. Since the output can be very large, mod your output by 10^9+7.Input:
2
0 0
2 1
Output:
1
2
Explanation: The Pascal Triangle has first value as 1. (0th row 0th element). The pascal triangle has next set of values as 1 1. (1th row). On the next level, pascal triangle has values 1 2 1. (2nd row). Its 1st element is 2. (0 based)., I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args)throws IOException {
BufferedReader br= new BufferedReader(new InputStreamReader(System.in));
int T=Integer.parseInt(br.readLine());
for(int k=0;k<T;k++){
String[] str=br.readLine().split(" ");
int i=Integer.parseInt(str[0]);
int j=Integer.parseInt(str[1]);
System.out.print(nCr(i,j));
System.out.println();
}
}
public static long fact(int n){
long mul=1;
int m=1000000007;
for(int i=1;i<=n;i++){
mul=((mul%m)*i%m)%m;
}
return mul;
}
public static long nCr(int n,int r){
int m=1000000007;
long mul=(fact(r)%m*fact(n-r)%m)%m;
return ((long)((fact(n)%m)*(pow(mul,m-2,m)%m))%m);
}
public static long pow(long a,long b,long m){
long res=1;
while(b>0){
if(b%2!=0)
res=(long)((res*(a%m))%m);
a=((a%m)*(a%m))%m;
b=b>>1;
}
return res;
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: In a pascal triangle starting from top to bottom. Find the value of the element in a given row and column position i.e. ith row and at jth position.The first line of input contains an integer T denoting the number of test cases. Each test case will have two integers indexes of the ith row and jth column provided in the new line.
Constraints:
1 <= T <= 100
0 <= j <= i <= 500
Print the answer for each testcase in separated line. Since the output can be very large, mod your output by 10^9+7.Input:
2
0 0
2 1
Output:
1
2
Explanation: The Pascal Triangle has first value as 1. (0th row 0th element). The pascal triangle has next set of values as 1 1. (1th row). On the next level, pascal triangle has values 1 2 1. (2nd row). Its 1st element is 2. (0 based)., I have written this Solution Code:
t = int(input())
mod = 10**9 + 7
while (t > 0):
n,k = map(int,input().split())
factOfRow = 1
factOfCol = 1
factRowMinusCol = 1
for i in range(1,n+1):
factOfRow *= i
for i in range(1,k+1):
factOfCol *= i
for i in range(1,n - k+1):
factRowMinusCol *= i
fact = factOfRow // (factOfCol * factRowMinusCol)
print(fact %mod)
t = t -1
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: In a pascal triangle starting from top to bottom. Find the value of the element in a given row and column position i.e. ith row and at jth position.The first line of input contains an integer T denoting the number of test cases. Each test case will have two integers indexes of the ith row and jth column provided in the new line.
Constraints:
1 <= T <= 100
0 <= j <= i <= 500
Print the answer for each testcase in separated line. Since the output can be very large, mod your output by 10^9+7.Input:
2
0 0
2 1
Output:
1
2
Explanation: The Pascal Triangle has first value as 1. (0th row 0th element). The pascal triangle has next set of values as 1 1. (1th row). On the next level, pascal triangle has values 1 2 1. (2nd row). Its 1st element is 2. (0 based)., I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int modInverse(long long a, long long m)
{
long m0 = m;
long y = 0, x = 1;
while (a > 1)
{
int q = a / m;
int t = m;
m = a % m, a = t;
t = y;
y = x - q * y;
x = t;
}
if (x < 0)
x += m0;
return x;
}
int main()
{
long long mod=1000000007;
int t;
cin>>t;
while(t--){
int i,j;
cin>>i>>j;
i++;j++;
long long C=1;
long long x=1;
for (int p = 1; p < j; p++)
{
C = C * (i - p);
x=x*p;
C=C%mod;
x=x%mod;
}
C=C*modInverse(x,mod);
C=C%mod;
cout<<C<<endl;
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of n pairs of integers. Your task is to sort the array on the basis of the first element of pairs in descending order. If the first element is equal in two or more pairs then give preference to the pair that has a greater second element value.<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>SortPair()</b> that takes the array of pairs and the integer N(size of the array) as a parameter.
<b>Custom Input <b/>
The first line of input will be a single integer N. The next line of input contains 2*N space-separated integers where unique adjacent elements are pairs. Custom input for 1st sample:-
4
1 2 3 4 5 6 7 8
<b>Constraints:-</b>
1<=N<=10<sup>3</sup>
1<=value<=10<sup>5</sup>Return the sorted array of pairs.Sample Input 1:
4
(1, 2), (3, 4), (5, 6), (7, 8)
Sample Output 1:
(7, 8), (5, 6), (3, 4), (1, 2)
Sample Input 2:
3
(1, 1), (2, 2), (3, 3)
Sample Output 2:
(3, 3), (2, 2), (1, 1)
Sample Input 3:
3
(1, 1), (1, 2), (3, 3)
Sample Output 3:
(3, 3), (1, 2), (1, 1)
<b>Explanation :</b>
(1,2) and (1,1) have the same first element. But (1,2) has a greater second element so (1,2) comes before (1,1) in a sorted array.
, I have written this Solution Code: def SortPair(items,n):
items.sort(reverse = True)
return items, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of n pairs of integers. Your task is to sort the array on the basis of the first element of pairs in descending order. If the first element is equal in two or more pairs then give preference to the pair that has a greater second element value.<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>SortPair()</b> that takes the array of pairs and the integer N(size of the array) as a parameter.
<b>Custom Input <b/>
The first line of input will be a single integer N. The next line of input contains 2*N space-separated integers where unique adjacent elements are pairs. Custom input for 1st sample:-
4
1 2 3 4 5 6 7 8
<b>Constraints:-</b>
1<=N<=10<sup>3</sup>
1<=value<=10<sup>5</sup>Return the sorted array of pairs.Sample Input 1:
4
(1, 2), (3, 4), (5, 6), (7, 8)
Sample Output 1:
(7, 8), (5, 6), (3, 4), (1, 2)
Sample Input 2:
3
(1, 1), (2, 2), (3, 3)
Sample Output 2:
(3, 3), (2, 2), (1, 1)
Sample Input 3:
3
(1, 1), (1, 2), (3, 3)
Sample Output 3:
(3, 3), (1, 2), (1, 1)
<b>Explanation :</b>
(1,2) and (1,1) have the same first element. But (1,2) has a greater second element so (1,2) comes before (1,1) in a sorted array.
, I have written this Solution Code:
static Pair[] SortPair(Pair arr[], int n)
{
// Comparator to sort the pair according to second element
Arrays.sort(arr, new Comparator<Pair>() {
@Override public int compare(Pair p1, Pair p2)
{
if(p1.x==p2.x){
return p2.y-p1.y;
}
return p2.x-p1.x;
}
});
return arr;
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given an array of integers. Consider absolute difference between all the pairs of the the elements. You need to find Kth smallest absolute difference. If the size of the array is N then value of K will be less than N and more than or equal to 1.The first line of input contains number of test cases T.
The first line of each test case contains a two integers N and K denoting the number of elements in the array A and difference you need to output. The second line of each test case contains N space separated integers denoting the elements of the array A
Constraints:
1<= T <= 10
2 <= N <= 100000
1 <= K < N < 100000
0 <= A[i] <= 100000For each test case, output Kth smallest absolute difference.Input :
1
6 2
1 3 4 1 3 8
Output :
0
Explanation :
Test case 1: First smallest difference is 0, between the pair (1, 1) and second smallest absolute difference difference is also 0 between the pairs (3, 3)., I have written this Solution Code: import java.util.*;
import java.io.*;
import java.lang.*;
class Main{
public static void main(String[] args)throws IOException {
BufferedReader read = new BufferedReader(new InputStreamReader(System.in));
int t = Integer.parseInt(read.readLine().trim());
while (t-- > 0) {
String str[] = read.readLine().trim().split(" ");
int n = Integer.parseInt(str[0]);
int k = Integer.parseInt(str[1]);
int arr[] = new int[n];
str = read.readLine().trim().split(" ");
for (int i = 0; i < n; i++)
arr[i] = Integer.parseInt(str[i]);
System.out.println(Math.abs(small(arr, k)));
}
}
public static int small(int arr[], int k) {
Arrays.sort(arr);
int l = 0, r = arr[arr.length - 1] - arr[0];
while (r > l) {
int mid = l + (r - l) / 2;
if (count(arr, mid) < k) {
l = mid + 1;
} else {
r = mid;
}
}
return r;
}
public static int count(int arr[], int mid) {
int ans = 0, j = 0;
for (int i = 1; i < arr.length; ++i) {
while (j < i && arr[i] - arr[j] > mid) {
++j;
}
ans += i - j;
}
return ans;
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a matrix <b>A</b> of dimensions <b>n x m</b>. The task is to perform <b>boundary traversal</b> on the matrix in clockwise manner.The first line of input contains T denoting the number of testcases. T testcases follow. Each testcase has two lines of input. The first line contains dimensions of the matrix A, n and m. The second line contains n*m elements separated by spaces.
Constraints:
1 <= T <= 100
1 <= n, m <= 30
0 <= A[i][j] <= 100For each testcase, in a new line, print the boundary traversal of the matrix A.Input:
4
4 4
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
3 4
12 11 10 9 8 7 6 5 4 3 2 1
1 4
1 2 3 4
4 1
1 2 3 4
Output:
1 2 3 4 8 12 16 15 14 13 9 5
12 11 10 9 5 1 2 3 4 8
1 2 3 4
1 2 3 4
Explanation:
Testcase1: The matrix is:
1 2 3 4
5 6 7 8
9 10 11 12
13 14 15 16
The boundary traversal is 1 2 3 4 8 12 16 15 14 13 9 5
Testcase 2: Boundary Traversal will be 12 11 10 9 5 1 2 3 4 8.
Testcase 3: Boundary Traversal will be 1 2 3 4.
Testcase 4: Boundary Traversal will be 1 2 3 4., I have written this Solution Code:
def boundaryTraversal(matrix, N, M): #N = 3, M = 4
start_row = 0
start_col = 0
count = 0
if N != 1 and M != 1:
total = 2 * (N - 1) + 2 * (M - 1)
else:
total = max(M, N)
#First row
for i in range(start_col, M, 1): #0,1, 2, 3
count += 1
print(matrix[start_row][i], end = " ")
if count == total:
return
start_row += 1
#Last column
for i in range(start_row, N, 1): # 1, 2
count += 1
print(matrix[i][M - 1], end = " ")
if count == total:
return
M -= 1
#Last Row
for i in range(M - 1, start_col - 1, -1): # 2, 1, 0
count += 1
print(matrix[N - 1][i], end = " ")
if count == total:
return
#First Column
N -= 1 # 2
for i in range(N - 1, start_row - 1, -1):
count += 1
print(matrix[i][start_col], end = " ")
start_col += 1
testcase = int(input().strip())
for test in range(testcase):
dim = input().strip().split(" ")
N = int(dim[0])
M = int(dim[1])
matrix = []
elements = input().strip().split(" ") #N * M
start = 0
for i in range(N):
matrix.append(elements[start : start + M]) # (0, M - 1) | (M, 2*m-1)
start = start + M
boundaryTraversal(matrix, N, M)
print(), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a matrix <b>A</b> of dimensions <b>n x m</b>. The task is to perform <b>boundary traversal</b> on the matrix in clockwise manner.The first line of input contains T denoting the number of testcases. T testcases follow. Each testcase has two lines of input. The first line contains dimensions of the matrix A, n and m. The second line contains n*m elements separated by spaces.
Constraints:
1 <= T <= 100
1 <= n, m <= 30
0 <= A[i][j] <= 100For each testcase, in a new line, print the boundary traversal of the matrix A.Input:
4
4 4
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
3 4
12 11 10 9 8 7 6 5 4 3 2 1
1 4
1 2 3 4
4 1
1 2 3 4
Output:
1 2 3 4 8 12 16 15 14 13 9 5
12 11 10 9 5 1 2 3 4 8
1 2 3 4
1 2 3 4
Explanation:
Testcase1: The matrix is:
1 2 3 4
5 6 7 8
9 10 11 12
13 14 15 16
The boundary traversal is 1 2 3 4 8 12 16 15 14 13 9 5
Testcase 2: Boundary Traversal will be 12 11 10 9 5 1 2 3 4 8.
Testcase 3: Boundary Traversal will be 1 2 3 4.
Testcase 4: Boundary Traversal will be 1 2 3 4., I have written this Solution Code: import java.io.*;
import java.lang.*;
import java.util.*;
class Main
{
public static void main(String args[])
{
Scanner sc = new Scanner(System.in);
int t = sc.nextInt();
while(t-- > 0)
{
int n1 = sc.nextInt();
int m1 = sc.nextInt();
int arr1[][] = new int[n1][m1];
for(int i = 0; i < n1; i++)
{
for(int j = 0; j < m1; j++)
arr1[i][j] = sc.nextInt();
}
boundaryTraversal(n1, m1,arr1);
System.out.println();
}
}
static void boundaryTraversal( int n1, int m1, int arr1[][])
{
// base cases
if(n1 == 1)
{
int i = 0;
while(i < m1)
System.out.print(arr1[0][i++] + " ");
}
else if(m1 == 1)
{
int i = 0;
while(i < n1)
System.out.print(arr1[i++][0]+" ");
}
else
{
// traversing the first row
for(int j=0;j<m1;j++)
{
System.out.print(arr1[0][j]+" ");
}
// traversing the last column
for(int j=1;j<n1;j++)
{
System.out.print(arr1[j][m1-1]+ " ");
}
// traversing the last row
for(int j=m1-2;j>=0;j--)
{
System.out.print(arr1[n1-1][j]+" ");
}
// traversing the first column
for(int j=n1-2;j>=1;j--)
{
System.out.print(arr1[j][0]+" ");
}
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a linked list of N nodes. The task is to reverse the list by changing links between nodes (i.e if the list is 1->2->3->4 then it becomes 1<-2<-3<-4) and return the head of the modified 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> ReverseLinkedList</b> that takes head node as parameter.
Constraints:
1 <=N <= 1000
1 <= Node.data<= 100Return the head of the modified linked list.Input-1:
6
1 2 3 4 5 6
Output-1:
6 5 4 3 2 1
Explanation:
After reversing the list, elements are as 6->5->4->3->2->1.
Input-2:
5
1 2 8 4 5
Output-2:
5 4 8 2 1, I have written this Solution Code: public static Node ReverseLinkedList(Node head) {
Node prev = null;
Node current = head;
Node next = null;
while (current != null) {
next = current.next;
current.next = prev;
prev = current;
current = next;
}
head = prev;
return head;
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a positive integer N, your task is to print a right-angle triangle pattern of consecutive numbers of height N.
See the example for a better understanding.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>pattern()</b> that takes integer n as a parameter.
Constraint:
1 <= N <= 100Print a right angle triangle of numbers of height N.Sample Input:
5
Sample Output:
1
1 2
1 2 3
1 2 3 4
1 2 3 4 5
Sample Input:
2
Sample Output:
1
1 2, I have written this Solution Code: static void pattern(int n){
for(int i=1;i<=n;i++){
for(int j=1;j<=i;j++){
System.out.print(j + " ");
}
System.out.println();
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a positive integer N, your task is to print a right-angle triangle pattern of consecutive numbers of height N.
See the example for a better understanding.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>pattern()</b> that takes integer n as a parameter.
Constraint:
1 <= N <= 100Print a right angle triangle of numbers of height N.Sample Input:
5
Sample Output:
1
1 2
1 2 3
1 2 3 4
1 2 3 4 5
Sample Input:
2
Sample Output:
1
1 2, I have written this Solution Code:
void patternPrinting(int n){
for(int i=1;i<=n;i++){
for(int j=1;j<=i;j++){
printf("%d ",j);
}
printf("\n");
}
}
, In this Programming Language: C, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a positive integer N, your task is to print a right-angle triangle pattern of consecutive numbers of height N.
See the example for a better understanding.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>pattern()</b> that takes integer n as a parameter.
Constraint:
1 <= N <= 100Print a right angle triangle of numbers of height N.Sample Input:
5
Sample Output:
1
1 2
1 2 3
1 2 3 4
1 2 3 4 5
Sample Input:
2
Sample Output:
1
1 2, I have written this Solution Code: function pattern(n) {
// write code herenum
for(let i = 1;i<=n;i++){
let str = ''
for(let k = 1; k <= i;k++){
if(k === 1) {
str += `${k}`
}else{
str += ` ${k}`
}
}
console.log(str)
}
}, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a positive integer N, your task is to print a right-angle triangle pattern of consecutive numbers of height N.
See the example for a better understanding.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>pattern()</b> that takes integer n as a parameter.
Constraint:
1 <= N <= 100Print a right angle triangle of numbers of height N.Sample Input:
5
Sample Output:
1
1 2
1 2 3
1 2 3 4
1 2 3 4 5
Sample Input:
2
Sample Output:
1
1 2, I have written this Solution Code:
void patternPrinting(int n){
for(int i=1;i<=n;i++){
for(int j=1;j<=i;j++){
printf("%d ",j);
}
printf("\n");
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a positive integer N, your task is to print a right-angle triangle pattern of consecutive numbers of height N.
See the example for a better understanding.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>pattern()</b> that takes integer n as a parameter.
Constraint:
1 <= N <= 100Print a right angle triangle of numbers of height N.Sample Input:
5
Sample Output:
1
1 2
1 2 3
1 2 3 4
1 2 3 4 5
Sample Input:
2
Sample Output:
1
1 2, I have written this Solution Code: def patternPrinting(n):
for i in range(1,n+1):
for j in range (1,i+1):
print(j,end=' ')
print()
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Mohit has an array of N integers containing all elements from 1 to N, somehow he lost one element from the array.
Given N-1 elements your task is to find the missing one.The first line of input contains a single integer N, the next line contains N-1 space-separated integers.
<b>Constraints:-</b>
1 ≤ N ≤ 1000
1 ≤ elements ≤ NPrint the missing elementSample Input:-
3
3 1
Sample Output:
2
Sample Input:-
5
1 4 5 2
Sample Output:-
3, I have written this Solution Code: def getMissingNo(arr, n):
total = (n+1)*(n)//2
sum_of_A = sum(arr)
return total - sum_of_A
N = int(input())
arr = list(map(int,input().split()))
one = getMissingNo(arr,N)
print(one), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Mohit has an array of N integers containing all elements from 1 to N, somehow he lost one element from the array.
Given N-1 elements your task is to find the missing one.The first line of input contains a single integer N, the next line contains N-1 space-separated integers.
<b>Constraints:-</b>
1 ≤ N ≤ 1000
1 ≤ elements ≤ NPrint the missing elementSample Input:-
3
3 1
Sample Output:
2
Sample Input:-
5
1 4 5 2
Sample Output:-
3, 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-1];
for(int i=0;i<n-1;i++){
a[i]=sc.nextInt();
}
boolean present = false;
for(int i=1;i<=n;i++){
present=false;
for(int j=0;j<n-1;j++){
if(a[j]==i){present=true;}
}
if(present==false){
System.out.print(i);
return;
}
}
}
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Mohit has an array of N integers containing all elements from 1 to N, somehow he lost one element from the array.
Given N-1 elements your task is to find the missing one.The first line of input contains a single integer N, the next line contains N-1 space-separated integers.
<b>Constraints:-</b>
1 ≤ N ≤ 1000
1 ≤ elements ≤ NPrint the missing elementSample Input:-
3
3 1
Sample Output:
2
Sample Input:-
5
1 4 5 2
Sample Output:-
3, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main(){
int n;
cin>>n;
int a[n-1];
for(int i=0;i<n-1;i++){
cin>>a[i];
}
sort(a,a+n-1);
for(int i=1;i<n;i++){
if(i!=a[i-1]){cout<<i<<endl;return 0;}
}
cout<<n;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You have a sequence p1, p2, p3..., pn which is a permutation of {1, 2, 3, ..., n}. You can do the following operation at most 1 time:
Choose 2 indices i and j. Swap (pi, pj).
Can you sort the permutation.The first line of the input contains an integer n, the number of elements in the permutation.
The second line contains p1, p2, ..., pn.
Constraints
2 <= n <= 100Output "YES" if it is possible to sort the permutation, else output "NO".Sample Input
5
5 2 3 4 1
Sample Output
YES
Explanation: We can swap p1, p5.
Sample Input:
5
5 1 2 4 3
Sample output:
NO, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void inputArr(BufferedReader read, int[] arr,int arrSize) throws IOException {
String[] str = read.readLine().trim().split(" ");
for(int i = 0; i < arrSize; i++){
arr[i] = Integer.parseInt(str[i]);
}
}
public static void main (String[] args) throws IOException {
BufferedReader read = new BufferedReader(new InputStreamReader(System.in));
int arrSize = Integer.parseInt(read.readLine());
int[] arr = new int[arrSize];
inputArr(read, arr, arrSize);
int count = 0;
for(int i = 0; i < arrSize; i++)
if(arr[i] != i+1)
count++;
if(count>2)
System.out.println("NO");
else
System.out.println("YES");
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You have a sequence p1, p2, p3..., pn which is a permutation of {1, 2, 3, ..., n}. You can do the following operation at most 1 time:
Choose 2 indices i and j. Swap (pi, pj).
Can you sort the permutation.The first line of the input contains an integer n, the number of elements in the permutation.
The second line contains p1, p2, ..., pn.
Constraints
2 <= n <= 100Output "YES" if it is possible to sort the permutation, else output "NO".Sample Input
5
5 2 3 4 1
Sample Output
YES
Explanation: We can swap p1, p5.
Sample Input:
5
5 1 2 4 3
Sample output:
NO, I have written this Solution Code: n=int(input())
c=0
li=list(map(int,input().strip().split()))
for i in range(1,n+1):
if i!=li[i-1]:
c+=1
if c<=2:
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 have a sequence p1, p2, p3..., pn which is a permutation of {1, 2, 3, ..., n}. You can do the following operation at most 1 time:
Choose 2 indices i and j. Swap (pi, pj).
Can you sort the permutation.The first line of the input contains an integer n, the number of elements in the permutation.
The second line contains p1, p2, ..., pn.
Constraints
2 <= n <= 100Output "YES" if it is possible to sort the permutation, else output "NO".Sample Input
5
5 2 3 4 1
Sample Output
YES
Explanation: We can swap p1, p5.
Sample Input:
5
5 1 2 4 3
Sample output:
NO, I have written this Solution Code: #include<bits/stdc++.h>
using namespace std;
int main(){
int n;
cin>>n;
int a;
int cnt=0;
for(int i=1;i<=n;i++){
cin>>a;
if(a!=i){cnt++;}
}
if(cnt<=2){cout<<"YES";}
else{cout<<"NO";}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: 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: You are given a cubic dice with 6 faces. All the individual faces have a numbers printed on them. The numbers are in the range of 1 to 6, like any ordinary dice.
Given a number on one face of the dice , you need to print the number on the opposite face .
NOTE : In a normal dice sum of numbers on opposite faces is 7 .The first line of the input contains a single integer T, denoting the number of test cases. Then T test case follows. Each test case contains a single line of the input containing a positive integer N.
Constraints:
1 <= T <= 100
1 <= N <= 6For each testcase, print the number that is on the opposite side of the given face.Input:
2
6
2
Output:
1
5
Explanation:
Testcase 1: For dice facing number 6 opposite face will have the number 1., I have written this Solution Code: def get_opposite_face(n):
return 7-n
t = int(input())
for n in range(t):
print(get_opposite_face(int(input()))), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a cubic dice with 6 faces. All the individual faces have a numbers printed on them. The numbers are in the range of 1 to 6, like any ordinary dice.
Given a number on one face of the dice , you need to print the number on the opposite face .
NOTE : In a normal dice sum of numbers on opposite faces is 7 .The first line of the input contains a single integer T, denoting the number of test cases. Then T test case follows. Each test case contains a single line of the input containing a positive integer N.
Constraints:
1 <= T <= 100
1 <= N <= 6For each testcase, print the number that is on the opposite side of the given face.Input:
2
6
2
Output:
1
5
Explanation:
Testcase 1: For dice facing number 6 opposite face will have the number 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;
cout << 7-n << endl;
}
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a cubic dice with 6 faces. All the individual faces have a numbers printed on them. The numbers are in the range of 1 to 6, like any ordinary dice.
Given a number on one face of the dice , you need to print the number on the opposite face .
NOTE : In a normal dice sum of numbers on opposite faces is 7 .The first line of the input contains a single integer T, denoting the number of test cases. Then T test case follows. Each test case contains a single line of the input containing a positive integer N.
Constraints:
1 <= T <= 100
1 <= N <= 6For each testcase, print the number that is on the opposite side of the given face.Input:
2
6
2
Output:
1
5
Explanation:
Testcase 1: For dice facing number 6 opposite face will have the number 1., I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args)throws IOException {
BufferedReader br =new BufferedReader(new InputStreamReader(System.in));
int t=Integer.parseInt(br.readLine());
while(t-->0){
int n=Integer.parseInt(br.readLine());
System.out.println(6-n+1);
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a year(an integer variable) as input, find if it is a leap year or not.
Note: Leap year is the year that is multiple of 4. But, if the year is a multiple of 100 and not a multiple of 400, then it is not a leap year.<b>User Task:</b>
Complete the function <b>LeapYear()</b> that takes integer n as a parameter.
<b>Constraint:</b>
1 <= n <= 5000If it is a leap year then print <b>YES</b> and if it is not a leap year, then print <b>NO</b>Sample Input:
2000
Sample Output:
YES
Sample Input:
2003
Sample Output:
NO
Sample Input:
1900
Sample Output:
NO, I have written this Solution Code: n = int(input())
if (n%4==0 and n%100!=0 or n%400==0):
print("YES")
elif n==0:
print("YES")
else:
print("NO"), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a year(an integer variable) as input, find if it is a leap year or not.
Note: Leap year is the year that is multiple of 4. But, if the year is a multiple of 100 and not a multiple of 400, then it is not a leap year.<b>User Task:</b>
Complete the function <b>LeapYear()</b> that takes integer n as a parameter.
<b>Constraint:</b>
1 <= n <= 5000If it is a leap year then print <b>YES</b> and if it is not a leap year, then print <b>NO</b>Sample Input:
2000
Sample Output:
YES
Sample Input:
2003
Sample Output:
NO
Sample Input:
1900
Sample Output:
NO, I have written this Solution Code:
import java.util.Scanner;
class Main {
public static void main (String[] args)
{
//Capture the user's input
Scanner scanner = new Scanner(System.in);
//Storing the captured value in a variable
int n = scanner.nextInt();
LeapYear(n);
}
static void LeapYear(int year){
if(year%400==0 || (year%100 != 0 && year%4==0)){System.out.println("YES");}
else {
System.out.println("NO");}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: According to the latest government guidelines, the mass gathering of more than or equal to X people are not allowed but Sara organizes a party in her home and invited Y of her friends. As per the government guidelines your task is to tell Sara if she is in trouble or not.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>candies()</b> that takes integers X, and Y as arguments.
Constraints:-
1 <= X <= 1000
1 <= Y <= 1000Return 1 if Sara is in trouble else return 0.Sample Input:-
X = 3, Y = 5
Sample Output:-
1
Explanation:-
Total people in Sara's home = 5, no. of people allowed = 3
Sample Input:-
X = 4, Y = 2
Sample Output:-
0, I have written this Solution Code: static int candies(int X, int Y){
if(X<=Y){return 1;}
return 0;
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: According to the latest government guidelines, the mass gathering of more than or equal to X people are not allowed but Sara organizes a party in her home and invited Y of her friends. As per the government guidelines your task is to tell Sara if she is in trouble or not.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>candies()</b> that takes integers X, and Y as arguments.
Constraints:-
1 <= X <= 1000
1 <= Y <= 1000Return 1 if Sara is in trouble else return 0.Sample Input:-
X = 3, Y = 5
Sample Output:-
1
Explanation:-
Total people in Sara's home = 5, no. of people allowed = 3
Sample Input:-
X = 4, Y = 2
Sample Output:-
0, I have written this Solution Code: int candies(int X, int Y){
if(X<=Y){return 1;}
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: According to the latest government guidelines, the mass gathering of more than or equal to X people are not allowed but Sara organizes a party in her home and invited Y of her friends. As per the government guidelines your task is to tell Sara if she is in trouble or not.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>candies()</b> that takes integers X, and Y as arguments.
Constraints:-
1 <= X <= 1000
1 <= Y <= 1000Return 1 if Sara is in trouble else return 0.Sample Input:-
X = 3, Y = 5
Sample Output:-
1
Explanation:-
Total people in Sara's home = 5, no. of people allowed = 3
Sample Input:-
X = 4, Y = 2
Sample Output:-
0, I have written this Solution Code: def candies(X,Y):
if(X<=Y):
return 1
return 0
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: According to the latest government guidelines, the mass gathering of more than or equal to X people are not allowed but Sara organizes a party in her home and invited Y of her friends. As per the government guidelines your task is to tell Sara if she is in trouble or not.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>candies()</b> that takes integers X, and Y as arguments.
Constraints:-
1 <= X <= 1000
1 <= Y <= 1000Return 1 if Sara is in trouble else return 0.Sample Input:-
X = 3, Y = 5
Sample Output:-
1
Explanation:-
Total people in Sara's home = 5, no. of people allowed = 3
Sample Input:-
X = 4, Y = 2
Sample Output:-
0, I have written this Solution Code: int candies(int X, int Y){
if(X<=Y){return 1;}
return 0;
}, In this Programming Language: C, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Take an integer as input and print it.The first line contains integer as input.
<b>Constraints</b>
1 <= N <= 10Print the input integer in a single lineSample Input:-
2
Sample Output:-
2
Sample Input:-
4
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 printVariable(int variable){
System.out.println(variable);
}
public static void main (String[] args) throws java.lang.Exception
{
Scanner sc = new Scanner(System.in);
int num = sc.nextInt();
printVariable(num);
}
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a string s, your task is to divide the string in as many parts as it is possible such that a character can only be present in at most 1 part.Input contains a single string S.
Constraints:-
1 < = N < = 100000
Note:-
String will contain only lowercase english alphabetsPrint the maximum number of partitions.Sample Input:-
abcdaefghgheklmnol
Sample Output:-
4
Explanation:-
abcda efghghe k lmnol
Sample Input:-
newtonschool
Sample Output:-
2
Explanation:-
newtonschoo l, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws IOException {
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
String str=br.readLine();
int n=str.length();
int count=0;
int t,index;
for(int i=0;i<n;)
{
index=str.lastIndexOf(str.charAt(i));
++i;
if(index==i-1)
{
++count;
continue;
}
while(i<index)
{
t=str.lastIndexOf(str.charAt(i));
++i;
if(t==i-1) continue;
if(t>index) index=t;
}
++count;
i=index+1;
}
System.out.print(count);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a string s, your task is to divide the string in as many parts as it is possible such that a character can only be present in at most 1 part.Input contains a single string S.
Constraints:-
1 < = N < = 100000
Note:-
String will contain only lowercase english alphabetsPrint the maximum number of partitions.Sample Input:-
abcdaefghgheklmnol
Sample Output:-
4
Explanation:-
abcda efghghe k lmnol
Sample Input:-
newtonschool
Sample Output:-
2
Explanation:-
newtonschoo l, 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 200001
#define MOD 1000000007
#define read(type) readInt<type>()
#define int long long
#define out(x) cout<<x<<'\n'
#define out1(x) cout<<x<<" "
#define END cout<<'\n'
void fast(){
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
}
int m[100001];
signed main(){
string s;
cin>>s;
int last[26];
FOR(i,26){last[i]=0;}
FOR(i,s.length()){last[s[i]-'a']=i;}
int j = 0, ans=0;
for (int i = 0; i < s.length(); ++i) {
j = max(j, last[s[i] - 'a']);
if (i == j) {
ans++;
}
}
out(ans);
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: When learning a new language, we first learn to output some message. Here, we'll start with the famous "Hello World" message. There is no input, you just have to print "Hello World".No InputHello WorldExplanation:
Hello World is printed., I have written this Solution Code: a="Hello World"
print(a), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: When learning a new language, we first learn to output some message. Here, we'll start with the famous "Hello World" message. There is no input, you just have to print "Hello World".No InputHello WorldExplanation:
Hello World is printed., I have written this Solution Code: import java.util.*;
import java.io.*;
class Main{
public static void main(String args[]){
System.out.println("Hello World");
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Ram is studying in Class V and has four subjects, each subject carry 100 marks. He passed with flying colors in his exam, but when his neighbour asked how much percentage did he got in exam, he got stuck in calculation. Ram is a good student but he forgot how to calculate percentage. Help Ram to get him out of this problem.
Given four numbers a , b , c and d denoting the marks in four subjects . Calculate the overall percentage (floor value ) Ram got in exam .First line contains four variables a, b, c and d.
<b>Constraints</b>
1<= a, b, c, d <= 100
Print single line containing the percentage.Sample Input 1:
25 25 25 25
Sample Output 1:
25
Sample Input 2:
75 25 75 25
Sample Output 2:
50, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args)throws Exception {
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
String str[]=br.readLine().split(" ");
int a[]=new int[str.length];
int sum=0;
for(int i=0;i<str.length;i++)
{
a[i]=Integer.parseInt(str[i]);
sum=sum+a[i];
}
System.out.println(sum/4);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Ram is studying in Class V and has four subjects, each subject carry 100 marks. He passed with flying colors in his exam, but when his neighbour asked how much percentage did he got in exam, he got stuck in calculation. Ram is a good student but he forgot how to calculate percentage. Help Ram to get him out of this problem.
Given four numbers a , b , c and d denoting the marks in four subjects . Calculate the overall percentage (floor value ) Ram got in exam .First line contains four variables a, b, c and d.
<b>Constraints</b>
1<= a, b, c, d <= 100
Print single line containing the percentage.Sample Input 1:
25 25 25 25
Sample Output 1:
25
Sample Input 2:
75 25 75 25
Sample Output 2:
50, I have written this Solution Code: a,b,c,d = map(int,input().split())
print((a+b+c+d)*100//400), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Al is given task to build a skyscraper of N floors. He can build 2**i (2 to the power i, where i can be any non-negative integer) floors in one day. Report the minimum number of days required to build the skyscraper.First and only line of input contains a single integer N.
Constraints :
1 <= N <= 1000000000000Print a single integer, the minimum number of days required to build the skyscraper.Sample Input:
5
Sample Output:
2
Sample Input:
1
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 buf =new BufferedReader(new InputStreamReader(System.in));
long t=Long.parseLong(buf.readLine().trim());
int ans=0;
for(int i=45;i>=0;i--)
{
if(((t>>i)&1)==1)
ans++;
}
System.out.println(ans);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Al is given task to build a skyscraper of N floors. He can build 2**i (2 to the power i, where i can be any non-negative integer) floors in one day. Report the minimum number of days required to build the skyscraper.First and only line of input contains a single integer N.
Constraints :
1 <= N <= 1000000000000Print a single integer, the minimum number of days required to build the skyscraper.Sample Input:
5
Sample Output:
2
Sample Input:
1
Sample Output:
1, I have written this Solution Code: n = int(input())
print(bin(n).count('1')), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Al is given task to build a skyscraper of N floors. He can build 2**i (2 to the power i, where i can be any non-negative integer) floors in one day. Report the minimum number of days required to build the skyscraper.First and only line of input contains a single integer N.
Constraints :
1 <= N <= 1000000000000Print a single integer, the minimum number of days required to build the skyscraper.Sample Input:
5
Sample Output:
2
Sample Input:
1
Sample Output:
1, I have written this Solution Code: #include<bits/stdc++.h>
using namespace std;
#define pu push_back
#define fi first
#define se second
#define mp make_pair
#define int long long
#define pii pair<int,int>
#define mm (s+e)/2
#define all(x) x.begin(), x.end()
#define For(i, st, en) for(int i=st; i<en; i++)
#define tr(x) for(auto it=x.begin(); it!=x.end(); it++)
#define fast std::ios::sync_with_stdio(false);cin.tie(NULL);
#define sz 200000
signed main()
{
int n;
cin>>n;
int cnt=0;
while(n>0)
{
int p=n%2LL;
cnt+=p;
n/=2LL;
}
cout<<cnt<<endl;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a matrix Mat of m rows and n columns. The matrix is boolean so the elements of the matrix can only be either 0 or 1.
Now, if any row of the matrix contains a 1, then you need to fill that whole row with 1. After doing the mentioned operation, you need to print the modified matrix.The first line of input contains T denoting the number of test cases. T test cases follow.
The first line of each test case contains m and n denotes the number of rows and a number of columns.
Then next m lines contain n elements denoting the elements of the matrix.
Constraints:
1 ≤ T ≤ 20
1 ≤ m, n ≤ 700
Mat[I][j] ∈ {0,1}For each testcase, in a new line, print the modified matrix.Input:
1
5 4
1 0 0 0
0 0 0 0
0 1 0 0
0 0 0 0
0 0 0 1
Output:
1 1 1 1
0 0 0 0
1 1 1 1
0 0 0 0
1 1 1 1
Explanation:
Rows = 5 and columns = 4
The given matrix is
1 0 0 0
0 0 0 0
0 1 0 0
0 0 0 0
0 0 0 1
Evidently, the first row contains a 1 so fill the whole row with 1. The third row also contains a 1 so that row will be filled too. Finally, the last row contains a 1 and therefore it needs to be filled with 1 too.
The final matrix is
1 1 1 1
0 0 0 0
1 1 1 1
0 0 0 0
1 1 1 1, I have written this Solution Code: t=int(input())
while t!=0:
m,n=input().split()
m,n=int(m),int(n)
for i in range(m):
arr=input().strip()
if '1' in arr:
arr='1 '*n
else:
arr='0 '*n
print(arr)
t-=1, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a matrix Mat of m rows and n columns. The matrix is boolean so the elements of the matrix can only be either 0 or 1.
Now, if any row of the matrix contains a 1, then you need to fill that whole row with 1. After doing the mentioned operation, you need to print the modified matrix.The first line of input contains T denoting the number of test cases. T test cases follow.
The first line of each test case contains m and n denotes the number of rows and a number of columns.
Then next m lines contain n elements denoting the elements of the matrix.
Constraints:
1 ≤ T ≤ 20
1 ≤ m, n ≤ 700
Mat[I][j] ∈ {0,1}For each testcase, in a new line, print the modified matrix.Input:
1
5 4
1 0 0 0
0 0 0 0
0 1 0 0
0 0 0 0
0 0 0 1
Output:
1 1 1 1
0 0 0 0
1 1 1 1
0 0 0 0
1 1 1 1
Explanation:
Rows = 5 and columns = 4
The given matrix is
1 0 0 0
0 0 0 0
0 1 0 0
0 0 0 0
0 0 0 1
Evidently, the first row contains a 1 so fill the whole row with 1. The third row also contains a 1 so that row will be filled too. Finally, the last row contains a 1 and therefore it needs to be filled with 1 too.
The final matrix is
1 1 1 1
0 0 0 0
1 1 1 1
0 0 0 0
1 1 1 1, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define N 1000
int a[N][N];
// Driver code
int main()
{
int t;
cin>>t;
while(t--){
int n,m;
cin>>n>>m;
bool b[n];
for(int i=0;i<n;i++){
b[i]=false;
}
for(int i=0;i<n;i++){
for(int j=0;j<m;j++){
cin>>a[i][j];
if(a[i][j]==1){
b[i]=true;
}
}
}
for(int i=0;i<n;i++){
if(b[i]){
for(int j=0;j<m;j++){
cout<<1<<" ";
}}
else{
for(int j=0;j<m;j++){
cout<<0<<" ";
}
}
cout<<endl;
}
}}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a matrix Mat of m rows and n columns. The matrix is boolean so the elements of the matrix can only be either 0 or 1.
Now, if any row of the matrix contains a 1, then you need to fill that whole row with 1. After doing the mentioned operation, you need to print the modified matrix.The first line of input contains T denoting the number of test cases. T test cases follow.
The first line of each test case contains m and n denotes the number of rows and a number of columns.
Then next m lines contain n elements denoting the elements of the matrix.
Constraints:
1 ≤ T ≤ 20
1 ≤ m, n ≤ 700
Mat[I][j] ∈ {0,1}For each testcase, in a new line, print the modified matrix.Input:
1
5 4
1 0 0 0
0 0 0 0
0 1 0 0
0 0 0 0
0 0 0 1
Output:
1 1 1 1
0 0 0 0
1 1 1 1
0 0 0 0
1 1 1 1
Explanation:
Rows = 5 and columns = 4
The given matrix is
1 0 0 0
0 0 0 0
0 1 0 0
0 0 0 0
0 0 0 1
Evidently, the first row contains a 1 so fill the whole row with 1. The third row also contains a 1 so that row will be filled too. Finally, the last row contains a 1 and therefore it needs to be filled with 1 too.
The final matrix is
1 1 1 1
0 0 0 0
1 1 1 1
0 0 0 0
1 1 1 1, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main(String[] args) throws Exception{
InputStreamReader isr = new InputStreamReader(System.in);
BufferedReader bf = new BufferedReader(isr);
int t = Integer.parseInt(bf.readLine());
while (t-- > 0){
String inputs[] = bf.readLine().split(" ");
int m = Integer.parseInt(inputs[0]);
int n = Integer.parseInt(inputs[1]);
String[] matrix = new String[m];
for(int i=0; i<m; i++){
matrix[i] = bf.readLine();
}
StringBuffer ones = new StringBuffer("");
StringBuffer zeros = new StringBuffer("");
for(int i=0; i<n; i++){
ones.append("1 ");
zeros.append("0 ");
}
for(int i=0; i<m; i++){
if(matrix[i].contains("1")){
System.out.println(ones);
}else{
System.out.println(zeros);
}
}
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given two numbers n and p. You need to find n raised to the power p.<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>RecursivePower</b> that takes the integer n and p as a parameter.
Constraints:
1 <= T <= 10
1 <= n, p <= 9Return n^p.Sample Input:
3
2 3
9 9
2 9
Sample Output:
8
387420489
512
Explanation:
Test case 2: 387420489 is the value obtained when 9 is raised to the power of 9.
Test case 3: 512 is the value obtained when 2 is raised to the power of 9, I have written this Solution Code:
static int Power(int n,int p)
{
if(p==0)
return 1;
return n*Power(n,p-1);
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given an array of integers. Consider absolute difference between all the pairs of the the elements. You need to find Kth smallest absolute difference. If the size of the array is N then value of K will be less than N and more than or equal to 1.The first line of input contains number of test cases T.
The first line of each test case contains a two integers N and K denoting the number of elements in the array A and difference you need to output. The second line of each test case contains N space separated integers denoting the elements of the array A
Constraints:
1<= T <= 10
2 <= N <= 100000
1 <= K < N < 100000
0 <= A[i] <= 100000For each test case, output Kth smallest absolute difference.Input :
1
6 2
1 3 4 1 3 8
Output :
0
Explanation :
Test case 1: First smallest difference is 0, between the pair (1, 1) and second smallest absolute difference difference is also 0 between the pairs (3, 3)., I have written this Solution Code: import java.util.*;
import java.io.*;
import java.lang.*;
class Main{
public static void main(String[] args)throws IOException {
BufferedReader read = new BufferedReader(new InputStreamReader(System.in));
int t = Integer.parseInt(read.readLine().trim());
while (t-- > 0) {
String str[] = read.readLine().trim().split(" ");
int n = Integer.parseInt(str[0]);
int k = Integer.parseInt(str[1]);
int arr[] = new int[n];
str = read.readLine().trim().split(" ");
for (int i = 0; i < n; i++)
arr[i] = Integer.parseInt(str[i]);
System.out.println(Math.abs(small(arr, k)));
}
}
public static int small(int arr[], int k) {
Arrays.sort(arr);
int l = 0, r = arr[arr.length - 1] - arr[0];
while (r > l) {
int mid = l + (r - l) / 2;
if (count(arr, mid) < k) {
l = mid + 1;
} else {
r = mid;
}
}
return r;
}
public static int count(int arr[], int mid) {
int ans = 0, j = 0;
for (int i = 1; i < arr.length; ++i) {
while (j < i && arr[i] - arr[j] > mid) {
++j;
}
ans += i - j;
}
return ans;
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer array A of N integers, find the pair of integers in the array which have minimum XOR value. Report the minimum XOR value.First line denotes N, the size of the array.
Next line denotes N space-separated array elements.
Constraints:
2 <= N <= 100000
0 <= A[i] <= 10^7Print a single integer denoting minimum xor valueSample Input
4
0 2 5 7
Sample Output
2
Explanation:
0 xor 2 = 2
Sample Input
4
0 4 7 9
Sample Output
3, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) {
int ans = 0;
int mini = Integer.MAX_VALUE;
Scanner sc = new Scanner(System.in);
int array_size = sc.nextInt();
int N[] = new int[array_size];
for (int i = 0; i < array_size; i++) {
if(mini == 0){
break;
}
N[i] = sc.nextInt();
for (int j = i - 1; j >= 0; j--) {
ans = N[i] ^ N[j];
if (mini > ans) {
mini = ans;
}
}
}
System.out.println(mini);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer array A of N integers, find the pair of integers in the array which have minimum XOR value. Report the minimum XOR value.First line denotes N, the size of the array.
Next line denotes N space-separated array elements.
Constraints:
2 <= N <= 100000
0 <= A[i] <= 10^7Print a single integer denoting minimum xor valueSample Input
4
0 2 5 7
Sample Output
2
Explanation:
0 xor 2 = 2
Sample Input
4
0 4 7 9
Sample Output
3, I have written this Solution Code: a=int(input())
lis = list(map(int,input().split()))
val=666666789
for i in range(a+1):
for j in range(i+1,a):
temp = lis[i]^lis[j]
if(temp<val):
val = temp
if(val==0):
break
print(val), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer array A of N integers, find the pair of integers in the array which have minimum XOR value. Report the minimum XOR value.First line denotes N, the size of the array.
Next line denotes N space-separated array elements.
Constraints:
2 <= N <= 100000
0 <= A[i] <= 10^7Print a single integer denoting minimum xor valueSample Input
4
0 2 5 7
Sample Output
2
Explanation:
0 xor 2 = 2
Sample Input
4
0 4 7 9
Sample Output
3, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
// Function to find minimum XOR pair
int minXOR(int arr[], int n)
{
// Sort given array
sort(arr, arr + n);
int minXor = INT_MAX;
int val = 0;
// calculate min xor of consecutive pairs
for (int i = 0; i < n - 1; i++) {
val = arr[i] ^ arr[i + 1];
minXor = min(minXor, val);
}
return minXor;
}
// Driver program
int main()
{
int n;
cin>>n;
int arr[n];
for(int i=0;i<n;i++)
{
cin>>arr[i];
}
cout << minXOR(arr, n) << endl;
return 0;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Nutan and Tusla are both students at Newton School. They are both among the best students in the class. In order to know who is better among them, a game was organised. The game consisted of L rounds, where L is an odd integer. The student winning more rounds than the other was declared the winner.
You would be given a string of odd length L in which each character is 'N' or 'T'. If the i<sup>th</sup> character is 'N', then the i<sup>th</sup> round was won by Nutan, else if the character is 'T' it was won by Tusla. Print "Nutan'' if Nutan has won more rounds than Tusla, else print "Tusla'' if Tusla has won more rounds than Nutan.
Note: You have to print everything without quotes.The first line of the input contains a single integer L — the number of rounds (1 ≤ L ≤ 100 and L is odd).
The second line contains a string S of length L. Each character of S is either 'N' or 'T'.Print "Nutan" or "Tusla" according to the input.Sample Input:
3
NNT
Sample Output:
Nutan
Explanation:
Nutan has won two games while Tusla has only won a single game, so the overall winner is Nutan., I have written this Solution Code: import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;
import static java.lang.System.out;
public class Main {
static class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader() {
br = new BufferedReader(new InputStreamReader(System.in));
}
String next() {
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
long nextLong() {
return Long.parseLong(next());
}
double nextDouble() {
return Double.parseDouble(next());
}
String nextLine() {
String str = "";
try {
str = br.readLine();
} catch (IOException e) {
e.printStackTrace();
}
return str;
}
}
public static void main(String[] args) {
FastReader reader = new FastReader();
int n = reader.nextInt();
String S = reader.next();
int ncount = 0;
int tcount = 0;
for (char c : S.toCharArray()) {
if (c == 'N') ncount++;
else tcount++;
}
if (ncount > tcount) {
out.print("Nutan\n");
} else {
out.print("Tusla\n");
}
out.flush();
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Nutan and Tusla are both students at Newton School. They are both among the best students in the class. In order to know who is better among them, a game was organised. The game consisted of L rounds, where L is an odd integer. The student winning more rounds than the other was declared the winner.
You would be given a string of odd length L in which each character is 'N' or 'T'. If the i<sup>th</sup> character is 'N', then the i<sup>th</sup> round was won by Nutan, else if the character is 'T' it was won by Tusla. Print "Nutan'' if Nutan has won more rounds than Tusla, else print "Tusla'' if Tusla has won more rounds than Nutan.
Note: You have to print everything without quotes.The first line of the input contains a single integer L — the number of rounds (1 ≤ L ≤ 100 and L is odd).
The second line contains a string S of length L. Each character of S is either 'N' or 'T'.Print "Nutan" or "Tusla" according to the input.Sample Input:
3
NNT
Sample Output:
Nutan
Explanation:
Nutan has won two games while Tusla has only won a single game, so the overall winner is Nutan., I have written this Solution Code: n = int(input())
s = input()
a1 = s.count('N')
a2 = s.count('T')
if(a1 > a2):
print("Nutan")
else:
print('Tusla'), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Nutan and Tusla are both students at Newton School. They are both among the best students in the class. In order to know who is better among them, a game was organised. The game consisted of L rounds, where L is an odd integer. The student winning more rounds than the other was declared the winner.
You would be given a string of odd length L in which each character is 'N' or 'T'. If the i<sup>th</sup> character is 'N', then the i<sup>th</sup> round was won by Nutan, else if the character is 'T' it was won by Tusla. Print "Nutan'' if Nutan has won more rounds than Tusla, else print "Tusla'' if Tusla has won more rounds than Nutan.
Note: You have to print everything without quotes.The first line of the input contains a single integer L — the number of rounds (1 ≤ L ≤ 100 and L is odd).
The second line contains a string S of length L. Each character of S is either 'N' or 'T'.Print "Nutan" or "Tusla" according to the input.Sample Input:
3
NNT
Sample Output:
Nutan
Explanation:
Nutan has won two games while Tusla has only won a single game, so the overall winner is Nutan., I have written this Solution Code: //Author: Xzirium
//Time and Date: 02:18:28 24 March 2022
//Optional FAST
//#pragma GCC optimize("Ofast")
//#pragma GCC optimize("unroll-loops")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,fma,abm,mmx,avx,avx2,tune=native")
//Required Libraries
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/detail/standard_policies.hpp>
//Required namespaces
using namespace std;
using namespace __gnu_pbds;
//Required defines
#define endl '\n'
#define READ(X) cin>>X;
#define READV(X) long long X; cin>>X;
#define READAR(A,N) long long A[N]; for(long long i=0;i<N;i++) {cin>>A[i];}
#define rz(A,N) A.resize(N);
#define sz(X) (long long)(X.size())
#define pb push_back
#define pf push_front
#define fi first
#define se second
#define FORI(a,b,c) for(long long a=b;a<c;a++)
#define FORD(a,b,c) for(long long a=b;a>c;a--)
//Required typedefs
template <typename T> using ordered_set = tree<T,null_type,less<T>,rb_tree_tag,tree_order_statistics_node_update>;
template <typename T> using ordered_set1 = tree<T,null_type,greater<T>,rb_tree_tag,tree_order_statistics_node_update>;
typedef long long ll;
typedef long double ld;
typedef pair<int,int> pii;
typedef pair<long long,long long> pll;
//Required Constants
const long long inf=(long long)1e18;
const long long MOD=(long long)(1e9+7);
const long long INIT=(long long)(1e6+1);
const long double PI=3.14159265358979;
// Required random number generators
// mt19937 gen_rand_int(chrono::steady_clock::now().time_since_epoch().count());
// mt19937_64 gen_rand_ll(chrono::steady_clock::now().time_since_epoch().count());
//Required Functions
ll power(ll b, ll e)
{
ll r = 1ll;
for(; e > 0; e /= 2, (b *= b) %= MOD)
if(e % 2) (r *= b) %= MOD;
return r;
}
ll modInverse(ll a)
{
return power(a,MOD-2);
}
//Work
int main()
{
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
//-----------------------------------------------------------------------------------------------------------//
READV(N);
string S;
cin>>S;
ll n=0,t=0;
FORI(i,0,N)
{
if(S[i]=='N')
{
n++;
}
else if(S[i]=='T')
{
t++;
}
}
if(n>t)
{
cout<<"Nutan"<<endl;
}
else
{
cout<<"Tusla"<<endl;
}
//-----------------------------------------------------------------------------------------------------------//
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: There are n boys and m toys. Your task is to distribute the toys so that as many boys as possible will get a toy.
Each boy has a desired toy size, and they will accept any toy whose size is close enough to the desired size.
So if the desired toy size of a particular boy is 'a' and a particular toy has size 'b', then boy will only accept the toy if |b-a| <= k.The first input line has three integers n, m, and k: the number of boys, the number of toys, and the maximum allowed difference.
The next line contains n integers a[1], a[2],…, a[n]: the desired toy size of each boy. If the desired toy size of a boy is x, he will accept any toy whose size is between x−k and x+k.
The last line contains m integers b[1], b[2],…, b[m]: the size of each toy.
<b>Constraints</b>
1 ≤ n,m ≤ 200000
0 ≤ k ≤ 10<sup>9</sup>
1 ≤ a[i],b[i] ≤ 10<sup>9</sup>Print one integer: the number of boys who will get a toy.Sample Input
4 3 5
60 45 80 60
30 60 75
Sample Output
2
Explanation: One possible way can give second toy to first boy and third toy to third boy.
Sample Input:
10 10 0
37 62 56 69 34 46 10 86 16 49
50 95 47 43 9 62 83 71 71 7
Sample Output:
1, I have written this Solution Code: import java.io.FileInputStream;
import java.io.FileNotFoundException;
import java.io.IOException;
import java.io.InputStream;
import java.math.BigInteger;
import java.util.Arrays;
import java.util.HashMap;
import java.util.InputMismatchException;
import static java.lang.Math.pow;
class InputReader {
private boolean finished = false;
private InputStream stream;
private byte[] buf = new byte[1024];
private int curChar;
private int numChars;
private SpaceCharFilter filter;
public InputReader(InputStream stream) {
this.stream = stream;
}
public static InputReader getInputReader(boolean readFromTextFile) throws FileNotFoundException {
return ((readFromTextFile) ? new InputReader(new FileInputStream("src/input.txt"))
: new InputReader(System.in));
}
public int read() {
if (numChars == -1) {
throw new InputMismatchException();
}
if (curChar >= numChars) {
curChar = 0;
try {
numChars = stream.read(buf);
} catch (IOException e) {
throw new InputMismatchException();
}
if (numChars <= 0) {
return -1;
}
}
return buf[curChar++];
}
public int peek() {
if (numChars == -1) {
return -1;
}
if (curChar >= numChars) {
curChar = 0;
try {
numChars = stream.read(buf);
} catch (IOException e) {
return -1;
}
if (numChars <= 0) {
return -1;
}
}
return buf[curChar];
}
public int nextInt() {
int c = read();
while (isSpaceChar(c)) {
c = read();
}
int sgn = 1;
if (c == '-') {
sgn = -1;
c = read();
}
int res = 0;
do {
if (c < '0' || c > '9') {
throw new InputMismatchException();
}
res *= 10;
res += c - '0';
c = read();
} while (!isSpaceChar(c));
return res * sgn;
}
public long nextLong() {
int c = read();
while (isSpaceChar(c)) {
c = read();
}
int sgn = 1;
if (c == '-') {
sgn = -1;
c = read();
}
long res = 0;
do {
if (c < '0' || c > '9') {
throw new InputMismatchException();
}
res *= 10;
res += c - '0';
c = read();
} while (!isSpaceChar(c));
return res * sgn;
}
public String nextString() {
int c = read();
while (isSpaceChar(c)) {
c = read();
}
StringBuilder res = new StringBuilder();
do {
if (Character.isValidCodePoint(c)) {
res.appendCodePoint(c);
}
c = read();
} while (!isSpaceChar(c));
return res.toString();
}
public boolean isSpaceChar(int c) {
if (filter != null) {
return filter.isSpaceChar(c);
}
return isWhitespace(c);
}
public static boolean isWhitespace(int c) {
return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
}
private String readLine0() {
StringBuilder buf = new StringBuilder();
int c = read();
while (c != '\n' && c != -1) {
if (c != '\r') {
buf.appendCodePoint(c);
}
c = read();
}
return buf.toString();
}
public String readLine() {
String s = readLine0();
while (s.trim().length() == 0) {
s = readLine0();
}
return s;
}
public String readLine(boolean ignoreEmptyLines) {
if (ignoreEmptyLines) {
return readLine();
} else {
return readLine0();
}
}
public BigInteger readBigInteger() {
try {
return new BigInteger(nextString());
} catch (NumberFormatException e) {
throw new InputMismatchException();
}
}
public char nextCharacter() {
int c = read();
while (isSpaceChar(c)) {
c = read();
}
return (char) c;
}
public double nextDouble() {
int c = read();
while (isSpaceChar(c)) {
c = read();
}
int sgn = 1;
if (c == '-') {
sgn = -1;
c = read();
}
double res = 0;
while (!isSpaceChar(c) && c != '.') {
if (c == 'e' || c == 'E') {
return res * pow(10, nextInt());
}
if (c < '0' || c > '9') {
throw new InputMismatchException();
}
res *= 10;
res += c - '0';
c = read();
}
if (c == '.') {
c = read();
double m = 1;
while (!isSpaceChar(c)) {
if (c == 'e' || c == 'E') {
return res * pow(10, nextInt());
}
if (c < '0' || c > '9') {
throw new InputMismatchException();
}
m /= 10;
res += (c - '0') * m;
c = read();
}
}
return res * sgn;
}
public boolean isExhausted() {
int value;
while (isSpaceChar(value = peek()) && value != -1) {
read();
}
return value == -1;
}
public String next() {
return nextString();
}
public SpaceCharFilter getFilter() {
return filter;
}
public void setFilter(SpaceCharFilter filter) {
this.filter = filter;
}
public interface SpaceCharFilter {
public boolean isSpaceChar(int ch);
}
public int[] nextIntArray(int n) {
int[] array = new int[n];
for (int i = 0; i < n; ++i) array[i] = nextInt();
return array;
}
public int[] nextSortedIntArray(int n) {
int array[] = nextIntArray(n);
Arrays.sort(array);
return array;
}
public int[] nextSumIntArray(int n) {
int[] array = new int[n];
array[0] = nextInt();
for (int i = 1; i < n; ++i) array[i] = array[i - 1] + nextInt();
return array;
}
public long[] nextLongArray(int n) {
long[] array = new long[n];
for (int i = 0; i < n; ++i) array[i] = nextLong();
return array;
}
public long[] nextSumLongArray(int n) {
long[] array = new long[n];
array[0] = nextInt();
for (int i = 1; i < n; ++i) array[i] = array[i - 1] + nextInt();
return array;
}
public long[] nextSortedLongArray(int n) {
long array[] = nextLongArray(n);
Arrays.sort(array);
return array;
}
}
class Main {
public static void main (String[] args) throws FileNotFoundException {
InputReader sc = InputReader.getInputReader(false);
int n = sc.nextInt();
int m = sc.nextInt();
long k = sc.nextLong();
long[] boyArray = new long[n];
long[] toyArray = new long[m];
for (int i = 0; i < n; i++) {
boyArray[i] = sc.nextLong();
}
for (int i = 0; i < m; i++) {
toyArray[i] = sc.nextLong();
}
Arrays.sort(boyArray);
Arrays.sort(toyArray);
int cnt = 0;
int l = 0 , h = n-1;
for (int i = 0; i < m; i++) {
while( h >= l){
if(toyArray[i] < boyArray[l]-k || toyArray[i] > boyArray[h]+k){
break;
}
else if(toyArray[i] >= boyArray[l]-k && toyArray[i] <= boyArray[l]+k){
l++;
cnt++;
break;
}
else if(toyArray[i] >= boyArray[h]-k && toyArray[i] <= boyArray[h]+k){
h--;
cnt++;
break;
}
else{
l++;
}
}
}
System.out.println(cnt);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: There are n boys and m toys. Your task is to distribute the toys so that as many boys as possible will get a toy.
Each boy has a desired toy size, and they will accept any toy whose size is close enough to the desired size.
So if the desired toy size of a particular boy is 'a' and a particular toy has size 'b', then boy will only accept the toy if |b-a| <= k.The first input line has three integers n, m, and k: the number of boys, the number of toys, and the maximum allowed difference.
The next line contains n integers a[1], a[2],…, a[n]: the desired toy size of each boy. If the desired toy size of a boy is x, he will accept any toy whose size is between x−k and x+k.
The last line contains m integers b[1], b[2],…, b[m]: the size of each toy.
<b>Constraints</b>
1 ≤ n,m ≤ 200000
0 ≤ k ≤ 10<sup>9</sup>
1 ≤ a[i],b[i] ≤ 10<sup>9</sup>Print one integer: the number of boys who will get a toy.Sample Input
4 3 5
60 45 80 60
30 60 75
Sample Output
2
Explanation: One possible way can give second toy to first boy and third toy to third boy.
Sample Input:
10 10 0
37 62 56 69 34 46 10 86 16 49
50 95 47 43 9 62 83 71 71 7
Sample Output:
1, I have written this Solution Code: n,m,k=[int(x) for x in input().split()[:3]]
a=[int(x) for x in input().split()[:n]]
b=[int(x) for x in input().split()[:m]]
a.sort()
b.sort()
j = 0
sum1 = 0
for i in a:
while(j<len(b)):
if(i-k<=b[j] and b[j]<=i+k):
j = j+1
sum1 +=1
break
elif i+k<b[j]:
break
else:
j +=1
if(j>=len(b)):
break
print(sum1), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: There are n boys and m toys. Your task is to distribute the toys so that as many boys as possible will get a toy.
Each boy has a desired toy size, and they will accept any toy whose size is close enough to the desired size.
So if the desired toy size of a particular boy is 'a' and a particular toy has size 'b', then boy will only accept the toy if |b-a| <= k.The first input line has three integers n, m, and k: the number of boys, the number of toys, and the maximum allowed difference.
The next line contains n integers a[1], a[2],…, a[n]: the desired toy size of each boy. If the desired toy size of a boy is x, he will accept any toy whose size is between x−k and x+k.
The last line contains m integers b[1], b[2],…, b[m]: the size of each toy.
<b>Constraints</b>
1 ≤ n,m ≤ 200000
0 ≤ k ≤ 10<sup>9</sup>
1 ≤ a[i],b[i] ≤ 10<sup>9</sup>Print one integer: the number of boys who will get a toy.Sample Input
4 3 5
60 45 80 60
30 60 75
Sample Output
2
Explanation: One possible way can give second toy to first boy and third toy to third boy.
Sample Input:
10 10 0
37 62 56 69 34 46 10 86 16 49
50 95 47 43 9 62 83 71 71 7
Sample Output:
1, I have written this Solution Code:
#include <bits/stdc++.h>
using namespace std;
#define MEM(a, b) memset(a, (b), sizeof(a))
#define FOREACH(it, l) for (auto it = l.begin(); it != l.end(); it++)
#define IN(A, B, C) assert( B <= A && A <= C)
#define MP make_pair
#define FOR(i,a) for(int i=0;i<a;i++)
#define FOR1(i,j,a) for(int i=j;i<a;i++)
#define EB emplace_back
#define INF (int)1e9
#define EPS 1e-9
#define PI 3.1415926535897932384626433832795
#define MOD 1000000007
const int N = 3e5+5;
#define read(type) readInt<type>()
#define max1 200010
#define out(x) cout<<x<<'\n'
#define out1(x) cout<<x<<" "
#define END cout<<'\n'
const double pi=acos(-1.0);
typedef pair<int, int> PII;
typedef vector<int> VI;
typedef vector<string> VS;
typedef vector<PII> VII;
typedef vector<VI> VVI;
typedef map<int,int> MPII;
typedef set<int> SETI;
typedef multiset<int> MSETI;
typedef long int li;
typedef unsigned long int uli;
typedef long long int ll;
typedef unsigned long long int ull;
const ll inf = 0x3f3f3f3f3f3f3f3f;
const ll mod = 998244353;
bool isPowerOfTwo (int x)
{
/* First x in the below expression is
for the case when x is 0 */
return x && (!(x&(x-1)));
}
void fast(){
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
}
ll power(ll x, ll y, ll p)
{
ll res = 1; // Initialize result
x = x % p; // Update x if it is more than or
// equal to p
while (y > 0)
{
// If y is odd, multiply x with result
if (y & 1)
res = (res*x) % p;
// y must be even now
y = y>>1; // y = y/2
x = (x*x) % p;
}
return res;
}
// Returns n^(-1) mod p
ll modInverse(ll n, ll p)
{
return power(n, p-2, p);
}
// Returns nCr % p using Fermat's little
// theorem.
int c[max1];
int main()
{
int n,m;
cin>>n>>m;
ll k;
cin>>k;
ll a[n],b[m];
FOR(i,n){
cin>>a[i];
}
FOR(i,m){
cin>>b[i];}
sort(a,a+n);
sort(b,b+m);
int i=0,j=0;
int cnt=0;
while(i!=n && j!=m){
if(abs(b[j]-a[i])<=k){cnt++;i++;j++;}
else{
if(b[j]>a[i]){i++;}
else{j++;}
}
}
out(cnt);
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Calculate inversion count of array of integers.
Inversion count of an array is quantisation of how much unsorted an array is. A sorted array has inversion count 0, while an unsorted array has maximum inversion count.
Formally speaking inversion count = number of pairs i, j such that i < j and a[i] > a[j].The first line contain integers N.
The second line of the input contains N singly spaces integers.
1 <= N <= 100000
1 <= A[i] <= 1000000000Output one integer the inversion count.Sample Input
5
1 1 3 2 2
Sample Output
2
Sample Input
5
5 4 3 2 1
Sample Output
10, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
static long count(int[] arr, int l, int h, int[] aux) {
if (l >= h) return 0;
int mid = (l +h) / 2;
long count = 0;
count += count(aux, l, mid, arr);
count += count(aux, mid + 1, h, arr);
count += merge(arr, l, mid, h, aux);
return count;
}
static long merge(int[] arr, int l, int mid, int h, int[] aux) {
long count = 0;
int i = l, j = mid + 1, k = l;
while (i <= mid || j <= h) {
if (i > mid) {
arr[k++] = aux[j++];
} else if (j > h) {
arr[k++] = aux[i++];
} else if (aux[i] <= aux[j]) {
arr[k++] = aux[i++];
} else {
arr[k++] = aux[j++];
count += mid + 1 - i;
}
}
return count;
}
public static void main (String[] args)throws IOException {
BufferedReader br =new BufferedReader(new InputStreamReader(System.in));
String str[];
str = br.readLine().split(" ");
int n = Integer.parseInt(str[0]);
str = br.readLine().split(" ");
int arr[] =new int[n];
for (int j = 0; j < n; j++) {
arr[j] = Integer.parseInt(str[j]);
}
int[] aux = arr.clone();
System.out.print(count(arr, 0, n - 1, aux));
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Calculate inversion count of array of integers.
Inversion count of an array is quantisation of how much unsorted an array is. A sorted array has inversion count 0, while an unsorted array has maximum inversion count.
Formally speaking inversion count = number of pairs i, j such that i < j and a[i] > a[j].The first line contain integers N.
The second line of the input contains N singly spaces integers.
1 <= N <= 100000
1 <= A[i] <= 1000000000Output one integer the inversion count.Sample Input
5
1 1 3 2 2
Sample Output
2
Sample Input
5
5 4 3 2 1
Sample Output
10, I have written this Solution Code: count=0
def implementMergeSort(arr,s,e):
global count
if e-s==1:
return
mid=(s+e)//2
implementMergeSort(arr,s,mid)
implementMergeSort(arr,mid,e)
count+=merge_sort_place(arr,s,mid,e)
return count
def merge_sort_place(arr,s,mid,e):
arr3=[]
i=s
j=mid
count=0
while i<mid and j<e:
if arr[i]>arr[j]:
arr3.append(arr[j])
j+=1
count+=(mid-i)
else:
arr3.append(arr[i])
i+=1
while (i<mid):
arr3.append(arr[i])
i+=1
while (j<e):
arr3.append(arr[j])
j+=1
for x in range(len(arr3)):
arr[s+x]=arr3[x]
return count
n=int(input())
arr=list(map(int,input().split()[:n]))
c=implementMergeSort(arr,0,len(arr))
print(c), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Calculate inversion count of array of integers.
Inversion count of an array is quantisation of how much unsorted an array is. A sorted array has inversion count 0, while an unsorted array has maximum inversion count.
Formally speaking inversion count = number of pairs i, j such that i < j and a[i] > a[j].The first line contain integers N.
The second line of the input contains N singly spaces integers.
1 <= N <= 100000
1 <= A[i] <= 1000000000Output one integer the inversion count.Sample Input
5
1 1 3 2 2
Sample Output
2
Sample Input
5
5 4 3 2 1
Sample Output
10, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
long long _mergeSort(long long arr[], int temp[], int left, int right);
long long merge(long long arr[], int temp[], int left, int mid, int right);
/* This function sorts the input array and returns the
number of inversions in the array */
long long mergeSort(long long arr[], int array_size)
{
int temp[array_size];
return _mergeSort(arr, temp, 0, array_size - 1);
}
/* An auxiliary recursive function that sorts the input array and
returns the number of inversions in the array. */
long long _mergeSort(long long arr[], int temp[], int left, int right)
{
long long mid, inv_count = 0;
if (right > left) {
/* Divide the array into two parts and
call _mergeSortAndCountInv()
for each of the parts */
mid = (right + left) / 2;
/* Inversion count will be sum of
inversions in left-part, right-part
and number of inversions in merging */
inv_count += _mergeSort(arr, temp, left, mid);
inv_count += _mergeSort(arr, temp, mid + 1, right);
/*Merge the two parts*/
inv_count += merge(arr, temp, left, mid + 1, right);
}
return inv_count;
}
/* This funt merges two sorted arrays
and returns inversion count in the arrays.*/
long long merge(long long arr[], int temp[], int left,
int mid, int right)
{
int i, j, k;
long long inv_count = 0;
i = left; /* i is index for left subarray*/
j = mid; /* j is index for right subarray*/
k = left; /* k is index for resultant merged subarray*/
while ((i <= mid - 1) && (j <= right)) {
if (arr[i] <= arr[j]) {
temp[k++] = arr[i++];
}
else {
temp[k++] = arr[j++];
/* this is tricky -- see above
explanation/diagram for merge()*/
inv_count = inv_count + (mid - i);
}
}
/* Copy the remaining elements of left subarray
(if there are any) to temp*/
while (i <= mid - 1)
temp[k++] = arr[i++];
/* Copy the remaining elements of right subarray
(if there are any) to temp*/
while (j <= right)
temp[k++] = arr[j++];
/*Copy back the merged elements to original array*/
for (i = left; i <= right; i++)
arr[i] = temp[i];
return inv_count;}
int main(){
int n;
cin>>n;
long long a[n];
for(int i=0;i<n;i++){
cin>>a[i];}
long long ans = mergeSort(a, n);
cout << ans; }
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a binary array A of size N. Now, an array P is calculated from A such that:
P[i] = A[i] + P[i-1]
(for 1 <= i <= N and P[0] = 0)
Now, you have to handle two types of queries:
1 L R - invert every element of A from L to R (If A[i] = 0, change it to 1 and vice versa)
2 L R - print the sum of array P from L to R
The array P is updated after every query of type 1.The first line contains two integers N and Q denoting the number of elements of the array and the number of queries respectively. The next line contains N boolean values representing the element of array A. The next Q lines contain three integers T L R denoting the queries.
1 <= N, Q <= 200000
0 <= A[i] <= 1
1 <= T <= 2
1 <= L <= R <= NFor every query of type 2, print a single integer denoting the sum of range from L to R in a separate line. Since the sum can be quite large, print sum modulo (10^9 + 7)Sample Input:
9 6
1 0 0 1 1 0 0 0 1
1 2 6
1 4 8
2 1 9
2 3 5
1 2 7
2 5 8
Sample Output:
41
12
8
Explanation:
Array P initially: [1, 1, 1, 2, 3, 3, 3, 3, 4]
After 1st query:
A: [1, 1, 1, 0, 0, 1, 0, 0, 1]
P: [1, 2, 3, 3, 3, 4, 4, 4, 5]
After 2nd query:
A: [1, 1, 1, 1, 1, 0, 1, 1, 1]
P: [1, 2, 3, 4, 5, 5, 6, 7, 8], I have written this Solution Code: #include "bits/stdc++.h"
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 n, q;
bool a[N], lz[4*N];
int sum[4*N][2], sumi[4*N][2];
void merge(int nd){
for(int i = 0; i < 2; i++){
sum[nd][i] = sum[nd << 1][i] + sum[nd << 1 | 1][i];
sumi[nd][i] = sumi[nd << 1][i] + sumi[nd << 1 | 1][i];
}
}
void modify(int nd){
swap(sum[nd][0], sum[nd][1]);
swap(sumi[nd][0], sumi[nd][1]);
}
void pushdown(int nd, int s, int e){
if(lz[nd] == 0) return;
modify(nd);
if(s != e){
lz[nd << 1] ^= lz[nd];
lz[nd << 1 | 1] ^= lz[nd];
}
lz[nd] = 0;
}
void build(int nd, int s, int e){
if(s == e){
sum[nd][a[s]] = n-s+1;
sumi[nd][a[s]]++;
return;
}
int md = (s + e) >> 1;
build(nd << 1, s, md);
build(nd << 1 | 1, md+1, e);
merge(nd);
}
void upd(int nd, int s, int e, int l, int r){
pushdown(nd, s, e);
if(s > e || s > r || e < l)
return;
if(s >= l && e <= r){
modify(nd);
if(s != e){
lz[nd << 1] ^= 1;
lz[nd << 1 | 1] ^= 1;
}
return;
}
int md = (s + e) >> 1;
upd(nd << 1, s, md, l, r);
upd(nd << 1 | 1, md+1, e, l, r);
merge(nd);
}
pi query(int nd, int s, int e, int l, int r){
pushdown(nd, s, e);
if(s > e || s > r || e < l)
return {0, 0};
if(s >= l && e <= r)
return {sum[nd][0], sumi[nd][0]};
int md = (s + e) >> 1;
pi p = query(nd << 1, s, md, l, r);
pi q = query(nd << 1 | 1, md+1, e, l, r);
return {p.fi+q.fi, p.se+q.se};
}
void solve(){
cin >> n >> q;
for(int i = 1; i <= n; i++)
cin >> a[i];
build(1, 1, n);
while(q--){
int t, l, r;
cin >> t >> l >> r;
if(t == 1)
upd(1, 1, n, l, r);
else{
pi x = query(1, 1, n, l, r);
pi y = query(1, 1, n, 1, l-1);
int ans = r*(r+1)/2 - (l-1)*l/2;
ans = ans - (x.fi - x.se*(n-r)) - y.se*(r-l+1);
cout << ans % mod << endl;
}
}
}
void testcases(){
int tt = 1;
//cin >> tt;
while(tt--){
solve();
}
}
signed main() {
IOS;
clock_t start = clock();
testcases();
cerr << (double)(clock() - start)*1000/CLOCKS_PER_SEC << " ms" << 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, 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: 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 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 a year(an integer variable) as input, find if it is a leap year or not.
Note: Leap year is the year that is multiple of 4. But, multiples of 100 which are not multiples of 400 are not leap years.The input contains a single integer N
<b>Constraint:</b>
1 <= n <= 10<sup>4</sup>Print "YES" if the year is a leap year else print "NO".Sample Input:
2000
Sample Output:
YES
Sample Input:
2003
Sample Output:
NO
Sample Input:
1900
Sample Output:
NO, I have written this Solution Code:
function LeapYear(year){
// write code here
// return the output using return keyword
// do not use console.log here
if ((0 != year % 4) || ((0 == year % 100) && (0 != year % 400))) {
return 0;
} else {
return 1
}
}, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a year(an integer variable) as input, find if it is a leap year or not.
Note: Leap year is the year that is multiple of 4. But, multiples of 100 which are not multiples of 400 are not leap years.The input contains a single integer N
<b>Constraint:</b>
1 <= n <= 10<sup>4</sup>Print "YES" if the year is a leap year else print "NO".Sample Input:
2000
Sample Output:
YES
Sample Input:
2003
Sample Output:
NO
Sample Input:
1900
Sample Output:
NO, I have written this Solution Code: year = int(input())
if year % 4 == 0 and not year % 100 == 0 or year % 400 == 0:
print("YES")
else:
print("NO"), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a year(an integer variable) as input, find if it is a leap year or not.
Note: Leap year is the year that is multiple of 4. But, multiples of 100 which are not multiples of 400 are not leap years.The input contains a single integer N
<b>Constraint:</b>
1 <= n <= 10<sup>4</sup>Print "YES" if the year is a leap year else print "NO".Sample Input:
2000
Sample Output:
YES
Sample Input:
2003
Sample Output:
NO
Sample Input:
1900
Sample Output:
NO, I have written this Solution Code: import java.util.Scanner;
class Main {
public static void main (String[] args)
{
//Capture the user's input
Scanner scanner = new Scanner(System.in);
//Storing the captured value in a variable
int side = scanner.nextInt();
int area = LeapYear(side);
if(area==1){
System.out.println("YES");}
else{
System.out.println("NO");}
}
static int LeapYear(int year){
if(year%400==0){return 1;}
if(year%100 != 0 && year%4==0){return 1;}
else {
return 0;}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a string S, your task is to print the string S.You don't have to worry about taking input, you just have to complete the function <b>printString</b>Print the string S.Sample Input 1:-
NewtonSchool
Sample Output 1:-
NewtonSchool
Sample Input 2:-
Hello
Sample Output 2:-
Hello, I have written this Solution Code: static void printString(String stringVariable){
System.out.println(stringVariable);
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a string S, your task is to print the string S.You don't have to worry about taking input, you just have to complete the function <b>printString</b>Print the string S.Sample Input 1:-
NewtonSchool
Sample Output 1:-
NewtonSchool
Sample Input 2:-
Hello
Sample Output 2:-
Hello, I have written this Solution Code: S=input()
print(S), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a string S, your task is to print the string S.You don't have to worry about taking input, you just have to complete the function <b>printString</b>Print the string S.Sample Input 1:-
NewtonSchool
Sample Output 1:-
NewtonSchool
Sample Input 2:-
Hello
Sample Output 2:-
Hello, I have written this Solution Code: void printString(string s){
cout<<s;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given an array P of size N. You have to find the maximum value of <b>P<sub>i</sub> % P<sub>j</sub></b> for all possible i, j pairs.First line contains a single integer N.
The second line of the input contains N space seperated integers.
Constraints:
2 <= N <= 10<sup>5</sup>
1 <= P<sub>i</sub> <= 10<sup>9</sup>Print the maximum value of P<sub>i</sub> % P<sub>j</sub> for all possible i, j pairs in the array.Sample Input:
5
8 6 7 9 4
Sample Output:
8
Explanation:
i = 1, j = 4 (1-based indexing) will give the maximum possible result over all (i, j) pairs., I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
static int getMaxValue(int arr[], int arr_size)
{
int i, first, second;
if (arr_size < 2)
{
return 0;
}
first = second = Integer.MIN_VALUE;
for (i = 0; i < arr_size; i++)
{
if (arr[i] > first)
{
second = first;
first = arr[i];
}
else if (arr[i] > second && arr[i] != first)
{
second = arr[i];
}
}
if (second == Integer.MIN_VALUE)
{
return 0;
}
else
{
return second;
}
}
public static void main (String[] args) throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int n = Integer.parseInt(br.readLine());
String[] num = br.readLine().split(" ");
int arr[] = new int[n];
for(int i=0;i<n;i++) arr[i] =Integer.parseInt(num[i]);
int max = Integer.MIN_VALUE;
System.out.println(getMaxValue(arr, n));
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given an array P of size N. You have to find the maximum value of <b>P<sub>i</sub> % P<sub>j</sub></b> for all possible i, j pairs.First line contains a single integer N.
The second line of the input contains N space seperated integers.
Constraints:
2 <= N <= 10<sup>5</sup>
1 <= P<sub>i</sub> <= 10<sup>9</sup>Print the maximum value of P<sub>i</sub> % P<sub>j</sub> for all possible i, j pairs in the array.Sample Input:
5
8 6 7 9 4
Sample Output:
8
Explanation:
i = 1, j = 4 (1-based indexing) will give the maximum possible result over all (i, j) pairs., I have written this Solution Code: n=int(input())
arr=list(map(int,input().split()))
arr.sort()
a=0
b=0
for i in range(n):
if arr[i]>b:
a=b
b=arr[i]
print(a%b), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given an array P of size N. You have to find the maximum value of <b>P<sub>i</sub> % P<sub>j</sub></b> for all possible i, j pairs.First line contains a single integer N.
The second line of the input contains N space seperated integers.
Constraints:
2 <= N <= 10<sup>5</sup>
1 <= P<sub>i</sub> <= 10<sup>9</sup>Print the maximum value of P<sub>i</sub> % P<sub>j</sub> for all possible i, j pairs in the array.Sample Input:
5
8 6 7 9 4
Sample Output:
8
Explanation:
i = 1, j = 4 (1-based indexing) will give the maximum possible result over all (i, j) pairs., I have written this Solution Code: #include<bits/stdc++.h>
using namespace std;
#define fast ios_base::sync_with_stdio(false); cin.tie(NULL);
#define int long long
#define pb push_back
#define ff first
#define ss second
#define endl '\n'
#define all(a) a.begin(), a.end()
#define rall(a) a.rbegin(), a.rend()
using T = pair<int, int>;
typedef long long ll;
const int mod = 1e9 + 7;
const int INF = 1e9;
void solve() {
int n;
cin >> n;
vector<int> p(n);
set<int> s;
for(auto &i : p) cin >> i, s.insert(i);
if(s.size() == 1){
cout << 0;
}
else{
auto it = s.rbegin();
it++;
cout << (*it);
}
}
signed main(){
fast
int t = 1;
// cin >> t;
for(int i = 1; i <= t; i++){
solve();
if(i != t) cout << endl;
}
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Ram has recently purchased an empty rectangular grid of NxM dimension. He wants to fill the grid but in a special way. His way of filling is, He choose any block which is not filled yet and fills it and the points obtained in particular step is the number of filled neighbouring block. Neighbour block denotes the block which shares some common side between them. When Ram fills all the blocks then he will stop and calculate the total number of points. Ram wants to know the maximum points he can get?The first line of the input contains a single Integer T, denoting number of test cases.
The next T lines contains two space separated integer N, M i.e dimensions of grid.
<b>Constraints</b>
1 ≤ T ≤ 100
1 ≤ N, M ≤ 1000Print the maximum number of points Ram can get.Sample Input
1
2 2
Sample Output
4
Explanation
Ram can obtain total score 4 in the following way.
First he filled top right block, point = 0;
Then he filled bottom left block, point = 0;
Then when he fill either top left and bottom right, the points obtained in both the cases will be 2 hence maximum 4 points are possible., I have written this Solution Code: t = int(input())
for _ in range(t):
a, b = list(map(int, input().split()))
count = 0
count+= 2*(a-1)*(b-1)
count+= b-1
count+= a-1
print(count), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Ram has recently purchased an empty rectangular grid of NxM dimension. He wants to fill the grid but in a special way. His way of filling is, He choose any block which is not filled yet and fills it and the points obtained in particular step is the number of filled neighbouring block. Neighbour block denotes the block which shares some common side between them. When Ram fills all the blocks then he will stop and calculate the total number of points. Ram wants to know the maximum points he can get?The first line of the input contains a single Integer T, denoting number of test cases.
The next T lines contains two space separated integer N, M i.e dimensions of grid.
<b>Constraints</b>
1 ≤ T ≤ 100
1 ≤ N, M ≤ 1000Print the maximum number of points Ram can get.Sample Input
1
2 2
Sample Output
4
Explanation
Ram can obtain total score 4 in the following way.
First he filled top right block, point = 0;
Then he filled bottom left block, point = 0;
Then when he fill either top left and bottom right, the points obtained in both the cases will be 2 hence maximum 4 points are possible., I have written this Solution Code: /**
* author: tourist1256
* created: 2022-06-24 16:26:02
**/
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace std;
using namespace __gnu_pbds;
template <class T>
using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
template <class key, class value, class cmp = std::less<key>>
using ordered_map = tree<key, value, cmp, rb_tree_tag, tree_order_statistics_node_update>;
// find_by_order(k) returns iterator to kth element starting from 0;
// order_of_key(k) returns count of elements strictly smaller than k;
#ifdef LOCAL
#define debug(...) cerr << "[" << #__VA_ARGS__ << "]:", debug_out(__VA_ARGS__)
#else
#define debug(...) 2351
#endif
#define int long long
mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
inline int64_t random_long(long long l = LLONG_MIN, long long r = LLONG_MAX) {
uniform_int_distribution<int64_t> generator(l, r);
return generator(rng);
}
int32_t main() {
ios::sync_with_stdio(false);
cin.tie(0);
int tt;
cin >> tt;
while (tt--) {
int n, m;
cin >> n >> m;
int array[n][m];
array[0][0] = 1;
int ans = 0;
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
if (i > 0 && array[i - 1][j] == 1) {
ans++;
}
if (j > 0 && array[i][j - 1] == 1) {
ans++;
}
array[i][j] = 1;
}
}
cout << ans << endl;
}
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Ram has recently purchased an empty rectangular grid of NxM dimension. He wants to fill the grid but in a special way. His way of filling is, He choose any block which is not filled yet and fills it and the points obtained in particular step is the number of filled neighbouring block. Neighbour block denotes the block which shares some common side between them. When Ram fills all the blocks then he will stop and calculate the total number of points. Ram wants to know the maximum points he can get?The first line of the input contains a single Integer T, denoting number of test cases.
The next T lines contains two space separated integer N, M i.e dimensions of grid.
<b>Constraints</b>
1 ≤ T ≤ 100
1 ≤ N, M ≤ 1000Print the maximum number of points Ram can get.Sample Input
1
2 2
Sample Output
4
Explanation
Ram can obtain total score 4 in the following way.
First he filled top right block, point = 0;
Then he filled bottom left block, point = 0;
Then when he fill either top left and bottom right, the points obtained in both the cases will be 2 hence maximum 4 points are possible., I have written this Solution Code: import java.util.*;
import java.lang.*;
import java.io.*;
class Main
{
static class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader()
{
br = new BufferedReader(
new InputStreamReader(System.in));
}
String next()
{
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
}
catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() { return Integer.parseInt(next()); }
long nextLong() { return Long.parseLong(next()); }
double nextDouble()
{
return Double.parseDouble(next());
}
String nextLine()
{
String str = "";
try {
str = br.readLine();
}
catch (IOException e) {
e.printStackTrace();
}
return str;
}
}
static boolean prime[]= new boolean[10000001];
public static void sieve()
{
for(int i=0;i<=10000000;i++)
prime[i] = true;
prime[0] = prime[1] = false;
for(int p = 2; p*p <=10000000; p++)
{
// If prime[p] is not changed, then it is a prime
if(prime[p] == true)
{
// Update all multiples of p
for(int i = p*p; i <= 10000000; i += p)
prime[i] = false;
}
}
}
static ArrayList<Integer> primeFactors(int n)
{
ArrayList<Integer> pflist=new ArrayList<>();
int c = 2;
while (n > 1) {
if (n % c == 0) {
pflist.add(c);
n /= c;
}
else
c++;
}
return pflist;
}
public static void main (String[] args) throws java.lang.Exception
{
// your code goes here
FastReader sc = new FastReader();
BufferedWriter out = new BufferedWriter(new OutputStreamWriter(System.out));
int cases = sc.nextInt();
//sieve();
while(cases-->0)
{
int n=sc.nextInt();
int m=sc.nextInt();
out.write((2*m*n-m-n)+"\n");
}
out.flush();
}
public static void reverse(int[] array)
{
int n=array.length;
for(int i=0;i<n/2;i++)
{
int temp=array[i];
array[i]=array[n-i-1];
array[n-i-1]=temp;
}
}
}
class Pair implements Comparable<Pair> {
int first,second;
public Pair(int first,int second)
{
this.first =first;
this.second=second;
}
public int compareTo(Pair b)
{
//first element in descending order
if (this.first!=b.first)
return (this.first>b.first)?-1:1;
else
return this.second<b.second?-1:1;
//second element in incresing order
}
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Straight and Simple.
Given N numbers, A[1], A[2],. , A[N], find their average.
Refer <a href="https://en.wikipedia.org/wiki/Average">here</a> for better understanding of average.The first line of the input contains a single integer N.
The second line of the input contains N singly spaced integers, A[1]...A[N].
Constraints
1 <= N <= 300000
0 <= A[i] <= 10<sup>18</sup> (for i = 1 to N)If the average is X, report <b>floor(X)</b>.Sample Input
5
1 2 3 4 6
Sample Output
3
Explanation:
(1 + 2 + 3 + 4 + 6) / 5 = 3.2. floor(3.2) = 3.
Sample Input
5
3 60 9 28 30
Sample Output
26, I have written this Solution Code: import java.util.*;
import java.lang.*;
import java.io.*;
import java.math.BigInteger;
class Main{
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
long x;
BigInteger sum = new BigInteger("0");
for(int i=0;i<n;i++){
x=sc.nextLong();
sum= sum.add(BigInteger.valueOf(x));
}
sum=sum.divide(BigInteger.valueOf(n));
System.out.print(sum);
}}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Straight and Simple.
Given N numbers, A[1], A[2],. , A[N], find their average.
Refer <a href="https://en.wikipedia.org/wiki/Average">here</a> for better understanding of average.The first line of the input contains a single integer N.
The second line of the input contains N singly spaced integers, A[1]...A[N].
Constraints
1 <= N <= 300000
0 <= A[i] <= 10<sup>18</sup> (for i = 1 to N)If the average is X, report <b>floor(X)</b>.Sample Input
5
1 2 3 4 6
Sample Output
3
Explanation:
(1 + 2 + 3 + 4 + 6) / 5 = 3.2. floor(3.2) = 3.
Sample Input
5
3 60 9 28 30
Sample Output
26, 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, cur = 0, rem = 0;
cin >> n;
for(int i = 1; i <= n; i++){
int p; cin >> p;
cur += (p + rem)/n;
rem = (p + rem)%n;
}
cout << cur;
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Straight and Simple.
Given N numbers, A[1], A[2],. , A[N], find their average.
Refer <a href="https://en.wikipedia.org/wiki/Average">here</a> for better understanding of average.The first line of the input contains a single integer N.
The second line of the input contains N singly spaced integers, A[1]...A[N].
Constraints
1 <= N <= 300000
0 <= A[i] <= 10<sup>18</sup> (for i = 1 to N)If the average is X, report <b>floor(X)</b>.Sample Input
5
1 2 3 4 6
Sample Output
3
Explanation:
(1 + 2 + 3 + 4 + 6) / 5 = 3.2. floor(3.2) = 3.
Sample Input
5
3 60 9 28 30
Sample Output
26, I have written this Solution Code: n = int(input())
a =list
a=list(map(int,input().split()))
sum=0
for i in range (0,n):
sum=sum+a[i]
print(int(sum//n))
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Hi, it's Monica!
Monica looks at her FRIENDS circle and wonders if her circle is bigger than yours. Please let her know if her friends' circle is bigger than yours, given she has a friends' circle of size 6.The first and the only line of input contains a single integer N, the size of your friends' circle.
Constraints
1 <= N <= 10Output "Yes" if the size of Monica's friends circle has more friends than yours, else output "No".Sample Input
3
Sample Output
Yes
Sample Input
10
Sample Output
No, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) {
Cf cf = new Cf();
cf.solve();
}
static class Cf {
InputReader in = new InputReader(System.in);
OutputWriter out = new OutputWriter(System.out);
int mod = (int)1e9+7;
public void solve() {
int t = in.readInt();
if(t>=6) {
out.printLine("No");
}else {
out.printLine("Yes");
}
}
public long findPower(long x,long n) {
long ans = 1;
long nn = n;
while(nn>0) {
if(nn%2==1) {
ans = (ans*x) % mod;
nn-=1;
}else {
x = (x*x)%mod;
nn/=2;
}
}
return ans%mod;
}
public static int log2(int x) {
return (int) (Math.log(x) / Math.log(2));
}
private static class InputReader {
private InputStream stream;
private byte[] buf = new byte[1024];
private int curChar;
private int numChars;
private SpaceCharFilter filter;
public InputReader(InputStream stream) {
this.stream = stream;
}
public int read() {
if (numChars == -1) {
throw new InputMismatchException();
}
if (curChar >= numChars) {
curChar = 0;
try {
numChars = stream.read(buf);
} catch (IOException e) {
throw new InputMismatchException();
}
if (numChars <= 0) {
return -1;
}
}
return buf[curChar++];
}
public int readInt() {
int c = read();
while (isSpaceChar(c)) {
c = read();
}
int sgn = 1;
if (c == '-') {
sgn = -1;
c = read();
}
int res = 0;
do {
if (c < '0' || c > '9') {
throw new InputMismatchException();
}
res *= 10;
res += c - '0';
c = read();
} while (!isSpaceChar(c));
return res * sgn;
}
public String readString() {
int c = read();
while (isSpaceChar(c)) {
c = read();
}
StringBuilder res = new StringBuilder();
do {
res.appendCodePoint(c);
c = read();
} while (!isSpaceChar(c));
return res.toString();
}
public double readDouble() {
int c = read();
while (isSpaceChar(c)) {
c = read();
}
int sgn = 1;
if (c == '-') {
sgn = -1;
c = read();
}
double res = 0;
while (!isSpaceChar(c) && c != '.') {
if (c == 'e' || c == 'E') {
return res * Math.pow(10, readInt());
}
if (c < '0' || c > '9') {
throw new InputMismatchException();
}
res *= 10;
res += c - '0';
c = read();
}
if (c == '.') {
c = read();
double m = 1;
while (!isSpaceChar(c)) {
if (c == 'e' || c == 'E') {
return res * Math.pow(10, readInt());
}
if (c < '0' || c > '9') {
throw new InputMismatchException();
}
m /= 10;
res += (c - '0') * m;
c = read();
}
}
return res * sgn;
}
public long readLong() {
int c = read();
while (isSpaceChar(c)) {
c = read();
}
int sgn = 1;
if (c == '-') {
sgn = -1;
c = read();
}
long res = 0;
do {
if (c < '0' || c > '9') {
throw new InputMismatchException();
}
res *= 10;
res += c - '0';
c = read();
} while (!isSpaceChar(c));
return res * sgn;
}
public boolean isSpaceChar(int c) {
if (filter != null) {
return filter.isSpaceChar(c);
}
return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
}
public String next() {
return readString();
}
public interface SpaceCharFilter {
public boolean isSpaceChar(int ch);
}
}
private static class OutputWriter {
private final PrintWriter writer;
public OutputWriter(OutputStream outputStream) {
writer = new PrintWriter(new BufferedWriter(new OutputStreamWriter(outputStream)));
}
public OutputWriter(Writer writer) {
this.writer = new PrintWriter(writer);
}
public void print(Object... objects) {
for (int i = 0; i < objects.length; i++) {
if (i != 0) {
writer.print(' ');
}
writer.print(objects[i]);
}
writer.flush();
}
public void printLine(Object... objects) {
print(objects);
writer.println();
writer.flush();
}
public void close() {
writer.close();
}
public void flush() {
writer.flush();
}
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Hi, it's Monica!
Monica looks at her FRIENDS circle and wonders if her circle is bigger than yours. Please let her know if her friends' circle is bigger than yours, given she has a friends' circle of size 6.The first and the only line of input contains a single integer N, the size of your friends' circle.
Constraints
1 <= N <= 10Output "Yes" if the size of Monica's friends circle has more friends than yours, else output "No".Sample Input
3
Sample Output
Yes
Sample Input
10
Sample Output
No, I have written this Solution Code: #pragma GCC optimize ("Ofast")
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define VV vector
#define pb push_back
#define bitc __builtin_popcountll
#define m_p make_pair
#define infi 1e18+1
#define eps 0.000000000001
#define fastio ios_base::sync_with_stdio(false);cin.tie(NULL);
string char_to_str(char c){string tem(1,c);return tem;}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
template<class T>//usage rand<long long>()
T rand() {
return uniform_int_distribution<T>()(rng);
}
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
template<class T>
using oset = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
// string to integer stoi()
// string to long long stoll()
// string.substr(position,length);
// integer to string to_string();
//////////////
template<class C> void mini(C&a4, C b4){a4=min(a4,b4);}
typedef unsigned long long ull;
auto clk=clock();
#define all(x) x.begin(),x.end()
#define S second
#define F first
#define sz(x) ((long long)x.size())
#define int long long
#define f80 __float128
#define mod 1000000007ll
#define pii pair<int,int>
/////////////
signed main(){
#ifdef ANIKET_GOYAL
freopen("inputf.in","r",stdin);
freopen("outputf.in","w",stdout);
#endif
int n;
cin>>n;
if(n<6)
cout<<"Yes";
else
cout<<"No";
#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: Hi, it's Monica!
Monica looks at her FRIENDS circle and wonders if her circle is bigger than yours. Please let her know if her friends' circle is bigger than yours, given she has a friends' circle of size 6.The first and the only line of input contains a single integer N, the size of your friends' circle.
Constraints
1 <= N <= 10Output "Yes" if the size of Monica's friends circle has more friends than yours, else output "No".Sample Input
3
Sample Output
Yes
Sample Input
10
Sample Output
No, I have written this Solution Code: n=int(input())
if(n>=6):
print("No")
else:
print("Yes"), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer N, your task is to print all the even integer from 1 to 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 functions <b>For_Loop()</b> that take the integer n as a parameter.
</b>Constraints:</b>
1 <= n <= 100
<b>Note:</b>
<i>But there is a catch here, given user function has already code in it which may or may not be correct, now you need to figure out these and correct them if it is required</i>Print all the even numbers from 1 to n. (print all the numbers in the same line, space-separated)Sample Input:-
5
Sample Output:-
2 4
Sample Input:-
6
Sample Output:-
2 4 6, I have written this Solution Code: def For_Loop(n):
string = ""
for i in range(1, n+1):
if i % 2 == 0:
string += "%s " % i
return string
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
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