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For this Question: You're given a string S of lowercase letters of the english alphabet. Find whether you can choose some characters of the string S in any order to create the string "red".The first and the only line of input contains the string S.
Constraints
1 <= |S| <= 100
All the characters in S are lowercase letters of the english alphabet.Output "Yes" (without quotes) if you can create the string "red", else output "No" (without quotes).Sample Input
damngrey
Sample Output
Yes
Explanation: We choose character at position 6, then position 7, then position 1.
Sample Input
newtonschool
Sample Output
No, 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();
boolean r=false,e=false,d=false;
for(int i=0;i<str.length();i++){
char ch = str.charAt(i);
if(ch=='r')r=true;
if(ch=='e')e=true;
if(ch=='d')d=true;
}
String ans = (r && e && d)?"Yes":"No";
System.out.print(ans);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You're given a string S of lowercase letters of the english alphabet. Find whether you can choose some characters of the string S in any order to create the string "red".The first and the only line of input contains the string S.
Constraints
1 <= |S| <= 100
All the characters in S are lowercase letters of the english alphabet.Output "Yes" (without quotes) if you can create the string "red", else output "No" (without quotes).Sample Input
damngrey
Sample Output
Yes
Explanation: We choose character at position 6, then position 7, then position 1.
Sample Input
newtonschool
Sample Output
No, I have written this Solution Code: S = input()
if ("r" in S and "e" in S and "d" in S ) :
print("Yes")
else:
print("No"), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You're given a string S of lowercase letters of the english alphabet. Find whether you can choose some characters of the string S in any order to create the string "red".The first and the only line of input contains the string S.
Constraints
1 <= |S| <= 100
All the characters in S are lowercase letters of the english alphabet.Output "Yes" (without quotes) if you can create the string "red", else output "No" (without quotes).Sample Input
damngrey
Sample Output
Yes
Explanation: We choose character at position 6, then position 7, then position 1.
Sample Input
newtonschool
Sample Output
No, 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(){
string s; cin>>s;
map<char, int> m;
For(i, 0, sz(s)){
m[s[i]]++;
}
if(m['r'] && m['e'] && m['d']){
cout<<"Yes";
}
else{
cout<<"No";
}
}
signed main()
{
fast
#ifdef SWAPNIL07
freopen("input.txt","r",stdin);
freopen("output.txt","w",stdout);
#endif
int t=1;
// cin>>t;
while(t--){
solve();
cout<<"\n";
}
return 0;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a N x M integer matrix A and Q queries of form X1 Y1 X2 Y2. Print the sum of A[i][j], for X1 <= i <= X2 and Y1 <= j <= Y2.The first line contains two integer N and M, denoting the size of the matrix.
Next N line contains M integers denoting elements of the matrix.
Next line contains a single integer Q, denoting the number of queries.
Next Q lines lines four integers X1 Y1 X2 Y2, denoting the query as mentioned in problem statement
1 <= N, M <= 100
1 <= A[i][j] <= 100
1 <= Q <= 100000
1 <= X1 <= X2 <= N
1 <= Y1 <= Y2 <= MPrint Q lines containing the answer to each query.Sample Input:
2 2
1 5
2 3
3
1 1 1 1
1 1 1 2
1 1 2 2
Sample Output:
1
6
11
Explanation:
Q1: 1
Q2: 1 + 5 = 6
Q3: 1 + 5 + 2 + 3 = 11, I have written this Solution Code: import java.io.*; // for handling input/output
import java.util.*; // contains Collections framework
// don't change the name of this class
// you can add inner classes if needed
class Main {
public
static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
int m = sc.nextInt();
int[][] a = new int[n + 1][m + 1];
for (int i = 0; i <= n; i++) {
for (int j = 0; j <= m; j++) {
a[i][j] = 0;
}
}
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= m; j++) {
a[i][j] = sc.nextInt();
a[i][j] += a[i - 1][j] + a[i][j - 1] - a[i - 1][j - 1];
}
}
int q = sc.nextInt();
while (q-- > 0) {
int x1 = sc.nextInt();
int y1 = sc.nextInt();
int x2 = sc.nextInt();
int y2 = sc.nextInt();
int sum = 0;
sum = a[x2][y2] - a[x1 - 1][y2] - a[x2][y1 - 1] + a[x1 - 1][y1 - 1];
System.out.println(sum);
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a N x M integer matrix A and Q queries of form X1 Y1 X2 Y2. Print the sum of A[i][j], for X1 <= i <= X2 and Y1 <= j <= Y2.The first line contains two integer N and M, denoting the size of the matrix.
Next N line contains M integers denoting elements of the matrix.
Next line contains a single integer Q, denoting the number of queries.
Next Q lines lines four integers X1 Y1 X2 Y2, denoting the query as mentioned in problem statement
1 <= N, M <= 100
1 <= A[i][j] <= 100
1 <= Q <= 100000
1 <= X1 <= X2 <= N
1 <= Y1 <= Y2 <= MPrint Q lines containing the answer to each query.Sample Input:
2 2
1 5
2 3
3
1 1 1 1
1 1 1 2
1 1 2 2
Sample Output:
1
6
11
Explanation:
Q1: 1
Q2: 1 + 5 = 6
Q3: 1 + 5 + 2 + 3 = 11, I have written this Solution Code: #include<bits/stdc++.h>
#define int long long
#define ll long long
#define pb push_back
#define endl '\n'
#define pii pair<int,int>
#define vi vector<int>
#define all(a) (a).begin(),(a).end()
#define F first
#define S second
#define sz(x) (int)x.size()
#define hell 1000000007
#define rep(i,a,b) for(int i=a;i<b;i++)
#define dep(i,a,b) for(int i=a;i>=b;i--)
#define lbnd lower_bound
#define ubnd upper_bound
#define bs binary_search
#define mp make_pair
using namespace std;
const int N = 1e2 + 5;
const int ten6 = 1e6;
const int inf = 1e9 + 9;
int a[N][N];
void solve(){
int n, m; cin >> n >> m;
for(int i = 1; i <= n; i++){
for(int j = 1; j <= m; j++){
cin >> a[i][j];
a[i][j] += a[i-1][j] + a[i][j-1] - a[i-1][j-1];
}
}
int q; cin >> q;
while(q--){
int x1, x2, y1, y2;
cin >> x1 >> y1 >> x2 >> y2;
cout << a[x2][y2] - a[x1-1][y2] - a[x2][y1-1] + a[x1-1][y1-1] << endl;
}
}
void testcases(){
int tt = 1;
//cin >> tt;
while(tt--){
solve();
}
}
signed main()
{
ios_base::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
clock_t start = clock();
testcases();
cerr << (double)(clock() - start)*1000/CLOCKS_PER_SEC << " ms: ";
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number N, find the value of the below Equation for the given number.
Equation: -
N
∑ {(X - 1)<sup>3</sup> + 3(X + 1)<sup>2</sup> + 2}
X = 1<b>User task:</b>
Since this is a functional problem, you don’t have to worry about the input, you just have to complete the function <b>equationSum()</b>, where you will get N as a parameter.
<b>Constraints:</b>
1 <= N <= 100Return the sum of equation.Sample Input:-
1
Sample Output:-
14
Sample Input:-
2
Sample Output:-
44, I have written this Solution Code: static int equationSum(int n)
{
int sum=n*(n+1);
sum=sum/2;
sum=sum*sum;
sum+=9*((n*(n+1))/2);
sum+=4*(n);
return sum;
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number N, find the value of the below Equation for the given number.
Equation: -
N
∑ {(X - 1)<sup>3</sup> + 3(X + 1)<sup>2</sup> + 2}
X = 1<b>User task:</b>
Since this is a functional problem, you don’t have to worry about the input, you just have to complete the function <b>equationSum()</b>, where you will get N as a parameter.
<b>Constraints:</b>
1 <= N <= 100Return the sum of equation.Sample Input:-
1
Sample Output:-
14
Sample Input:-
2
Sample Output:-
44, I have written this Solution Code: def equationSum(N) :
re=(N*(N+1))//2
re=re*re
re=re+9*((N*(N+1))//2)
re=re+4*N
return re
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number N, find the value of the below Equation for the given number.
Equation: -
N
∑ {(X - 1)<sup>3</sup> + 3(X + 1)<sup>2</sup> + 2}
X = 1<b>User task:</b>
Since this is a functional problem, you don’t have to worry about the input, you just have to complete the function <b>equationSum()</b>, where you will get N as a parameter.
<b>Constraints:</b>
1 <= N <= 100Return the sum of equation.Sample Input:-
1
Sample Output:-
14
Sample Input:-
2
Sample Output:-
44, I have written this Solution Code: int equationSum(int n){
int sum=n*(n+1);
sum=sum/2;
sum=sum*sum;
sum+=9*((n*(n+1))/2);
sum+=4*(n);
return sum;
}
, In this Programming Language: C, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number N, find the value of the below Equation for the given number.
Equation: -
N
∑ {(X - 1)<sup>3</sup> + 3(X + 1)<sup>2</sup> + 2}
X = 1<b>User task:</b>
Since this is a functional problem, you don’t have to worry about the input, you just have to complete the function <b>equationSum()</b>, where you will get N as a parameter.
<b>Constraints:</b>
1 <= N <= 100Return the sum of equation.Sample Input:-
1
Sample Output:-
14
Sample Input:-
2
Sample Output:-
44, I have written this Solution Code: int equationSum(int n){
int sum=n*(n+1);
sum=sum/2;
sum=sum*sum;
sum+=9*((n*(n+1))/2);
sum+=4*(n);
return sum;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A[], of size N containing positive integers. You need to print the elements of an array in increasing order.The first line of the input denotes the number of test cases 'T'. The first line of the test case is the size of the array and the second line consists of array elements.
For Python Users just complete the given function.
<b>Constraints:</b>
1 ≤ T ≤ 100
1 ≤ N ≤ 1000
1 ≤ A[i] ≤ 1000For each testcase print the sorted array in a new line.Input:
2
5
4 1 3 9 7
10
10 9 8 7 6 5 4 3 2 1
Output:
1 3 4 7 9
1 2 3 4 5 6 7 8 9 10
<b>Explanation:</b>
Testcase 1: 1 3 4 7 9 are in sorted form.
Testcase 2: For the given input, 1 2 3 4 5 6 7 8 9 10 are in sorted form., I have written this Solution Code: #include "bits/stdc++.h"
#pragma GCC optimize "03"
using namespace std;
#define int long long int
#define ld long double
#define pi pair<int, int>
#define pb push_back
#define fi first
#define se second
#define IOS ios::sync_with_stdio(false); cin.tie(0); cout.tie(0)
#ifndef LOCAL
#define endl '\n'
#endif
const int N = 2e5 + 5;
const int mod = 1e9 + 7;
const int inf = 1e9 + 9;
int a[N];
signed main() {
IOS;
int t; cin >> t;
while(t--){
int n; cin >> n;
for(int i = 1; i <= n; i++)
cin >> a[i];
sort(a + 1, a + n + 1);
for(int i = 1; i <= n; i++)
cout << a[i] << " ";
cout << endl;
}
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A[], of size N containing positive integers. You need to print the elements of an array in increasing order.The first line of the input denotes the number of test cases 'T'. The first line of the test case is the size of the array and the second line consists of array elements.
For Python Users just complete the given function.
<b>Constraints:</b>
1 ≤ T ≤ 100
1 ≤ N ≤ 1000
1 ≤ A[i] ≤ 1000For each testcase print the sorted array in a new line.Input:
2
5
4 1 3 9 7
10
10 9 8 7 6 5 4 3 2 1
Output:
1 3 4 7 9
1 2 3 4 5 6 7 8 9 10
<b>Explanation:</b>
Testcase 1: 1 3 4 7 9 are in sorted form.
Testcase 2: For the given input, 1 2 3 4 5 6 7 8 9 10 are in sorted form., I have written this Solution Code: // arr is unsorted array
// n is the number of elements in the array
function bubbleSort(arr, n) {
// write code here
// do not console.log the answer
// return sorted array
return arr.sort((a, b) => a - b)
}, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A[], of size N containing positive integers. You need to print the elements of an array in increasing order.The first line of the input denotes the number of test cases 'T'. The first line of the test case is the size of the array and the second line consists of array elements.
For Python Users just complete the given function.
<b>Constraints:</b>
1 ≤ T ≤ 100
1 ≤ N ≤ 1000
1 ≤ A[i] ≤ 1000For each testcase print the sorted array in a new line.Input:
2
5
4 1 3 9 7
10
10 9 8 7 6 5 4 3 2 1
Output:
1 3 4 7 9
1 2 3 4 5 6 7 8 9 10
<b>Explanation:</b>
Testcase 1: 1 3 4 7 9 are in sorted form.
Testcase 2: For the given input, 1 2 3 4 5 6 7 8 9 10 are in sorted form., I have written this Solution Code:
import java.util.*;
import java.lang.*;
import java.io.*;
class Main
{
public static void main (String[] args) throws java.lang.Exception
{
Scanner sc = new Scanner(System.in);
int t = sc.nextInt();
while(t-->0){
int n = sc.nextInt();
int a[] = new int[n];
for(int i=0;i<n;i++){
a[i]=sc.nextInt();
}
int temp;
for(int i=1;i<n;i++){
if(a[i]<a[i-1]){
for(int j=i;j>0;j--){
if(a[j]<a[j-1]){
temp=a[j];
a[j]=a[j-1];
a[j-1]=temp;
}
else{
break;
}
}
}
}
for(int i=0;i<n;i++){
System.out.print(a[i]+" ");
}
System.out.println();
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A[], of size N containing positive integers. You need to print the elements of an array in increasing order.The first line of the input denotes the number of test cases 'T'. The first line of the test case is the size of the array and the second line consists of array elements.
For Python Users just complete the given function.
<b>Constraints:</b>
1 ≤ T ≤ 100
1 ≤ N ≤ 1000
1 ≤ A[i] ≤ 1000For each testcase print the sorted array in a new line.Input:
2
5
4 1 3 9 7
10
10 9 8 7 6 5 4 3 2 1
Output:
1 3 4 7 9
1 2 3 4 5 6 7 8 9 10
<b>Explanation:</b>
Testcase 1: 1 3 4 7 9 are in sorted form.
Testcase 2: For the given input, 1 2 3 4 5 6 7 8 9 10 are in sorted form., I have written this Solution Code: def bubbleSort(arr):
arr.sort()
return arr
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: While Solo was enjoying her daddy's party, her little sister Tono gave her the following problem to solve:
Given an array A of N integers, find the number of occurrences of the maximum integer in the array.
As Solo is small and enjoying the party, please help her solve this problem.The first line of the input contains an integer N, the size of the array A.
The next line contains N space separated integers, the elements of the array A.
Constraints
1 <= N <= 100
1 <= A[i] <= 100Output a single integer, the number of occurrences of the maximum integer in the array A.Sample Input
5
1 2 3 2 1
Sample Output
1
Explanation: The maximum integer is 3 and it occurs once in the array.
Sample Input
5
5 5 5 5 5
Sample Output
5, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws IOException {
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
int n=Integer.parseInt(br.readLine());
int[] arr=new int[n];
String[] s=br.readLine().trim().split(" ");
for(int i=0;i<arr.length;i++)
arr[i]=Integer.parseInt(s[i]);
int max=Integer.MIN_VALUE,c=0;
for(int i=0;i<arr.length;i++){
if(max==arr[i])
c++;
if(max<arr[i]){
max=arr[i];
c=1;
}
}
System.out.println(c);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: While Solo was enjoying her daddy's party, her little sister Tono gave her the following problem to solve:
Given an array A of N integers, find the number of occurrences of the maximum integer in the array.
As Solo is small and enjoying the party, please help her solve this problem.The first line of the input contains an integer N, the size of the array A.
The next line contains N space separated integers, the elements of the array A.
Constraints
1 <= N <= 100
1 <= A[i] <= 100Output a single integer, the number of occurrences of the maximum integer in the array A.Sample Input
5
1 2 3 2 1
Sample Output
1
Explanation: The maximum integer is 3 and it occurs once in the array.
Sample Input
5
5 5 5 5 5
Sample Output
5, I have written this Solution Code: n=int(input())
a=list(map(int,input().split()))
print(a.count(max(a))), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: While Solo was enjoying her daddy's party, her little sister Tono gave her the following problem to solve:
Given an array A of N integers, find the number of occurrences of the maximum integer in the array.
As Solo is small and enjoying the party, please help her solve this problem.The first line of the input contains an integer N, the size of the array A.
The next line contains N space separated integers, the elements of the array A.
Constraints
1 <= N <= 100
1 <= A[i] <= 100Output a single integer, the number of occurrences of the maximum integer in the array A.Sample Input
5
1 2 3 2 1
Sample Output
1
Explanation: The maximum integer is 3 and it occurs once in the array.
Sample Input
5
5 5 5 5 5
Sample Output
5, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define sd(x) scanf("%d", &x)
#define sz(v) (int) v.size()
#define pr(v) For(i, 0, sz(v)) {cout<<v[i]<<" ";} cout<<endl;
#define slld(x) scanf("%lld", &x)
#define all(x) x.begin(), x.end()
#define For(i, st, en) for(int i=st; i<en; i++)
#define tr(x) for(auto it=x.begin(); it!=x.end(); it++)
#define fast std::ios::sync_with_stdio(false);cin.tie(NULL);
#define pb push_back
#define ll long long
#define ld long double
#define int long long
#define double long double
#define mp make_pair
#define F first
#define S second
typedef pair<int, int> pii;
typedef vector<int> vi;
#define pi 3.141592653589793238
const int MOD = 1e9+7;
const int INF = 1LL<<60;
const int N = 2e5+5;
// it's swapnil07 ;)
#ifdef SWAPNIL07
#define trace(...) __f(#__VA_ARGS__, __VA_ARGS__)
template <typename Arg1>
void __f(const char* name, Arg1&& arg1){
cout << name << " : " << arg1 << endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args){
const char* comma = strchr(names + 1, ',');cout.write(names, comma - names) << " : " << arg1<<" | ";__f(comma+1, args...);
}
int begtime = clock();
#define end_routine() cout << "\n\nTime elapsed: " << (clock() - begtime)*1000/CLOCKS_PER_SEC << " ms\n\n";
#else
#define endl '\n'
#define trace(...)
#define end_routine()
#endif
void solve(){
int n; cin>>n;
vector<int> a(101, 0);
For(i, 0, n){
int x; cin>>x;
a[x]++;
}
for(int i=100; i>=1; i--){
if(a[i]){
cout<<a[i];
return;
}
}
}
signed main()
{
fast
#ifdef SWAPNIL07
freopen("input.txt","r",stdin);
freopen("output.txt","w",stdout);
#endif
int t=1;
// cin>>t;
while(t--){
solve();
}
return 0;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Numbers are awesome, larger numbers are more awesome!
Given an array A of size N, you need to find the maximum sum that can be obtained from the elements of the array (the selected elements need not be contiguous). You may even decide to take no element to get a sum of 0.The first line of the input contains the size of the array.
The next line contains N (white-space separated) integers denoting the elements of the array.
<b>Constraints:</b>
1 ≤ N ≤ 10<sup>4</sup>
-10<sup>7</sup> ≤ A[i] ≤10<sup>7</sup>For each test case, output one integer representing the maximum value of the sum that can be obtained using the various elements of the array.Input 1:
5
1 2 1 -1 1
output 1:
5
input 2:
5
0 0 -1 0 0
output 2:
0
<b>Explanation 1:</b>
In order to maximize the sum among [ 1 2 1 -1 1], we need to only consider [ 1 2 1 1] and neglect the [-1]., I have written this Solution Code: #include "bits/stdc++.h"
#pragma GCC optimize "03"
using namespace std;
#define int long long int
#define ld long double
#define pi pair<int, int>
#define pb push_back
#define fi first
#define se second
#define IOS ios::sync_with_stdio(false); cin.tie(0); cout.tie(0)
#ifndef LOCAL
#define endl '\n'
#endif
const int N = 2e5 + 5;
const int mod = 1e9 + 7;
const int inf = 1e9 + 9;
signed main() {
IOS;
int n; cin >> n;
int sum = 0;
for(int i = 1; i <= n; i++){
int p; cin >> p;
if(p > 0)
sum += p;
}
cout << sum;
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Numbers are awesome, larger numbers are more awesome!
Given an array A of size N, you need to find the maximum sum that can be obtained from the elements of the array (the selected elements need not be contiguous). You may even decide to take no element to get a sum of 0.The first line of the input contains the size of the array.
The next line contains N (white-space separated) integers denoting the elements of the array.
<b>Constraints:</b>
1 ≤ N ≤ 10<sup>4</sup>
-10<sup>7</sup> ≤ A[i] ≤10<sup>7</sup>For each test case, output one integer representing the maximum value of the sum that can be obtained using the various elements of the array.Input 1:
5
1 2 1 -1 1
output 1:
5
input 2:
5
0 0 -1 0 0
output 2:
0
<b>Explanation 1:</b>
In order to maximize the sum among [ 1 2 1 -1 1], we need to only consider [ 1 2 1 1] and neglect the [-1]., I have written this Solution Code: n=int(input())
li = list(map(int,input().strip().split()))
sum=0
for i in li:
if i>0:
sum+=i
print(sum), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Numbers are awesome, larger numbers are more awesome!
Given an array A of size N, you need to find the maximum sum that can be obtained from the elements of the array (the selected elements need not be contiguous). You may even decide to take no element to get a sum of 0.The first line of the input contains the size of the array.
The next line contains N (white-space separated) integers denoting the elements of the array.
<b>Constraints:</b>
1 ≤ N ≤ 10<sup>4</sup>
-10<sup>7</sup> ≤ A[i] ≤10<sup>7</sup>For each test case, output one integer representing the maximum value of the sum that can be obtained using the various elements of the array.Input 1:
5
1 2 1 -1 1
output 1:
5
input 2:
5
0 0 -1 0 0
output 2:
0
<b>Explanation 1:</b>
In order to maximize the sum among [ 1 2 1 -1 1], we need to only consider [ 1 2 1 1] and neglect the [-1]., I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws IOException {
InputStreamReader ir = new InputStreamReader(System.in);
BufferedReader br = new BufferedReader(ir);
int n = Integer.parseInt(br.readLine());
String str[] = br.readLine().split(" ");
long arr[] = new long[n];
long sum=0;
for(int i=0;i<n;i++){
arr[i] = Integer.parseInt(str[i]);
if(arr[i]>0){
sum+=arr[i];
}
}
System.out.print(sum);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A[] of size N containing non-negative integers. You have to move all 0's to the end of array while maintaining the relative order of the non-zero elements.
Note:
You must do this in-place without making a copy of the array.
Minimize the total number of operations.The first line of input line contains T, denoting the number of testcases. Each test cases contains 2 lines. First-line contains N, size of array. Second-line contains elements of array separated by space.
Constraints:
1 <= T <= 100
1 <= N <= 10^5
1 <= A[i] <= 10^5
<b>Sum of N over all testcases does not exceed 10^6</b>
For each testcase you need to print the updated array.Sample Input:
2
5
0 1 0 3 12
8
0 0 0 0 1 2 3 4
Sample Output:
1 3 12 0 0
1 2 3 4 0 0 0 0, I have written this Solution Code:
import java.util.*;
import java.lang.*;
import java.io.*;
class Main
{
public static void main (String[] args) throws java.lang.Exception
{
Scanner sc = new Scanner(System.in);
int t;
t=sc.nextInt();
while(t-->0){
int n=sc.nextInt();
int a[] = new int[n];
for(int i=0;i<n;i++){
a[i]=sc.nextInt();
}
int j=-1;
for(int i=0;i<n;i++){
if(a[i]==0 && j==-1){
j=i;
}
if(j!=-1 && a[i]!=0){
a[j]=a[i];
a[i]=0;
j++;
}
}
for(int i=0;i<n;i++){
System.out.print(a[i]+" ");
}
System.out.println();
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A[] of size N containing non-negative integers. You have to move all 0's to the end of array while maintaining the relative order of the non-zero elements.
Note:
You must do this in-place without making a copy of the array.
Minimize the total number of operations.The first line of input line contains T, denoting the number of testcases. Each test cases contains 2 lines. First-line contains N, size of array. Second-line contains elements of array separated by space.
Constraints:
1 <= T <= 100
1 <= N <= 10^5
1 <= A[i] <= 10^5
<b>Sum of N over all testcases does not exceed 10^6</b>
For each testcase you need to print the updated array.Sample Input:
2
5
0 1 0 3 12
8
0 0 0 0 1 2 3 4
Sample Output:
1 3 12 0 0
1 2 3 4 0 0 0 0, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main(){
int t;
cin>>t;
while(t--){
int n,k;
cin>>n;
int a[n];
int cnt=0;
for(int i=0;i<n;i++){
cin>>a[i];\
if(a[i]==0){cnt++;}
}
for(int i=0;i<n;i++){
if(a[i]!=0){
cout<<a[i]<<" ";
}
}
while(cnt--){
cout<<0<<" ";
}
cout<<endl;
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a tree consisting of N nodes (from 1 to N) rooted at 1. You are given a node, your task is to calculate the sum of values of all nodes present at the same level that is of the given node.First line of input contains two space separated integers denoting the number of nodes N and the given node P. Next N-1 lines contains two space separated integers U and V, denoting an edge of the tree.
Last line of input contains N space separated integers denoting value of each node.
Constraints:-
2 <= N <= 10<sup>4</sup>
1 <= P <= N
1 <= U, V <= N
1 <= value[i] <= 10<sup>5</sup>Print the sum of all the nodes present at the same level as the node P.Sample Input:-
5 4
1 2
1 3
3 4
3 5
2 4 8 16 32
Sample Output:-
48
Explanation:-
1
/ \
2 3
/ \
4 5
Nodes present at same level as node 5 are:- 4 and 5, I have written this Solution Code: import java.io.*;
import java.util.*;
class Edge{
int s;
int d;
Edge(int s,int d){
this.s = s;
this.d = d;
}
}
class Tree{
LinkedList<Integer> node = new LinkedList<>();
}
class Main {
public static void print(Tree tree[]){
int i,n;
n = tree.length;
for(i=1;i<n;i++){
LinkedList mlist = tree[i].node;
System.out.print(i + " : ");
for(Object val : mlist)
System.out.print(val + " ");
System.out.println();
}
}
public static int calSum(Tree tree[],int data[],int p){
int i;
int n=data.length;
if(n == 0)
return 0;
if(n == 1)
return data[1];
int level[] = new int[n];
int visit[] = new int[n];
Queue<Integer> queue = new LinkedList<>();
queue.add(1);
visit[1] = 1;
while(queue.size() > 0){
int val = queue.peek();
queue.remove();
int size = tree[val].node.size();
for(i=0;i<size;i++){
int d = tree[val].node.get(i);
if(visit[d] == 0){
level[d] = level[val] + 1;
queue.add(d);
visit[d] = 1;
}
}
}
int sum = 0;
for(i=1;i<n;i++){
if(level[i] == level[p])
sum+=data[i];
}
return sum;
}
public static void main (String[] args) {
int i,n,p,s,d;
Scanner sc = new Scanner(System.in);
n = sc.nextInt();
p = sc.nextInt();
Tree tree[] = new Tree[n+1];
Edge edge[] = new Edge[n];
int data[] = new int[n+1];
for(i=1;i<=n;i++)
tree[i] = new Tree();
for(i=1;i<n;i++){
s = sc.nextInt();
d = sc.nextInt();
edge[i] = new Edge(s,d);
tree[s].node.add(d);
tree[d].node.add(s);
}
for(i=1;i<=n;i++)
data[i] = sc.nextInt();
int ans = calSum(tree,data,p);
System.out.println(ans);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a tree consisting of N nodes (from 1 to N) rooted at 1. You are given a node, your task is to calculate the sum of values of all nodes present at the same level that is of the given node.First line of input contains two space separated integers denoting the number of nodes N and the given node P. Next N-1 lines contains two space separated integers U and V, denoting an edge of the tree.
Last line of input contains N space separated integers denoting value of each node.
Constraints:-
2 <= N <= 10<sup>4</sup>
1 <= P <= N
1 <= U, V <= N
1 <= value[i] <= 10<sup>5</sup>Print the sum of all the nodes present at the same level as the node P.Sample Input:-
5 4
1 2
1 3
3 4
3 5
2 4 8 16 32
Sample Output:-
48
Explanation:-
1
/ \
2 3
/ \
4 5
Nodes present at same level as node 5 are:- 4 and 5, I have written this Solution Code: from collections import deque
N,node = map(int,input().strip().split())
dic = {}
for i in range(N-1):
n1,n2 = map(int,input().strip().split())
if dic.get(n1):
dic[n1].append(n2)
else:
dic[n1] = [n2]
if dic.get(n2):
dic[n2].append(n1)
else:
dic[n2] = [n1]
arr = [int(i) for i in input().strip().split()]
visited = set()
queue = deque()
queue.append(1)
def find_sum(node,queue):
if node==1:
return arr[0]
flag = False
while(queue):
length = len(queue)
while(length):
nodes = queue.popleft()
visited.add(nodes)
if dic.get(nodes):
for i in dic[nodes]:
if i not in visited:
queue.append(i)
if i==node:
flag = True
length -= 1
#print(queue)
if flag:
sumi = 0
for i in queue:
sumi += arr[i-1]
return sumi
print(find_sum(node,queue)), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a tree consisting of N nodes (from 1 to N) rooted at 1. You are given a node, your task is to calculate the sum of values of all nodes present at the same level that is of the given node.First line of input contains two space separated integers denoting the number of nodes N and the given node P. Next N-1 lines contains two space separated integers U and V, denoting an edge of the tree.
Last line of input contains N space separated integers denoting value of each node.
Constraints:-
2 <= N <= 10<sup>4</sup>
1 <= P <= N
1 <= U, V <= N
1 <= value[i] <= 10<sup>5</sup>Print the sum of all the nodes present at the same level as the node P.Sample Input:-
5 4
1 2
1 3
3 4
3 5
2 4 8 16 32
Sample Output:-
48
Explanation:-
1
/ \
2 3
/ \
4 5
Nodes present at same level as node 5 are:- 4 and 5, I have written this Solution Code:
#include <bits/stdc++.h>
using namespace std;
#define MEM(a, b) memset(a, (b), sizeof(a))
#define FOREACH(it, l) for (auto it = l.begin(); it != l.end(); it++)
#define IN(A, B, C) assert( B <= A && A <= C)
#define MP make_pair
#define FOR(i,a) for(int i=0;i<a;i++)
#define FOR1(i,j,a) for(int i=j;i<a;i++)
#define EB emplace_back
#define INF (int)1e9
#define EPS 1e-9
#define PI 3.1415926535897932384626433832795
#define max1 1000001
#define MOD 1000000007
#define read(type) readInt<type>()
#define out(x) cout<<x<<'\n'
#define out1(x) cout<<x<<" "
#define END cout<<'\n'
#define int long long
void fast(){
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
}
signed main()
{
int n;
cin>>n;
vector<int> v[n];
int P;
cin>>P;
int a,b;
for(int i=0;i<n-1;i++){
cin>>a>>b;
a--;b--;
v[a].EB(b);
v[b].EB(a);
}
P--;
int ans[n];
FOR(i,n){
cin>>ans[i];
}
queue<int> q,p;
q.push(0);
bool check[n];
FOR(i,n){check[i]=false;}
int sum=0;
check[0]=true;
if(P==0){
out(ans[0]);
return 0;
}
while(q.size()!=0){
bool win=false;
int s=q.size();
for(int i=0;i<s;i++){
a=q.front();
//out(a);
for(int j=0;j<v[a].size();j++){
if(check[v[a][j]]==false){
if(v[a][j]==P){win=true;}
p.push(v[a][j]);
check[v[a][j]]=true;
}
}
q.pop();
}
while(p.size()!=0){
q.push(p.front());
p.pop();
}
if(win){
a=q.size();
//out(a);
FOR(i,a){
sum+=ans[q.front()];
q.pop();
}
break;
}
}
out(sum);
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer N, find the sum of i*j for each pair (i, j) such that 1 <= i < j <= N.
Since the answer can be large, output it modulo 10<sup>9</sup>+7.The first and the only line of the input contains an integer N.
Constraints
2 <= N <= 200000Output a single integer, the answer to the above problem.Sample Input
4
Sample Output
35
Explanation:
1*2 + 1*3 + 1*4 + 2*3 + 2*4 + 3*4 = 2 + 3 + 4 + 6 + 8 + 12 = 35
Sample Input
87
Sample Output
7215142, I have written this Solution Code: M = 1000000007
n = int(input())
ans = 0;
for i in range(1,n):
result = (((n-i)*(i+1+n))/2)%M
ans = (ans + result*i)%M
print(int(ans)), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer N, find the sum of i*j for each pair (i, j) such that 1 <= i < j <= N.
Since the answer can be large, output it modulo 10<sup>9</sup>+7.The first and the only line of the input contains an integer N.
Constraints
2 <= N <= 200000Output a single integer, the answer to the above problem.Sample Input
4
Sample Output
35
Explanation:
1*2 + 1*3 + 1*4 + 2*3 + 2*4 + 3*4 = 2 + 3 + 4 + 6 + 8 + 12 = 35
Sample Input
87
Sample Output
7215142, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args)throws IOException {
BufferedReader bf=new BufferedReader(new InputStreamReader(System.in));
long n=Long.parseLong(bf.readLine());
long div=1000000007;
long totalSum=(n)*((n)+1)/2;
long sum=0;
for(long i=1;i<n;i++)
{
totalSum = totalSum - i;
sum = (sum + (totalSum*i))%div;
}
System.out.println(sum);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer N, find the sum of i*j for each pair (i, j) such that 1 <= i < j <= N.
Since the answer can be large, output it modulo 10<sup>9</sup>+7.The first and the only line of the input contains an integer N.
Constraints
2 <= N <= 200000Output a single integer, the answer to the above problem.Sample Input
4
Sample Output
35
Explanation:
1*2 + 1*3 + 1*4 + 2*3 + 2*4 + 3*4 = 2 + 3 + 4 + 6 + 8 + 12 = 35
Sample Input
87
Sample Output
7215142, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define sd(x) scanf("%d", &x)
#define sz(v) (int) v.size()
#define pr(v) For(i, 0, sz(v)) {cout<<v[i]<<" ";} cout<<endl;
#define slld(x) scanf("%lld", &x)
#define all(x) x.begin(), x.end()
#define For(i, st, en) for(int i=st; i<en; i++)
#define tr(x) for(auto it=x.begin(); it!=x.end(); it++)
#define fast std::ios::sync_with_stdio(false);cin.tie(NULL);
#define pb push_back
#define ll long long
#define ld long double
#define int long long
#define double long double
#define mp make_pair
#define F first
#define S second
typedef pair<int, int> pii;
typedef vector<int> vi;
#define pi 3.141592653589793238
const int MOD = 1e9+7;
const int INF = 1LL<<60;
const int N = 2e5+5;
// it's swapnil07 ;)
#ifdef SWAPNIL07
#define trace(...) __f(#__VA_ARGS__, __VA_ARGS__)
template <typename Arg1>
void __f(const char* name, Arg1&& arg1){
cout << name << " : " << arg1 << endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args){
const char* comma = strchr(names + 1, ',');cout.write(names, comma - names) << " : " << arg1<<" | ";__f(comma+1, args...);
}
int begtime = clock();
#define end_routine() cout << "\n\nTime elapsed: " << (clock() - begtime)*1000/CLOCKS_PER_SEC << " ms\n\n";
#else
#define endl '\n'
#define trace(...)
#define end_routine()
#endif
void solve(){
int n; cin>>n;
int s = 0;
For(i, 1, n+1){
s += i;
}
s %= MOD;
int ans = 0;
For(i, 1, n+1){
s -= i;
ans += (i*s);
ans %= MOD;
}
if(ans < 0)
ans += MOD;
cout<<ans;
}
signed main()
{
fast
#ifdef SWAPNIL07
freopen("input.txt","r",stdin);
freopen("output.txt","w",stdout);
#endif
int t=1;
// cin>>t;
while(t--){
solve();
// cout<<"\n";
}
return 0;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You'll be given a word as an input, you need to find out how many words can be made out from the letters used, word given as the input might have repeated letters, so use any appropriate method to identify those to find the actual number of combinations.
answer should be an integer, not a float.
import math needs to be called as a first step if you want to use factorial function.statistics50400statistics
now this has this dictionary {'s': 3, 't': 3, 'a': 1, 'i': 2, 'c': 1} if we look at separate letters, and each letter can be used as many times as it occurs in the original word, therefore a count for each letter needs to be maintained,
and now PnC should be put into use to derive a formula as to how many combinations are possible when the words are repeated, using that the solution could be found out., I have written this Solution Code: import math
s=input()
d = {}
for a in s:
if a in d:
d[a]= d[a]+1
else:
d[a]= 1
numer = math.factorial(len(s))
denom = 1
for key in d:
denom = denom*(math.factorial(d[key]))
sol = int(numer/denom)
print(sol), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a string S consisting of lowercase English letters. Determine whether adding some number of a's (possibly zero) at the beginning of S can make it a palindrome.The input consists of a single string S.
Constraints
1 ≤ ∣S∣ ≤ 10^6
S consists of lowercase English letters.If adding some number of a's (possibly zero) at the beginning of S can make it a palindrome, print Yes; otherwise, print No.<b>Sample Input 1</b>
kasaka
<b>Sample Output 1</b>
Yes
<b>Sample Input 2</b>
reveh
<b>Sample Output 2</b>
No, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main(void) {
int n, x, y;
string a;
cin >> a;
n = a.size();
x = 0;
for (int i = 0; i < n; i++) {
if (a[i] == 'a')x++;
else break;
}
y = 0;
for (int i = n - 1; i >= 0; i--) {
if (a[i] == 'a')y++;
else break;
}
if (x == n) {
cout << "Yes" << endl;
return 0;
}
if (x > y) {
cout << "No" << endl;
return 0;
}
for (int i = x; i < (n - y); i++) {
if (a[i] != a[x + n - y - i - 1]) {
cout << "No" << endl;
return 0;
}
}
cout << "Yes" << 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 of N integers, your task is to calculate the sum of bit difference in all pairs which can be formed.The first line of input contains a single integer N, the second line of input contains N space separated integers depicting values of the array.
Constraints:-
1 <= N <= 100000
0 <= Arr[i] <= 1000000000Print the sum of bit difference of all possible pairs.
Note:- Since the answer can be quite large print your answer modulo 10<sup>9</sup> + 7Sample Input:-
2
1 3
Sample Output:-
2
Explanation:-
(1, 1) = 0
(1, 3) = 1
(3, 1) = 1
(3, 3) = 0
Sample Input:-
2
1 2
Sample Output:-
4, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws IOException {
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
int n=Integer.parseInt(br.readLine());
String str[]=br.readLine().split(" ");
int a[]=new int[n];
for(int i=0;i<n;i++){
a[i]=Integer.parseInt(str[i]);
}
long res=0;
for (int i=0;i<32;i++){
long cnt=0;
for (int j=0;j<n;j++)
if ((a[j] & (1 << i)) == 0)
cnt++;
res=(res+(cnt*(n-cnt)*2))%1000000007;
}
System.out.println(res%1000000007);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of N integers, your task is to calculate the sum of bit difference in all pairs which can be formed.The first line of input contains a single integer N, the second line of input contains N space separated integers depicting values of the array.
Constraints:-
1 <= N <= 100000
0 <= Arr[i] <= 1000000000Print the sum of bit difference of all possible pairs.
Note:- Since the answer can be quite large print your answer modulo 10<sup>9</sup> + 7Sample Input:-
2
1 3
Sample Output:-
2
Explanation:-
(1, 1) = 0
(1, 3) = 1
(3, 1) = 1
(3, 3) = 0
Sample Input:-
2
1 2
Sample Output:-
4, I have written this Solution Code: def suBD(arr, n):
ans = 0 # Initialize result
for i in range(0, 64):
count = 0
for j in range(0, n):
if ( (arr[j] & (1 << i)) ):
count+= 1
ans += (count * (n - count)) * 2;
return (ans)%(10**9+7)
n=int(input())
arr = map(int,input().split())
arr=list(arr)
print(suBD(arr, n)), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of N integers, your task is to calculate the sum of bit difference in all pairs which can be formed.The first line of input contains a single integer N, the second line of input contains N space separated integers depicting values of the array.
Constraints:-
1 <= N <= 100000
0 <= Arr[i] <= 1000000000Print the sum of bit difference of all possible pairs.
Note:- Since the answer can be quite large print your answer modulo 10<sup>9</sup> + 7Sample Input:-
2
1 3
Sample Output:-
2
Explanation:-
(1, 1) = 0
(1, 3) = 1
(3, 1) = 1
(3, 3) = 0
Sample Input:-
2
1 2
Sample Output:-
4, I have written this Solution Code:
#include <bits/stdc++.h>
using namespace std;
#define MEM(a, b) memset(a, (b), sizeof(a))
#define FOREACH(it, l) for (auto it = l.begin(); it != l.end(); it++)
#define IN(A, B, C) assert( B <= A && A <= C)
#define MP make_pair
#define FOR(i,a) for(int i=0;i<a;i++)
#define FOR1(i,j,a) for(int i=j;i<a;i++)
#define EB emplace_back
#define INF (int)1e9
#define EPS 1e-9
#define PI 3.1415926535897932384626433832795
#define max1 101
#define MOD 1000000007
#define read(type) readInt<type>()
#define out(x) cout<<x<<'\n'
#define out1(x) cout<<x<<" "
#define END cout<<'\n'
#define int long long
#define sz(v) ((int)(v).size())
#define all(v) (v).begin(), (v).end()
void fast(){
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
}signed main(){
int N;
cin>>N;
int a[55];
int A[N];
FOR(i,N){
cin>>A[i];}
for(int i=0;i<55;i++){
a[i]=0;
}
int ans=1,p=2;
for(int i=0;i<55;i++){
for(int j=0;j<N;j++){
if(ans&A[j]){a[i]++;}
}
ans*=p;
// out(ans);
}
ans=0;
for(int i=0;i<55;i++){
ans+=(a[i]*(N-a[i])*2);
ans%=MOD;
}
out(ans);
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number N, you need to check whether the given number is <b>Palindrome</b> or not. A number is said to be Palindrome when it reads the same from backward as forward.User task:
Since this is a functional problem you don't have to worry about the input. You just have to complete the function <b>isPalindrome()</b> which contains N as a parameter.
<b>Constraints:</b>
1 <= N <= 9999You need to return "true" is the number is palindrome otherwise "false".Sample Input:
5
Sample Output:
true
Sample Input:
121
Sample Output:
true, I have written this Solution Code: static boolean isPalindrome(int N)
{
int sum = 0;
int rev = N;
while(N > 0)
{
int digit = N%10;
sum = sum*10+digit;
N = N/10;
}
if(rev == sum)
return true;
else return false;
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number N, you need to check whether the given number is <b>Palindrome</b> or not. A number is said to be Palindrome when it reads the same from backward as forward.User task:
Since this is a functional problem you don't have to worry about the input. You just have to complete the function <b>isPalindrome()</b> which contains N as a parameter.
<b>Constraints:</b>
1 <= N <= 9999You need to return "true" is the number is palindrome otherwise "false".Sample Input:
5
Sample Output:
true
Sample Input:
121
Sample Output:
true, I have written this Solution Code: def isPalindrome(N):
sum1 = 0
rev = N
while(N > 0):
digit = N%10
sum1 = sum1*10+digit
N = N//10
if(rev == sum1):
return True
return False, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Triangle Land is a place where there are infinite countries. Each country is identified by two integers (x, y) where x >= 0 and y >= 0. Due to some mysterious event (maybe some people from earth teleporting to triangle land), some of the countries are getting infected with coronavirus. When a country (x, y) is infected with coronavirus immediately countries (x, y-1) and (x+1, y-1) get the infection too (if they exist) and they further propagate the coronavirus. Given a series of countries that got the virus due to the mysterious event find the number of new countries that got infected after propagation of current event ends.First line of input contains a single integer N, number of mysterious events.
N lines follow each containing two integers x, y denoting the country in which the event occured.
Constraints:
1 <= N <= 100000
0 <= x, y <= 1000000000For each event print the number of new countries that got infected after propagation of current event ends.Sample Input
3
0 1
1 1
0 3
Sample Output
3
2
5
Explanation:
After 1st event: (0,0) (1,0) (0,1) got infected.
After 2nd event: (1,1) (2,0) got infected while (1,0) was already infected
After 3rd event: (0,3) (0,2) (1,2) (2,1) (3,0) got infected while (0,0) (1,0) (0,1) (1,1) (2,0) were already infected., I have written this Solution Code: #include<bits/stdc++.h>
using namespace std;
#define VV vector
#define pb push_back
#define bitc __builtin_popcountll
#define m_p make_pair
#define infi 1e18+1
#define eps 0.000000000001
#define fastio ios_base::sync_with_stdio(false);cin.tie(NULL);
string char_to_str(char c){string tem(1,c);return tem;}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
template<class T>//usage rand<long long>()
T rand() {
return uniform_int_distribution<T>()(rng);
}
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
typedef tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> oset;
// string to integer stoi()
// string to long long stoll()
// string.substr(position,length);
// integer to string to_string();
//////////////
auto clk=clock();
#define all(x) x.begin(),x.end()
#define S second
#define F first
#define sz(x) ((long long)x.size())
#define int long long
#define f80 __float128
#define pii pair<int,int>
/////////////
map<int, int> tr;
int getH(int x) {
if (tr.size() == 0) {
return 0;
}
map<int, int>::iterator it = tr.upper_bound(x);
if (it == tr.begin()) {
return 0;
}
--it;
return max(0ll, it->second - (x - it->first));
}
long long insert(int x, int y) {
int h = getH(x);
if (h >= y) {
return 0;
}
long long ret = y - h;
tr[x] = y;
++x, --y, h = max(h - 1, 0ll);
while (y > 0) {
map<int, int>::iterator it = tr.lower_bound(x);
if (it == tr.end() || it->first - x >= y) {
ret += (long long)y * (y + 1) / 2 - (long long)h * (h + 1) / 2;
y = 0;
} else {
int d = it->first - x;
ret += (long long)d * (y + y - d + 1) / 2;
int dh = min(d, h);
ret -= (long long)dh * (h + h - dh + 1) / 2;
if (it->second >= y - (it->first - x)) {
y = 0;
} else {
y = y - (it->first - x);
x = it->first,
h = it->second;
tr.erase(it);
}
}
}
return ret;
}
signed main() {
#ifdef ANIKET_GOYAL
freopen("inputf.in","r",stdin);
freopen("outputf.in","w",stdout);
#endif
int t;
cin>>t;
while (t--) {
int x, y;
cin>>x>>y;
++y;
cout << insert(x, y) << endl;
}
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a reference to the head of a sorted doubly-linked list and an integer k, your task is to insert the integer k in your doubly linked-list while maintaining the sort.<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>insertnew()</b> that takes head node of the linked list and the integer k as parameter.
Constraints:
1 <= T <= 100
1 <= Node.data <= 1000Return the head of the modified linked list.Sample Input
1
4
1
3
4
10
5
Sample Output
1 3 4 5 10, I have written this Solution Code: public static Node insertnew(Node head,int k) {
if(head == null){
Node temp = new Node(k);
return temp;
}
Node temp=head;
while(temp != null){
if (temp.val >= k){
Node x=new Node(k);
x.prev = temp.prev;
x.next = temp;
temp.prev = x;
if (x.prev == null){
return x;
}
else {
x.prev.next =x;
return head;
}
}
if (temp.next == null){
Node x=new Node(k);
x.prev = temp;
x.next = null;
temp.next = x;
break;
}
temp = temp.next;
}
return head;
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: There is a saying in Sara's village that "A number is called prime if the sum of all the factors of N is less than or equal to (3*N)/2 ". Given the number N, your task is to check if it is a myth or a fact for the number given.<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>mythOrFact()</b> that takes integer N as argument.
Constraints:-
2 <= N <= 1000Return 1 if it is a fact else return 0.Sample Input:-
2
Sample Output:-
1
Explanation:-
Sum = 2 + 1 = 3
3*(2) / 2 = 3
Sample Input:-
9
Sample Output:-
0, I have written this Solution Code:
int mythOrFact(int N){
int prime =1;
int cnt = N+1;
for(int i=2;i<N;i++){
if(N%i==0){cnt+=i;prime=0;}
}
int p = 3*N;
p/=2;
if((cnt<=p && prime==1) || (cnt>p && prime==0) ){return 1;}
return 0;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: There is a saying in Sara's village that "A number is called prime if the sum of all the factors of N is less than or equal to (3*N)/2 ". Given the number N, your task is to check if it is a myth or a fact for the number given.<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>mythOrFact()</b> that takes integer N as argument.
Constraints:-
2 <= N <= 1000Return 1 if it is a fact else return 0.Sample Input:-
2
Sample Output:-
1
Explanation:-
Sum = 2 + 1 = 3
3*(2) / 2 = 3
Sample Input:-
9
Sample Output:-
0, I have written this Solution Code:
int mythOrFact(int N){
int prime =1;
int cnt = N+1;
for(int i=2;i<N;i++){
if(N%i==0){cnt+=i;prime=0;}
}
int p = 3*N;
p/=2;
if((cnt<=p && prime==1) || (cnt>p && prime==0) ){return 1;}
return 0;
}
, In this Programming Language: C, Now tell me if this Code is compilable or not? | Compilable |
For this Question: There is a saying in Sara's village that "A number is called prime if the sum of all the factors of N is less than or equal to (3*N)/2 ". Given the number N, your task is to check if it is a myth or a fact for the number given.<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>mythOrFact()</b> that takes integer N as argument.
Constraints:-
2 <= N <= 1000Return 1 if it is a fact else return 0.Sample Input:-
2
Sample Output:-
1
Explanation:-
Sum = 2 + 1 = 3
3*(2) / 2 = 3
Sample Input:-
9
Sample Output:-
0, I have written this Solution Code: static int mythOrFact(int N){
int prime =1;
int cnt = N+1;
for(int i=2;i<N;i++){
if(N%i==0){cnt+=i;prime=0;}
}
int p = 3*N;
p/=2;
if((cnt<=p && prime==1) || (cnt>p && prime==0) ){return 1;}
return 0;
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: There is a saying in Sara's village that "A number is called prime if the sum of all the factors of N is less than or equal to (3*N)/2 ". Given the number N, your task is to check if it is a myth or a fact for the number given.<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>mythOrFact()</b> that takes integer N as argument.
Constraints:-
2 <= N <= 1000Return 1 if it is a fact else return 0.Sample Input:-
2
Sample Output:-
1
Explanation:-
Sum = 2 + 1 = 3
3*(2) / 2 = 3
Sample Input:-
9
Sample Output:-
0, I have written this Solution Code: def mythOrFact(N):
prime=1
cnt=N+1
for i in range (2,N):
if N%i==0:
prime=0
cnt=cnt+i
x = 3*N
x=x//2
if(cnt <= x and prime==1) or (cnt>x and prime==0):
return 1
return 0
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: In Newton City, there are N buildings in a line. A building is considered good if the height of the building is greater than or equal to all the buildings in its right side.
Given the height of N buildings in the line, your task is to print the good buildings in the same order as they are given.
Note:- The rightmost building will always be good as their is no building in its right sideThe first line of input contains a single integer N. The next line of input contains N space- separated integers depicting the values of the array.
Constraints:-
1 <= N <= 100000
1 <= Arr[i] <= 100000Print all the good buildings as they appear in the array.Sample Input:-
6
3 7 4 1 2 1
Sample Output:-
7 4 2 1
Explanation:-
There is no greater element than 7 4 2 1 on their right side.
Sample Input:-
5
1 2 3 4 5
Sample Output:-
5, 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));
br.readLine();
String[] line = br.readLine().split(" ");
Stack<Integer> st = new Stack<>();
int max = Integer.MIN_VALUE;
for (int i=line.length-1;i>0;--i){
int num = Integer.parseInt(line[i]);
if(num >= max){
st.push(num);
max = num;
}
}
while (!st.empty()){
System.out.print(st.pop()+" ");
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: In Newton City, there are N buildings in a line. A building is considered good if the height of the building is greater than or equal to all the buildings in its right side.
Given the height of N buildings in the line, your task is to print the good buildings in the same order as they are given.
Note:- The rightmost building will always be good as their is no building in its right sideThe first line of input contains a single integer N. The next line of input contains N space- separated integers depicting the values of the array.
Constraints:-
1 <= N <= 100000
1 <= Arr[i] <= 100000Print all the good buildings as they appear in the array.Sample Input:-
6
3 7 4 1 2 1
Sample Output:-
7 4 2 1
Explanation:-
There is no greater element than 7 4 2 1 on their right side.
Sample Input:-
5
1 2 3 4 5
Sample Output:-
5, I have written this Solution Code: N = int(input())
data = [int(x) for x in input().strip().split(' ')]
viewable = []
for i in range(N):
good = True
for j in range(i+1,N):
if data[j] > data[i]:
good = False
break
if good:
viewable.append(data[i])
print(*viewable), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: In Newton City, there are N buildings in a line. A building is considered good if the height of the building is greater than or equal to all the buildings in its right side.
Given the height of N buildings in the line, your task is to print the good buildings in the same order as they are given.
Note:- The rightmost building will always be good as their is no building in its right sideThe first line of input contains a single integer N. The next line of input contains N space- separated integers depicting the values of the array.
Constraints:-
1 <= N <= 100000
1 <= Arr[i] <= 100000Print all the good buildings as they appear in the array.Sample Input:-
6
3 7 4 1 2 1
Sample Output:-
7 4 2 1
Explanation:-
There is no greater element than 7 4 2 1 on their right side.
Sample Input:-
5
1 2 3 4 5
Sample Output:-
5, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define int long long
signed main(){
int n;
cin>>n;
int a[n];
for(int i=0;i<n;i++){
cin>>a[i];
}
vector<int> v;
int x = -1;
for(int i=n-1;i>=0;i--){
if(a[i]>=x){
x=a[i];
v.emplace_back(a[i]);
}
}
for(int i=v.size()-1;i>=0;i--){
cout<<v[i]<<" ";
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer N, your task is to return the sum of all of its divisors.<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>SumOfDivisors()</b> that takes the integer N as parameter.
Constraints:-
1<=N<=10^9Return the sum of all of the divisors.Sample Input:-
4
Sample Output:-
7
Sample Input:-
13
Sample Output:-
14, I have written this Solution Code: public static long SumOfDivisors(long N){
long sum=0;
long c=(long)Math.sqrt(N);
for(long i=1;i<=c;i++){
if(N%i==0){
sum+=i;
if(i*i!=N){sum+=N/i;}
}
}
return sum;
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer N, your task is to return the sum of all of its divisors.<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>SumOfDivisors()</b> that takes the integer N as parameter.
Constraints:-
1<=N<=10^9Return the sum of all of the divisors.Sample Input:-
4
Sample Output:-
7
Sample Input:-
13
Sample Output:-
14, I have written this Solution Code:
long long SumOfDivisors(long long N){
long long sum=0;
long sq=sqrt(N);
for(long i=1;i<=sq;i++){
if(N%i==0){
sum+=i;
if(i*i!=N){
sum+=N/i;
}
}
}
return sum;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer N, your task is to return the sum of all of its divisors.<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>SumOfDivisors()</b> that takes the integer N as parameter.
Constraints:-
1<=N<=10^9Return the sum of all of the divisors.Sample Input:-
4
Sample Output:-
7
Sample Input:-
13
Sample Output:-
14, I have written this Solution Code:
long long SumOfDivisors(long long N){
long long sum=0;
long sq=sqrt(N);
for(long i=1;i<=sq;i++){
if(N%i==0){
sum+=i;
if(i*i!=N){
sum+=N/i;
}
}
}
return sum;
}, In this Programming Language: C, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer N, your task is to return the sum of all of its divisors.<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>SumOfDivisors()</b> that takes the integer N as parameter.
Constraints:-
1<=N<=10^9Return the sum of all of the divisors.Sample Input:-
4
Sample Output:-
7
Sample Input:-
13
Sample Output:-
14, I have written this Solution Code:
def SumOfDivisors(num) :
# Final result of summation of divisors
result = 0
# find all divisors which divides 'num'
i = 1
while i<= (math.sqrt(num)) :
# if 'i' is divisor of 'num'
if (num % i == 0) :
# if both divisors are same then
# add it only once else add both
if (i == (num / i)) :
result = result + i;
else :
result = result + (i + num/i);
i = i + 1
# Add 1 to the result as 1 is also
# a divisor
return (result); , In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: N people are standing in line numbered 1 to N from left to right. Each person wants to know the height of the person to left of him having height less than him. If there are multiple such people he wants to know the height of the person closest to him.
If there is no such person report -1.The first line of input contains N, the size of the array.
The second line of input contains N space-separated integers.
Constraints
2 ≤ N ≤ 100000
0 ≤ Arr[i] ≤ 1000000000 (Height can be zero wierd people :p )The output should contain N space separated integers, the ith integer should be the height reported to ith person (-1 if no person to the left is found whose height is less).Sample Input 1
5
1 2 3 4 5
Sample Output 1
-1 1 2 3 4
Sample Input 2
2
1 1
Sample Output 2
-1 -1, I have written this Solution Code: import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.Stack;
import java.util.StringTokenizer;
public class Main {
final static long MOD = 1000000007;
public static void main(String args[]) {
FastScanner fs = new FastScanner();
PrintWriter out = new PrintWriter(System.out);
int n = fs.nextInt();
int[] a = fs.nextIntArray(n);
Stack<Integer> stck = new Stack<>();
for (int i = 0; i < n; i++) {
while (!stck.isEmpty() && stck.peek() >= a[i]) {
stck.pop();
}
if (stck.isEmpty()) {
out.print("-1 ");
stck.push(a[i]);
} else {
out.print(stck.peek() + " ");
stck.push(a[i]);
}
}
out.flush();
out.close();
}
static class FastScanner {
BufferedReader br;
StringTokenizer st;
public FastScanner() {
try {
br = new BufferedReader(new InputStreamReader(System.in));
st = new StringTokenizer("");
} catch (Exception e) {
e.printStackTrace();
}
}
public String next() {
if (st.hasMoreTokens())
return st.nextToken();
try {
st = new StringTokenizer(br.readLine());
} catch (Exception e) {
e.printStackTrace();
}
return st.nextToken();
}
public int nextInt() {
return Integer.parseInt(next());
}
public long nextLong() {
return Long.parseLong(next());
}
public double nextDouble() {
return Double.parseDouble(next());
}
public String nextLine() {
String line = "";
try {
line = br.readLine();
} catch (Exception e) {
e.printStackTrace();
}
return line;
}
public char nextChar() {
return next().charAt(0);
}
public Integer[] nextIntegerArray(int n) {
Integer[] a = new Integer[n];
for (int i = 0; i < n; i++)
a[i] = nextInt();
return a;
}
public int[] nextIntArray(int n) {
int[] a = new int[n];
for (int i = 0; i < n; i++)
a[i] = nextInt();
return a;
}
public long[] nextLongArray(int n) {
long[] a = new long[n];
for (int i = 0; i < n; i++)
a[i] = nextLong();
return a;
}
public char[] nextCharArray() {
return nextLine().toCharArray();
}
}
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: N people are standing in line numbered 1 to N from left to right. Each person wants to know the height of the person to left of him having height less than him. If there are multiple such people he wants to know the height of the person closest to him.
If there is no such person report -1.The first line of input contains N, the size of the array.
The second line of input contains N space-separated integers.
Constraints
2 ≤ N ≤ 100000
0 ≤ Arr[i] ≤ 1000000000 (Height can be zero wierd people :p )The output should contain N space separated integers, the ith integer should be the height reported to ith person (-1 if no person to the left is found whose height is less).Sample Input 1
5
1 2 3 4 5
Sample Output 1
-1 1 2 3 4
Sample Input 2
2
1 1
Sample Output 2
-1 -1, I have written this Solution Code: import sys
n=int(input())
myList = [int(x) for x in sys.stdin.readline().rstrip().split(' ')]
outputList = []
outputList.append(-1);
for i in range(1,len(myList),1):
flag=False
for j in range(i-1,-1,-1):
if myList[j]<myList[i]:
outputList.append(myList[j])
flag=True
break
if flag==False:
outputList.append(-1)
print(*outputList), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: N people are standing in line numbered 1 to N from left to right. Each person wants to know the height of the person to left of him having height less than him. If there are multiple such people he wants to know the height of the person closest to him.
If there is no such person report -1.The first line of input contains N, the size of the array.
The second line of input contains N space-separated integers.
Constraints
2 ≤ N ≤ 100000
0 ≤ Arr[i] ≤ 1000000000 (Height can be zero wierd people :p )The output should contain N space separated integers, the ith integer should be the height reported to ith person (-1 if no person to the left is found whose height is less).Sample Input 1
5
1 2 3 4 5
Sample Output 1
-1 1 2 3 4
Sample Input 2
2
1 1
Sample Output 2
-1 -1, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main(){
int n;
cin>>n;
stack <long long > s;
long long a;
for(int i=0;i<n;i++){
cin>>a;
while(!(s.empty())){
if(s.top()<a){break;}
s.pop();
}
if(s.empty()){cout<<-1<<" ";}
else{cout<<s.top()<<" ";}
s.push(a);
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Harry is very confused after knowing all about Sirius. To lighten Harry's mood Sirius asks a fun problem to him. Sirius puts K talking skulls equally spaced on the circumference of a circle. Now he enumerates them from 1 to K in clockwise order. Now given skull X and skull Z, he asks Harry to find number of possible Y such that angle XYZ is obtuse(more than 90 degrees).
Note:- X Y and Z must be different.The first line of the input contains single integer T, denoting number of test cases.
Each test case contains three integers K, X and Z.
Constraints
1 <= T <= 10
1 <= K <= 1000000000
1 <= X, Y <= KFor each testcase print number of possible Y in a new line.Sample Input
2
6 1 3
10 1 5
Sample Output
1
3
Explanation:
For first case, 2 is the suitable skull.
For second case, 2, 3 and 4 are the suitable skulls., I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws IOException{
BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
int t = Integer.parseInt(in.readLine());
while(t-- > 0){
String str[] = in.readLine().trim().split(" ");
int k = Integer.parseInt(str[0]);
int x = Integer.parseInt(str[1]);
int z = Integer.parseInt(str[2]);
int cnt = Math.abs(x-z);
int rem = k - cnt;
if(cnt < rem){
if(cnt != 0){
System.out.println(cnt-1);
}
else{
System.out.println(0);
}
}
else if(rem < cnt){
if(rem != 0){
System.out.println(rem-1);
}
else{
System.out.println(0);
}
}
else{
System.out.println(0);
}
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Harry is very confused after knowing all about Sirius. To lighten Harry's mood Sirius asks a fun problem to him. Sirius puts K talking skulls equally spaced on the circumference of a circle. Now he enumerates them from 1 to K in clockwise order. Now given skull X and skull Z, he asks Harry to find number of possible Y such that angle XYZ is obtuse(more than 90 degrees).
Note:- X Y and Z must be different.The first line of the input contains single integer T, denoting number of test cases.
Each test case contains three integers K, X and Z.
Constraints
1 <= T <= 10
1 <= K <= 1000000000
1 <= X, Y <= KFor each testcase print number of possible Y in a new line.Sample Input
2
6 1 3
10 1 5
Sample Output
1
3
Explanation:
For first case, 2 is the suitable skull.
For second case, 2, 3 and 4 are the suitable skulls., I have written this Solution Code: k = int(input())
for i in range(k):
z = 0
a = input().split()
a = [int(i) for i in a]
rad = 360/a[0]
dia = abs(a[2]-a[1])
ang = rad*dia
if(ang/2 > 90):
z = a[0]- dia - 1
elif(ang/2 < 90):
z = dia - 1
else:
z = 0
print(z)
a.clear(), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Harry is very confused after knowing all about Sirius. To lighten Harry's mood Sirius asks a fun problem to him. Sirius puts K talking skulls equally spaced on the circumference of a circle. Now he enumerates them from 1 to K in clockwise order. Now given skull X and skull Z, he asks Harry to find number of possible Y such that angle XYZ is obtuse(more than 90 degrees).
Note:- X Y and Z must be different.The first line of the input contains single integer T, denoting number of test cases.
Each test case contains three integers K, X and Z.
Constraints
1 <= T <= 10
1 <= K <= 1000000000
1 <= X, Y <= KFor each testcase print number of possible Y in a new line.Sample Input
2
6 1 3
10 1 5
Sample Output
1
3
Explanation:
For first case, 2 is the suitable skull.
For second case, 2, 3 and 4 are the suitable skulls., I have written this Solution Code: #pragma GCC optimize ("Ofast")
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define VV vector
#define pb push_back
#define bitc __builtin_popcountll
#define m_p make_pair
#define infi 1e18+1
#define eps 0.000000000001
#define fastio ios_base::sync_with_stdio(false);cin.tie(NULL);
string char_to_str(char c){string tem(1,c);return tem;}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
template<class T>//usage rand<long long>()
T rand() {
return uniform_int_distribution<T>()(rng);
}
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
template<class T>
using oset = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
// string to integer stoi()
// string to long long stoll()
// string.substr(position,length);
// integer to string to_string();
//////////////
auto clk=clock();
#define all(x) x.begin(),x.end()
#define S second
#define F first
#define sz(x) ((long long)x.size())
#define int long long
#define f80 __float128
#define pii pair<int,int>
/////////////
signed main()
{
fastio;
#ifdef ANIKET_GOYAL
freopen("inputf.in","r",stdin);
freopen("outputf.in","w",stdout);
#endif
int t,k,r,s,a,b;
cin>>t;
while(t--)
{
cin>>k>>r>>s;
b=max(r,s);
a=min(r,s);
if(k%2==0&&(b-a)==k/2)
cout<<0;
else
if((k%2!=0&&(b-a)<=k/2)||(k%2==0&&(b-a)<k/2))
cout<<b-a-1;
else if((b-a)>k/2)
cout<<k-(b-a)-1;
cout<<endl;
}
#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: Pino is super fond of candies. Today she has A chocolate candies, B orange candies, and C unknown candies. You are also given the happiness she gets after eating a particular candy.
She wants to have exactly X chocolate candies and Y orange candies. She can transform an unknown candy into a candy of any type (chocolate or orange).
Find the maximum happiness she can attain.The first line of input contains 5 integers X, Y, A, B, and C.
The second line contains an A integer corresponding to the happiness of the various chocolate candies.
The third line contains B integers corresponding to the happiness of the various orange candies.
The fourth line contains C integers corresponding to the happiness of the various unknown candies.
<b>Constraints:-</b>
1 <= A, B, C <= 100000
1 <= X <= A
1 <= Y <= B
1 <= happiness of any candy <= 1000000000 (10^9)Output a single integer, the maximum happiness Pino can achieve by eating the candies.Sample Input 1:-
1 2 2 2 1
2 4
5 1
3
Sample Output 1:-
12
Sample Input 2:-
2 2 2 2 2
8 6
9 1
2 1
Sample Output 2:-
25
<b>Explanation:-</b>
Pino eats the 2nd chocolate candy (happiness=4), then eats the 1st orange candy (happiness=4+5=9), then transforms the first unknown candy to orange candy and eats it (happiness=9+3=12), I have written this Solution Code: X, Y, A, B, C = [int(i) for i in input().split()]
t = X + Y
a = sorted([int(i) for i in input().split()])
b = sorted([int(i) for i in input().split()])
c = [int(i) for i in input().split()]
o = []
for i in range(A-1,-1,-1):
if (X) == 0:
break
else:
X -= 1
o.append(a[i])
for i in range(B-1,-1,-1):
if (Y) == 0:
break
else:
Y -= 1
o.append(b[i])
o.extend(c)
s = 0
o.sort()
for i in range(len(o)-1,-1,-1):
if t == 0:
break
else:
t -= 1
s += o[i]
print(s), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Pino is super fond of candies. Today she has A chocolate candies, B orange candies, and C unknown candies. You are also given the happiness she gets after eating a particular candy.
She wants to have exactly X chocolate candies and Y orange candies. She can transform an unknown candy into a candy of any type (chocolate or orange).
Find the maximum happiness she can attain.The first line of input contains 5 integers X, Y, A, B, and C.
The second line contains an A integer corresponding to the happiness of the various chocolate candies.
The third line contains B integers corresponding to the happiness of the various orange candies.
The fourth line contains C integers corresponding to the happiness of the various unknown candies.
<b>Constraints:-</b>
1 <= A, B, C <= 100000
1 <= X <= A
1 <= Y <= B
1 <= happiness of any candy <= 1000000000 (10^9)Output a single integer, the maximum happiness Pino can achieve by eating the candies.Sample Input 1:-
1 2 2 2 1
2 4
5 1
3
Sample Output 1:-
12
Sample Input 2:-
2 2 2 2 2
8 6
9 1
2 1
Sample Output 2:-
25
<b>Explanation:-</b>
Pino eats the 2nd chocolate candy (happiness=4), then eats the 1st orange candy (happiness=4+5=9), then transforms the first unknown candy to orange candy and eats it (happiness=9+3=12), I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define sd(x) scanf("%d", &x)
#define sz(v) (int) v.size()
#define pr(v) For(i, 0, sz(v)) {cout<<v[i]<<" ";} cout<<endl;
#define slld(x) scanf("%lld", &x)
#define all(x) x.begin(), x.end()
#define For(i, st, en) for(ll i=st; i<en; i++)
#define tr(x) for(auto it=x.begin(); it!=x.end(); it++)
#define fast std::ios::sync_with_stdio(false);cin.tie(NULL);
#define pb push_back
#define ll long long
#define int long long
#define mp make_pair
#define F first
#define S second
typedef pair<int, int> pii;
typedef vector<int> vi;
#define MOD 1000000007
#define INF 1000000000000000007LL
const int N = 100005;
// 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
int red[N], grn[N], col[N];
signed main()
{
fast
#ifdef SWAPNIL07
freopen("input.txt","r",stdin);
freopen("output.txt","w",stdout);
#endif
int x, y, a, b, c; cin>>x>>y>>a>>b>>c;
For(i, 0, a){
cin>>red[i];
}
For(i, 0, b){
cin>>grn[i];
}
For(i, 0, c){
cin>>col[i];
}
vector<int> vect;
sort(red, red+a); sort(grn, grn+b);
reverse(red, red+a); reverse(grn, grn+b);
For(i, 0, x){
vect.pb(red[i]);
}
For(i, 0, y){
vect.pb(grn[i]);
}
For(i, 0, c){
vect.pb(col[i]);
}
sort(all(vect));
reverse(all(vect));
int ans = 0;
For(i, 0, x+y){
ans+=vect[i];
}
cout<<ans;
return 0;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Pino is super fond of candies. Today she has A chocolate candies, B orange candies, and C unknown candies. You are also given the happiness she gets after eating a particular candy.
She wants to have exactly X chocolate candies and Y orange candies. She can transform an unknown candy into a candy of any type (chocolate or orange).
Find the maximum happiness she can attain.The first line of input contains 5 integers X, Y, A, B, and C.
The second line contains an A integer corresponding to the happiness of the various chocolate candies.
The third line contains B integers corresponding to the happiness of the various orange candies.
The fourth line contains C integers corresponding to the happiness of the various unknown candies.
<b>Constraints:-</b>
1 <= A, B, C <= 100000
1 <= X <= A
1 <= Y <= B
1 <= happiness of any candy <= 1000000000 (10^9)Output a single integer, the maximum happiness Pino can achieve by eating the candies.Sample Input 1:-
1 2 2 2 1
2 4
5 1
3
Sample Output 1:-
12
Sample Input 2:-
2 2 2 2 2
8 6
9 1
2 1
Sample Output 2:-
25
<b>Explanation:-</b>
Pino eats the 2nd chocolate candy (happiness=4), then eats the 1st orange candy (happiness=4+5=9), then transforms the first unknown candy to orange candy and eats it (happiness=9+3=12), I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args)throws IOException {
BufferedReader read = new BufferedReader(new InputStreamReader(System.in));
String str[] = read.readLine().split(" ");
int X = Integer.parseInt(str[0]);
int Y = Integer.parseInt(str[1]);
int A = Integer.parseInt(str[2]);
int B = Integer.parseInt(str[3]);
int C = Integer.parseInt(str[4]);
int arr[] = new int[X+Y+C];
int arrA[] = new int[A];
int arrB[] = new int[B];
int arrC[] = new int[C];
String strA[] = read.readLine().split(" ");
for(int k = 0; k < A; k++) {
arrA[k] = Integer.parseInt(strA[k]);
}
Arrays.sort(arrA);
String strB[] = read.readLine().split(" ");
for(int p = 0; p < B; p++) {
arrB[p] = Integer.parseInt(strB[p]);
}
Arrays.sort(arrB);
String strC[] = read.readLine().split(" ");
for(int q = 0; q < C; q++) {
arrC[q] = Integer.parseInt(strC[q]);
}
Arrays.sort(arrC);
System.arraycopy(arrA, arrA.length - X, arr, 0, X);
System.arraycopy(arrB, arrB.length - Y, arr, X, Y);
System.arraycopy(arrC, 0, arr, X+Y, C);
Arrays.sort(arr);
long happiness = 0;
int lastIndex = arr.length - 1;
int candies = 0;
for(int z = lastIndex; z >= 0; z--) {
happiness += arr[z];
candies++;
if(candies == X + Y) {
break;
}
}
System.out.print(happiness);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer <b>N</b>, you need to typecast this integer to String. If the typecasting is done successfully then we will print "<b>Nice Job</b>" otherwise "<b>Wrong answer</b>".User task:
Since this is a functional problem you don't have to worry about the input. You just have to complete the function <b>checkConvertion()</b>, which contains N as a parameter.You need to return the typecasted string value. The driver code will print "<b>Nice Job</b>" otherwise "<b>Wrong answer</b>".Sample Input:
5
Sample Output:
Nice Job
Sample Input:
6
Sample Output:
Nice Job, I have written this Solution Code: def checkConevrtion(a):
return str(a)
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer <b>N</b>, you need to typecast this integer to String. If the typecasting is done successfully then we will print "<b>Nice Job</b>" otherwise "<b>Wrong answer</b>".User task:
Since this is a functional problem you don't have to worry about the input. You just have to complete the function <b>checkConvertion()</b>, which contains N as a parameter.You need to return the typecasted string value. The driver code will print "<b>Nice Job</b>" otherwise "<b>Wrong answer</b>".Sample Input:
5
Sample Output:
Nice Job
Sample Input:
6
Sample Output:
Nice Job, I have written this Solution Code: static String checkConevrtion(int a)
{
return String.valueOf(a);
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Santa has been trying to solve the following task. Would you please help him so that he can go back to distributing gifts?
Given an integer N, print "AC" if N is 1. Otherwise, print "WA".
Note: You are supposed to print "AC" and "WA" without the quotes.The input consists of a single line containing the integer N.
<b> Constraints: </b>
0 ≤ N ≤ 1000Print the required answer.Sample Input 1
1
Sample Output 1
AC
Sample Input 2
0
Sample Output 2
WA
, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws IOException{
InputStreamReader r = new InputStreamReader(System.in);
BufferedReader br=new BufferedReader(r);
String num = br.readLine();
int N = Integer.parseInt(num);
String result = "";
if(N < 1000 && N > 0 && N == 1){
result = "AC";
} else {
result = "WA";
}
System.out.println(result);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Santa has been trying to solve the following task. Would you please help him so that he can go back to distributing gifts?
Given an integer N, print "AC" if N is 1. Otherwise, print "WA".
Note: You are supposed to print "AC" and "WA" without the quotes.The input consists of a single line containing the integer N.
<b> Constraints: </b>
0 ≤ N ≤ 1000Print the required answer.Sample Input 1
1
Sample Output 1
AC
Sample Input 2
0
Sample Output 2
WA
, I have written this Solution Code: N=int(input())
if N==1:
print("AC")
else:
print("WA"), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Santa has been trying to solve the following task. Would you please help him so that he can go back to distributing gifts?
Given an integer N, print "AC" if N is 1. Otherwise, print "WA".
Note: You are supposed to print "AC" and "WA" without the quotes.The input consists of a single line containing the integer N.
<b> Constraints: </b>
0 ≤ N ≤ 1000Print the required answer.Sample Input 1
1
Sample Output 1
AC
Sample Input 2
0
Sample Output 2
WA
, I have written this Solution Code: //Author: Xzirium
//Time and Date: 03:04:29 27 December 2021
//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()
{
#ifndef ONLINE_JUDGE
if (fopen("INPUT.txt", "r"))
{
freopen ("INPUT.txt" , "r" , stdin);
//freopen ("OUTPUT.txt" , "w" , stdout);
}
#endif
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
clock_t clk;
clk = clock();
//-----------------------------------------------------------------------------------------------------------//
READV(N);
if(N==1)
{
cout<<"AC"<<endl;
}
else
{
cout<<"WA"<<endl;
}
//-----------------------------------------------------------------------------------------------------------//
clk = clock() - clk;
cerr << fixed << setprecision(6) << "Time: " << ((double)clk)/CLOCKS_PER_SEC << endl;
return 0;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Find the number of prime numbers before N (Including that number too).The first line of the input contains the number of test cases T.
Next T lines contain the value N.
<b>Constraints</b>
1 <= T <= 1e5
1 <= N <= 1e5Print the number of primes number before that number.Sample Input 1:
2
3
11
Sample Output 1:
2
5, I have written this Solution Code: import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.Scanner;
import java.util.StringTokenizer;
class Main
{
static int [] booleanArray(int num)
{
boolean [] bool = new boolean[num+1];
int [] count = new int [num+1];
bool[0] = true;
bool[1] = true;
for(int i=2; i*i<=num; i++)
{
if(bool[i]==false)
{
for(int j=i*i; j<=num; j+=i)
bool[j] = true;
}
}
int counter = 0;
for(int i=1; i<=num; i++)
{
if(bool[i]==false)
{
counter = counter+1;
count[i] = counter;
}
else
{
count[i] = counter;
}
}
return count;
}
public static void main (String[] args) throws IOException {
InputStreamReader ak = new InputStreamReader(System.in);
BufferedReader hk = new BufferedReader(ak);
int[] v = booleanArray(100000);
int t = Integer.parseInt(hk.readLine());
for (int i = 1; i <= t; i++) {
int a = Integer.parseInt(hk.readLine());
System.out.println(v[a]);
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Find the number of prime numbers before N (Including that number too).The first line of the input contains the number of test cases T.
Next T lines contain the value N.
<b>Constraints</b>
1 <= T <= 1e5
1 <= N <= 1e5Print the number of primes number before that number.Sample Input 1:
2
3
11
Sample Output 1:
2
5, I have written this Solution Code: m=100001
prime=[True for i in range(m)]
p=2
while(p*p<=m):
if prime[p]:
for i in range(p*p,m,p):
prime[i]=False
p+=1
c=[0 for i in range(m)]
c[2]=1
for i in range(3,m):
if prime[i]:
c[i]=c[i-1]+1
else:
c[i]=c[i-1]
t=int(input())
while t>0:
n=int(input())
print(c[n])
t-=1, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Find the number of prime numbers before N (Including that number too).The first line of the input contains the number of test cases T.
Next T lines contain the value N.
<b>Constraints</b>
1 <= T <= 1e5
1 <= N <= 1e5Print the number of primes number before that number.Sample Input 1:
2
3
11
Sample Output 1:
2
5, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#ifdef LOCAL
#define debug(...) cerr << "[" << #__VA_ARGS__ << "]:", debug_out(__VA_ARGS__)
#else
#define debug(...) 2351
#endif
vector<bool> sieve(int n) {
vector<bool> is_prime(n + 1, true);
is_prime[0] = is_prime[1] = false;
for (int i = 2; i * i <= n; i++) {
if (is_prime[i]) {
for (int j = i * i; j <= n; j += i)
is_prime[j] = false;
}
}
return is_prime;
}
int main() {
vector<bool> prime = sieve(1e5 + 1);
vector<int> prefix(1e5 + 1, 0);
for (int i = 1; i <= 1e5; i++) {
if (prime[i]) {
prefix[i] = prefix[i - 1] + 1;
} else {
prefix[i] = prefix[i - 1];
}
}
int tt;
cin >> tt;
while (tt--) {
int n;
cin >> n;
cout << prefix[n] << "\n";
}
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Rick wants to give Morty a chapo (a super awesome treat :P).
The only condition for Morty to get a chapo is that he should be able to reach Rick's place. Both Rick and Morty live on the number line at integer points A and B respectively. There are N types of moves M<sub>1</sub>, M<sub>2</sub>,. , M<sub>N</sub>. Morty can only take steps of size M<sub>i</sub> (1 <= i <= N) in either of the two directions on the number line (any number of times), while Rick prefers to stay at his place.
Please let Morty know if he can ever reach Rick's place and get a chapo.The first line of the input contains three integers N, A, and B denoting the total number of step sizes, the position of Rick, and the position of Morty.
The next line contains N integers M<sub>i</sub> denoting the various step sizes that Morty can take.
Constraints
1 <= N <= 200000
1 <= M<sub>i</sub> <= 10<sup>9</sup>
-10<sup>9</sup> <= A, B <= 10<sup>9</sup>Output <b>Yes</b> if Morty can reach Rick's place, else output <b>No</b>Sample Input
3 19 2
3 5 4
Sample Output
Yes
Explanation: Morty lives at position 2, while Rick lives at position 19. Morty can take the following path to reach Rick's place (there are several other ways as well):
Move M<sub>2</sub> = 5 to the right.
Move M<sub>1</sub> = 4 to the right.
Move M<sub>1</sub> = 4 to the right.
Move M<sub>1</sub> = 4 to the right.
Sample Input 2
4 10 15
10 20 30 40
Sample Output 2
No, I have written this Solution Code: import java.io.*;
import java.util.*;
import static java.lang.Math.*;
public class Main implements Runnable {
private boolean console=false;
private long MOD = 1000_000_007L;
private int MAX = 1000_001;
private void solve1(){
int n=in.ni();
long a=in.nl(),b=in.nl();
long g = 0;
for(int i=0;i<n;++i){
g = gcd(g,in.nl());
}
if( Math.abs(a-b)%g==0){
out.printLn("Yes");
}else{
out.printLn("No");
}
}
private void solve() {
int testCases = 1;
while (testCases-->0){
solve1();
}
}
private void add(TreeMap<Integer, Integer> map, int key){
map.put(key,map.getOrDefault(key,0)+1);
}
private void remove(TreeMap<Integer,Integer> map,int key){
if(!map.containsKey(key))
return;
map.put(key,map.getOrDefault(key,0)-1);
if(map.get(key)==0)
map.remove(key);
}
@Override
public void run() {
long time = System.currentTimeMillis();
try {
init();
} catch (FileNotFoundException e) {
e.printStackTrace();
}
try {
solve();
out.flush();
System.err.println(System.currentTimeMillis()-time);
System.exit(0);
}catch (Exception e){
e.printStackTrace(); System.exit(1);
}
}
private FastInput in;
private FastOutput out;
public static void main(String[] args) throws Exception {
new Main().run();
}
private void init() throws FileNotFoundException {
InputStream inputStream = System.in;
OutputStream outputStream = System.out;
try {
if (!console && System.getProperty("user.name").equals("puneetkumar")) {
outputStream = new FileOutputStream("/Users/puneetkumar/output.txt");
inputStream = new FileInputStream("/Users/puneetkumar/input.txt");
}
} catch (Exception ignored) {
}
out = new FastOutput(outputStream);
in = new FastInput(inputStream);
}
private void maualAssert(int a,int b,int c){
if(a<b || a>c) throw new RuntimeException();
}
private void maualAssert(long a,long b,long c){
if(a<b || a>c) throw new RuntimeException();
}
private void sort(int[] arr) {
List<Integer> list = new ArrayList<>();
for (int object : arr) list.add(object);
Collections.sort(list);
for (int i = 0; i < list.size(); ++i) arr[i] = list.get(i);
}
private void sort(long[] arr) {
List<Long> list = new ArrayList<>();
for (long object : arr) list.add(object);
Collections.sort(list);
for (int i = 0; i < list.size(); ++i) arr[i] = list.get(i);
}
private long ModPow(long x, long y, long MOD) {
long res = 1L;
x = x % MOD;
while (y >= 1L) {
if ((y & 1L) > 0) res = (res * x) % MOD;
x = (x * x) % MOD;
y >>= 1L;
}
return res;
}
private int gcd(int a, int b) {
if (a == 0) return b;
return gcd(b % a, a);
}
private long gcd(long a, long b) {
if (a == 0) return b;
return gcd(b % a, a);
}
private int[] arrInt(int n){
int[] arr=new int[n];for(int i=0;i<n;++i)arr[i]=in.ni();
return arr;
}
private long[] arrLong(int n){
long[] arr=new long[n];for(int i=0;i<n;++i)arr[i]=in.nl();
return arr;
}
private int arrMax(int[] arr){
int ans = arr[0];
for(int i=1;i<arr.length;++i){
ans = max(ans,arr[i]);
}
return ans;
}
private long arrMax(long[] arr){
long ans = arr[0];
for(int i=1;i<arr.length;++i){
ans = max(ans,arr[i]);
}
return ans;
}
private int arrMin(int[] arr){
int ans = arr[0];
for(int i=1;i<arr.length;++i){
ans = max(ans,arr[i]);
}
return ans;
}
private long arrMin(long[] arr){
long ans = arr[0];
for(int i=1;i<arr.length;++i){
ans = max(ans,arr[i]);
}
return ans;
}
class FastInput { InputStream obj;
public FastInput(InputStream obj) {
this.obj = obj;
}
private byte inbuffer[] = new byte[1024];
private int lenbuffer = 0, ptrbuffer = 0;
private int readByte() { if (lenbuffer == -1) throw new InputMismatchException();
if (ptrbuffer >= lenbuffer) { ptrbuffer = 0;
try { lenbuffer = obj.read(inbuffer);
} catch (IOException e) { throw new InputMismatchException(); } }
if (lenbuffer <= 0) return -1;return inbuffer[ptrbuffer++]; }
String ns() { int b = skip();StringBuilder sb = new StringBuilder();
while (!(isSpaceChar(b)))
{ sb.appendCodePoint(b);b = readByte(); }return sb.toString();}
int ni() {
int num = 0, b;boolean minus = false;
while ((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-')) ;
if (b == '-') { minus = true;b = readByte(); }
while (true) { if (b >= '0' && b <= '9') { num = num * 10 + (b - '0'); } else {
return minus ? -num : num; }b = readByte(); }}
long nl() { long num = 0;int b;boolean minus = false;
while ((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-')) ;
if (b == '-') { minus = true;b = readByte(); }
while (true) { if (b >= '0' && b <= '9') { num = num * 10L + (b - '0'); } else {
return minus ? -num : num; }b = readByte(); } }
private boolean isSpaceChar(int c) {
return (!(c >= 33 && c <= 126));
}
int skip() { int b;while ((b = readByte()) != -1 && isSpaceChar(b)) ;return b; }
float nf() {return Float.parseFloat(ns());}
double nd() {return Double.parseDouble(ns());}
char nc() {return (char) skip();}
}
class FastOutput{
private final PrintWriter writer;
public FastOutput(OutputStream outputStream) {
writer = new PrintWriter(outputStream);
}
public PrintWriter getWriter(){
return writer;
}
public void print(Object obj){
writer.print(obj);
}
public void printLn(){
writer.println();
}
public void printLn(Object obj){
writer.print(obj);
printLn();
}
public void printSp(Object obj){
writer.print(obj+" ");
}
public void printArr(int[] arr){
for(int i:arr)
printSp(i);
printLn();
}
public void printArr(long[] arr){
for(long i:arr)
printSp(i);
printLn();
}
public void flush(){
writer.flush();
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Rick wants to give Morty a chapo (a super awesome treat :P).
The only condition for Morty to get a chapo is that he should be able to reach Rick's place. Both Rick and Morty live on the number line at integer points A and B respectively. There are N types of moves M<sub>1</sub>, M<sub>2</sub>,. , M<sub>N</sub>. Morty can only take steps of size M<sub>i</sub> (1 <= i <= N) in either of the two directions on the number line (any number of times), while Rick prefers to stay at his place.
Please let Morty know if he can ever reach Rick's place and get a chapo.The first line of the input contains three integers N, A, and B denoting the total number of step sizes, the position of Rick, and the position of Morty.
The next line contains N integers M<sub>i</sub> denoting the various step sizes that Morty can take.
Constraints
1 <= N <= 200000
1 <= M<sub>i</sub> <= 10<sup>9</sup>
-10<sup>9</sup> <= A, B <= 10<sup>9</sup>Output <b>Yes</b> if Morty can reach Rick's place, else output <b>No</b>Sample Input
3 19 2
3 5 4
Sample Output
Yes
Explanation: Morty lives at position 2, while Rick lives at position 19. Morty can take the following path to reach Rick's place (there are several other ways as well):
Move M<sub>2</sub> = 5 to the right.
Move M<sub>1</sub> = 4 to the right.
Move M<sub>1</sub> = 4 to the right.
Move M<sub>1</sub> = 4 to the right.
Sample Input 2
4 10 15
10 20 30 40
Sample Output 2
No, I have written this Solution Code: def gcd(a,b):
if a == 0:
return b
return gcd(b % a, a)
def lcm(a,b):
return (a / gcd(a,b))* b
n,x,y=map(int,input().split())
a=list(map(int,input().split()))
ans=a[0]
for i in range(1,len(a)):
ans=gcd(ans,a[i])
if abs(x-y)%ans==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: Rick wants to give Morty a chapo (a super awesome treat :P).
The only condition for Morty to get a chapo is that he should be able to reach Rick's place. Both Rick and Morty live on the number line at integer points A and B respectively. There are N types of moves M<sub>1</sub>, M<sub>2</sub>,. , M<sub>N</sub>. Morty can only take steps of size M<sub>i</sub> (1 <= i <= N) in either of the two directions on the number line (any number of times), while Rick prefers to stay at his place.
Please let Morty know if he can ever reach Rick's place and get a chapo.The first line of the input contains three integers N, A, and B denoting the total number of step sizes, the position of Rick, and the position of Morty.
The next line contains N integers M<sub>i</sub> denoting the various step sizes that Morty can take.
Constraints
1 <= N <= 200000
1 <= M<sub>i</sub> <= 10<sup>9</sup>
-10<sup>9</sup> <= A, B <= 10<sup>9</sup>Output <b>Yes</b> if Morty can reach Rick's place, else output <b>No</b>Sample Input
3 19 2
3 5 4
Sample Output
Yes
Explanation: Morty lives at position 2, while Rick lives at position 19. Morty can take the following path to reach Rick's place (there are several other ways as well):
Move M<sub>2</sub> = 5 to the right.
Move M<sub>1</sub> = 4 to the right.
Move M<sub>1</sub> = 4 to the right.
Move M<sub>1</sub> = 4 to the right.
Sample Input 2
4 10 15
10 20 30 40
Sample Output 2
No, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define sd(x) scanf("%d", &x)
#define sz(v) (int) v.size()
#define pr(v) For(i, 0, sz(v)) {cout<<v[i]<<" ";} cout<<endl;
#define slld(x) scanf("%lld", &x)
#define all(x) x.begin(), x.end()
#define For(i, st, en) for(ll i=st; i<en; i++)
#define tr(x) for(auto it=x.begin(); it!=x.end(); it++)
#define fast std::ios::sync_with_stdio(false);cin.tie(NULL);
#define pb push_back
#define ll long long
#define ld long double
#define int long long
#define double long double
#define mp make_pair
#define F first
#define S second
typedef pair<int, int> pii;
typedef vector<int> vi;
#define pi 3.141592653589793238
const int MOD = 1e9+7;
const int INF = 1LL<<60;
const int N = 2e5+5;
// it's swapnil07 ;)
#ifdef SWAPNIL07
#define trace(...) __f(#__VA_ARGS__, __VA_ARGS__)
template <typename Arg1>
void __f(const char* name, Arg1&& arg1){
cout << name << " : " << arg1 << endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args){
const char* comma = strchr(names + 1, ',');cout.write(names, comma - names) << " : " << arg1<<" | ";__f(comma+1, args...);
}
int begtime = clock();
#define end_routine() cout << "\n\nTime elapsed: " << (clock() - begtime)*1000/CLOCKS_PER_SEC << " ms\n\n";
#else
#define endl '\n'
#define trace(...)
#define end_routine()
#endif
void solve(){
int n, a, b; cin>>n>>a>>b;
a = abs(a-b);
int gv = 0;
For(i, 0, n){
int m; cin>>m;
gv = __gcd(gv, m);
}
if(a%gv == 0){
cout<<"Yes";
}
else{
cout<<"No";
}
}
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: For Christmas decoration, Jerry has set up N christmas lights in a row. The brightness of the i<sup>th</sup> light from the left is B[i]. A light at position i is powerful if its brightness is greater than or equal to the sum of brightness of all the lights to its left. Now, Tom wants the number of powerful lights to be maximum so he wants to rearrange the lights so that maximum number of lights are powerful. Find the maximum number of powerful lights, and make Tom and Jerry happy this Christmas.The first line of the input contains a single integer N.
The next line contain N integers denoting array B.
Constraints
1 <= N <= 100000
1 <= B[i] <= 10<sup>15</sup>Print a single integer denoting maximum number of powerful lights after rearrangement.Sample Input 1
5
10 1 3 23 3
Sample Output 1
4
Explanation: we can rearrange the lights as 1 3 10 23 3. In this lights at index 1, 2, 3 and 4 are powerful.
Sample Input 2
5
1 1 1 1 1
Sample Output 2
2, I have written this Solution Code: import java.io.*;
import java.util.*;
public class Main {
InputStream is;
PrintWriter out;
String INPUT = "";
void solve(int TC) throws Exception {
int n = ni();
long[] a = new long[n];
for(int i=0;i<n;i++) {
a[i] = nl();
}
long sum = 0, ans = 0;
Arrays.sort(a);
for(long i: a) {
if(i >= sum) {
++ ans;
sum += i;
}
}
pn(ans);
}
boolean TestCases = false;
public static void main(String[] args) throws Exception { new Main().run(); }
void hold(boolean b)throws Exception{if(!b)throw new Exception("Hold right there, Sparky!");}
static void dbg(Object... o){System.err.println(Arrays.deepToString(o));}
void run() throws Exception {
is = INPUT.isEmpty() ? System.in : new ByteArrayInputStream(INPUT.getBytes());
out = new PrintWriter(System.out);
long s = System.currentTimeMillis();
int T = TestCases ? ni() : 1;
for(int t=1;t<=T;t++) solve(t);
out.flush();
if(!INPUT.isEmpty())tr(System.currentTimeMillis()-s+"ms");
}
void p(Object o) { out.print(o); }
void pn(Object o) { out.println(o); }
void pni(Object o) { out.println(o);out.flush(); }
int ni() {
int num = 0, b;
boolean minus = false;
while((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-'));
if(b == '-') {
minus = true;
b = readByte();
}
while(true) {
if(b >= '0' && b <= '9'){
num = num * 10 + (b - '0');
} else {
return minus ? -num : num;
}
b = readByte();
}
}
long nl() {
long num = 0;
int b;
boolean minus = false;
while((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-'));
if(b == '-') {
minus = true;
b = readByte();
}
while(true) {
if(b >= '0' && b <= '9') {
num = num * 10 + (b - '0');
} else {
return minus ? -num : num;
}
b = readByte();
}
}
double nd() { return Double.parseDouble(ns()); }
char nc() { return (char)skip(); }
int BUF_SIZE = 1024 * 8;
byte[] inbuf = new byte[BUF_SIZE];
int lenbuf = 0, ptrbuf = 0;
int readByte() {
if(lenbuf == -1)throw new InputMismatchException();
if(ptrbuf >= lenbuf){
ptrbuf = 0;
try { lenbuf = is.read(inbuf); } catch (IOException e) { throw new InputMismatchException(); }
if(lenbuf <= 0)return -1;
} return inbuf[ptrbuf++];
}
boolean isSpaceChar(int c) { return !(c >= 33 && c <= 126); }
int skip() { int b; while((b = readByte()) != -1 && isSpaceChar(b)); return b; }
String ns() {
int b = skip();
StringBuilder sb = new StringBuilder();
while(!(isSpaceChar(b))) {
sb.appendCodePoint(b); b = readByte();
} return sb.toString();
}
char[] ns(int n) {
char[] buf = new char[n];
int b = skip(), p = 0;
while(p < n && !(isSpaceChar(b))){
buf[p++] = (char)b;
b = readByte();
} return n == p ? buf : Arrays.copyOf(buf, p);
}
void tr(Object... o) { if(INPUT.length() > 0)System.out.println(Arrays.deepToString(o)); }
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: For Christmas decoration, Jerry has set up N christmas lights in a row. The brightness of the i<sup>th</sup> light from the left is B[i]. A light at position i is powerful if its brightness is greater than or equal to the sum of brightness of all the lights to its left. Now, Tom wants the number of powerful lights to be maximum so he wants to rearrange the lights so that maximum number of lights are powerful. Find the maximum number of powerful lights, and make Tom and Jerry happy this Christmas.The first line of the input contains a single integer N.
The next line contain N integers denoting array B.
Constraints
1 <= N <= 100000
1 <= B[i] <= 10<sup>15</sup>Print a single integer denoting maximum number of powerful lights after rearrangement.Sample Input 1
5
10 1 3 23 3
Sample Output 1
4
Explanation: we can rearrange the lights as 1 3 10 23 3. In this lights at index 1, 2, 3 and 4 are powerful.
Sample Input 2
5
1 1 1 1 1
Sample Output 2
2, I have written this Solution Code: def lights(x):
sum = 0
count = 0
for i in x:
if i >= sum:
count += 1
sum += i
return count
if __name__ == '__main__':
n = int(input())
x = input().split()
for i in range(len(x)):
x[i] = int(x[i])
x.sort()
print(lights(x)), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: For Christmas decoration, Jerry has set up N christmas lights in a row. The brightness of the i<sup>th</sup> light from the left is B[i]. A light at position i is powerful if its brightness is greater than or equal to the sum of brightness of all the lights to its left. Now, Tom wants the number of powerful lights to be maximum so he wants to rearrange the lights so that maximum number of lights are powerful. Find the maximum number of powerful lights, and make Tom and Jerry happy this Christmas.The first line of the input contains a single integer N.
The next line contain N integers denoting array B.
Constraints
1 <= N <= 100000
1 <= B[i] <= 10<sup>15</sup>Print a single integer denoting maximum number of powerful lights after rearrangement.Sample Input 1
5
10 1 3 23 3
Sample Output 1
4
Explanation: we can rearrange the lights as 1 3 10 23 3. In this lights at index 1, 2, 3 and 4 are powerful.
Sample Input 2
5
1 1 1 1 1
Sample Output 2
2, I have written this Solution Code: #pragma GCC optimize ("Ofast")
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define VV vector
#define pb push_back
#define bitc __builtin_popcountll
#define m_p make_pair
#define infi 1e18+1
#define eps 0.000000000001
#define fastio ios_base::sync_with_stdio(false);cin.tie(NULL);
string char_to_str(char c){string tem(1,c);return tem;}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
template<class T>//usage rand<long long>()
T rand() {
return uniform_int_distribution<T>()(rng);
}
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
template<class T>
using oset = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
// string to integer stoi()
// string to long long stoll()
// string.substr(position,length);
// integer to string to_string();
//////////////
auto clk=clock();
#define all(x) x.begin(),x.end()
#define S second
#define F first
#define sz(x) ((long long)x.size())
#define int long long
#define f80 __float128
#define pii pair<int,int>
/////////////
signed main(){
fastio;
#ifdef ANIKET_GOYAL
freopen("inputf.in","r",stdin);
freopen("outputf.in","w",stdout);
#endif
int n;
cin>>n;
int a[n];
for(int i=0;i<n;++i){
cin>>a[i];
}
sort(a,a+n);
int ans=0;
int v=0;
for(int i=0;i<n;++i){
if(a[i]>=v){
++ans;
v+=a[i];
}
}
cout<<ans;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara has developed a new algorithm to find sprime :
For an integer n , To find all the sprime between 1 and n , she will start from the end n , mark it as sprime, and then mark all its factors (excluding itself) as not sprime. Then she will find the next greatest unmarked number less than the current sprime number , mark it as sprime, and mark all its factors (excluding itself) as not sprime. She will continue this process till all the numbers between 1 and n has been marked either sprime or not sprime .
Your task is to calculate the the number of sprimes that are also prime between 1 and n.The first line contains T the number of test cases.
Each of the next T lines contain an integer n.
Constraint:-
1 <= T <= 100
2 <= n <= 10000000Output T lines, one for each test case, containing the required answer.Sample Input :
3
2
4
7
Sample Output :
1
1
2
Explanation:-
For test 3:- 7 and 5 are the required primes
, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void sieve(boolean prime[], int n) {
int i,j;
for(i = 0; i <= n; i++)
prime[i] = true;
for(i = 2; i*i <= n; i++)
if(prime[i])
for(j = i*i; j<=n; j+=i)
prime[j] = false;
}
public static void main (String[] args) throws IOException {
int num = 10000005;
boolean prime[] = new boolean[num+1];
sieve(prime, num);
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int T = Integer.parseInt(br.readLine().trim());
while(T --> 0) {
int n = Integer.parseInt(br.readLine().trim());
int count = 0;
for(int i=(n/2)+1; i<=n; i++)
if(prime[i])
count++;
System.out.println(count);
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara has developed a new algorithm to find sprime :
For an integer n , To find all the sprime between 1 and n , she will start from the end n , mark it as sprime, and then mark all its factors (excluding itself) as not sprime. Then she will find the next greatest unmarked number less than the current sprime number , mark it as sprime, and mark all its factors (excluding itself) as not sprime. She will continue this process till all the numbers between 1 and n has been marked either sprime or not sprime .
Your task is to calculate the the number of sprimes that are also prime between 1 and n.The first line contains T the number of test cases.
Each of the next T lines contain an integer n.
Constraint:-
1 <= T <= 100
2 <= n <= 10000000Output T lines, one for each test case, containing the required answer.Sample Input :
3
2
4
7
Sample Output :
1
1
2
Explanation:-
For test 3:- 7 and 5 are the required primes
, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define max1 10000001
bool a[max1];
long b[max1];
void pre(){
b[0]=0;b[1]=0;
for(int i=0;i<max1;i++){
a[i]=false;
}
long cnt=0;
for(int i=2;i<max1;i++){
if(a[i]==false){
cnt++;
for(int j=i+i;j<=max1;j=j+i){a[j]=true;}
}
b[i]=cnt;
}
}
int main(){
pre();
int t;
cin>>t;
while(t--){
long n;
cin>>n;
cout<<(b[n]-b[(n)/2])<<endl;
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: <b> The only difference between the easy and the tricky version is the constraint on the number of queries Q. Solving the tricky version automatically solves the easy version.</b>
You are given an array A of N integers. You will be given Q queries with two integers L and R. For each query, you need to find the sum of all elements of the array excluding the elements in the range [L, R] (both inclusive) modulo 10<sup>9</sup>+7.The first line of the input contains an integer N, the size of the array.
The second line of the input contains N integers, the elements of the array A.
The third line of the input contains an integer Q.
The next Q lines contain two integers L and R.
Constraints
1 <= N <= 200000
1 <= A[i] <= 1000000000
1 <= Q <= 100000
1 <= L <= R <= NOutput Q lines, each having a single integer the answer to Q-th query modulo 10<sup>9</sup>+7.Sample Input
4
5 3 3 1
2
2 3
4 4
Sample Output
6
11
Explanation: The array A = [5, 3, 3, 1].
First Query: Sum = 5 + 1 = 6.
Second Query: Sum = 5 + 3 + 3 = 11, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
static class Reader {
final private int BUFFER_SIZE = 1 << 16;
private DataInputStream din;
private byte[] buffer;
private int bufferPointer, bytesRead;
public Reader()
{
din = new DataInputStream(System.in);
buffer = new byte[BUFFER_SIZE];
bufferPointer = bytesRead = 0;
}
public Reader(String file_name) throws IOException
{
din = new DataInputStream(
new FileInputStream(file_name));
buffer = new byte[BUFFER_SIZE];
bufferPointer = bytesRead = 0;
}
public String readLine() throws IOException
{
byte[] buf = new byte[64];
int cnt = 0, c;
while ((c = read()) != -1) {
if (c == '\n') {
if (cnt != 0) {
break;
}
else {
continue;
}
}
buf[cnt++] = (byte)c;
}
return new String(buf, 0, cnt);
}
public int nextInt() throws IOException
{
int ret = 0;
byte c = read();
while (c <= ' ') {
c = read();
}
boolean neg = (c == '-');
if (neg)
c = read();
do {
ret = ret * 10 + c - '0';
} while ((c = read()) >= '0' && c <= '9');
if (neg)
return -ret;
return ret;
}
public long nextLong() throws IOException
{
long ret = 0;
byte c = read();
while (c <= ' ')
c = read();
boolean neg = (c == '-');
if (neg)
c = read();
do {
ret = ret * 10 + c - '0';
} while ((c = read()) >= '0' && c <= '9');
if (neg)
return -ret;
return ret;
}
public double nextDouble() throws IOException
{
double ret = 0, div = 1;
byte c = read();
while (c <= ' ')
c = read();
boolean neg = (c == '-');
if (neg)
c = read();
do {
ret = ret * 10 + c - '0';
} while ((c = read()) >= '0' && c <= '9');
if (c == '.') {
while ((c = read()) >= '0' && c <= '9') {
ret += (c - '0') / (div *= 10);
}
}
if (neg)
return -ret;
return ret;
}
private void fillBuffer() throws IOException
{
bytesRead = din.read(buffer, bufferPointer = 0,
BUFFER_SIZE);
if (bytesRead == -1)
buffer[0] = -1;
}
private byte read() throws IOException
{
if (bufferPointer == bytesRead)
fillBuffer();
return buffer[bufferPointer++];
}
public void close() throws IOException
{
if (din == null)
return;
din.close();
}
}
public static void main (String[] args) throws IOException {
Reader sc = new Reader();
int arrSize = sc.nextInt();
long[] arr = new long[arrSize];
for(int i = 0; i < arrSize; i++)
{
arr[i] = sc.nextLong();
}
long[] prefix = new long[arrSize];
prefix[0] = arr[0];
for(int i = 1; i < arrSize; i++)
{
long val = 1000000007;
prefix[i - 1] %= val;
long prefixSum = prefix[i - 1] + arr[i];
prefixSum %= val;
prefix[i] = prefixSum;
}
long[] suffix = new long[arrSize];
suffix[arrSize - 1] = arr[arrSize - 1];
for(int i = arrSize - 2; i >= 0; i--)
{
long val = 1000000007;
suffix[i + 1] %= val;
long suffixSum = suffix[i + 1] + arr[i];
suffixSum %= val;
suffix[i] = suffixSum;
}
int query = sc.nextInt();
for(int x = 1; x <= query; x++)
{
int l = sc.nextInt();
int r = sc.nextInt();
long val = 1000000007;
long ans = 0;
ans += (l != 1 ? prefix[l-2] : 0);
ans %= val;
ans += (r != arrSize ? suffix[r] : 0);
ans %= val;
System.out.println(ans);
}
}
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: <b> The only difference between the easy and the tricky version is the constraint on the number of queries Q. Solving the tricky version automatically solves the easy version.</b>
You are given an array A of N integers. You will be given Q queries with two integers L and R. For each query, you need to find the sum of all elements of the array excluding the elements in the range [L, R] (both inclusive) modulo 10<sup>9</sup>+7.The first line of the input contains an integer N, the size of the array.
The second line of the input contains N integers, the elements of the array A.
The third line of the input contains an integer Q.
The next Q lines contain two integers L and R.
Constraints
1 <= N <= 200000
1 <= A[i] <= 1000000000
1 <= Q <= 100000
1 <= L <= R <= NOutput Q lines, each having a single integer the answer to Q-th query modulo 10<sup>9</sup>+7.Sample Input
4
5 3 3 1
2
2 3
4 4
Sample Output
6
11
Explanation: The array A = [5, 3, 3, 1].
First Query: Sum = 5 + 1 = 6.
Second Query: Sum = 5 + 3 + 3 = 11, I have written this Solution Code: n = int(input())
arr = list(map(int ,input().rstrip().split(" ")))
temp = 0
modified = []
for i in range(n):
temp = temp + arr[i]
modified.append(temp)
q = int(input())
for i in range(q):
s = list(map(int ,input().rstrip().split(" ")))
l = s[0]
r = s[1]
sum = 0
sum = ((modified[l-1]-arr[l-1])+(modified[n-1]-modified[r-1]))%1000000007
print(sum), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: <b> The only difference between the easy and the tricky version is the constraint on the number of queries Q. Solving the tricky version automatically solves the easy version.</b>
You are given an array A of N integers. You will be given Q queries with two integers L and R. For each query, you need to find the sum of all elements of the array excluding the elements in the range [L, R] (both inclusive) modulo 10<sup>9</sup>+7.The first line of the input contains an integer N, the size of the array.
The second line of the input contains N integers, the elements of the array A.
The third line of the input contains an integer Q.
The next Q lines contain two integers L and R.
Constraints
1 <= N <= 200000
1 <= A[i] <= 1000000000
1 <= Q <= 100000
1 <= L <= R <= NOutput Q lines, each having a single integer the answer to Q-th query modulo 10<sup>9</sup>+7.Sample Input
4
5 3 3 1
2
2 3
4 4
Sample Output
6
11
Explanation: The array A = [5, 3, 3, 1].
First Query: Sum = 5 + 1 = 6.
Second Query: Sum = 5 + 3 + 3 = 11, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define sd(x) scanf("%d", &x)
#define sz(v) (int) v.size()
#define pr(v) For(i, 0, sz(v)) {cout<<v[i]<<" ";} cout<<endl;
#define slld(x) scanf("%lld", &x)
#define all(x) x.begin(), x.end()
#define For(i, st, en) for(int i=st; i<en; i++)
#define tr(x) for(auto it=x.begin(); it!=x.end(); it++)
#define fast std::ios::sync_with_stdio(false);cin.tie(NULL);
#define pb push_back
#define ll long long
#define ld long double
#define int long long
#define double long double
#define mp make_pair
#define F first
#define S second
typedef pair<int, int> pii;
typedef vector<int> vi;
#define pi 3.141592653589793238
const int MOD = 1e9+7;
const int INF = 1LL<<60;
const int N = 2e5+5;
// it's swapnil07 ;)
#ifdef SWAPNIL07
#define trace(...) __f(#__VA_ARGS__, __VA_ARGS__)
template <typename Arg1>
void __f(const char* name, Arg1&& arg1){
cout << name << " : " << arg1 << endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args){
const char* comma = strchr(names + 1, ',');cout.write(names, comma - names) << " : " << arg1<<" | ";__f(comma+1, args...);
}
int begtime = clock();
#define end_routine() cout << "\n\nTime elapsed: " << (clock() - begtime)*1000/CLOCKS_PER_SEC << " ms\n\n";
#else
#define endl '\n'
#define trace(...)
#define end_routine()
#endif
void solve(){
int n; cin>>n;
vector<int> a(n+1);
vector<int> pre(n+1, 0);
int tot = 0;
For(i, 1, n+1){
cin>>a[i];
assert(a[i]>0LL && a[i]<=1000000000LL);
tot += a[i];
pre[i]=pre[i-1]+a[i];
}
int q; cin>>q;
while(q--){
int l, r; cin>>l>>r;
int s = tot - (pre[r]-pre[l-1]);
s %= MOD;
cout<<s<<"\n";
}
}
signed main()
{
fast
#ifdef SWAPNIL07
freopen("input.txt","r",stdin);
freopen("output.txt","w",stdout);
#endif
int t=1;
// cin>>t;
while(t--){
solve();
}
return 0;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer Q representing the number of queries and an integer K. There are two types of queries:
(i) 1 x : Add the number x to the stream
(ii) 2 : Print the sum of last K numbers of the stream. If there are less than K numbers in the stream, print the sum of current stream.
Process all the queries.First line contains two integers Q and K.
Next Q lines contains the queries.
Constraints
1 <= Q <= 10^5
1 <= x <= 10^5
1 <= K <= Q
There is atleast one query of 2nd type. For each query of type 2, print the sum of last K numbers of the stream on a new line.Sample Input 1:
5 2
1 4
2
1 1
1 3
2
Output
4
4
Explanation:
Initial Stream = {}
Add 4. Stream = {4}
Sum of last two elements = 4
Add 1. Stream = {4, 1}
Add 3. Stream = {4, 1, 3}
Sum of last two elements = 4
Sample Input 2:
3 1
1 1
2
2
Output
1
1
Explanation
Initial Stream = {}
Add 1. Stream = {1}
Sum of last element = 1
Sum of last element = 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 n,k;
String line = br.readLine();
String[] strs = line.trim().split("\\s+");
n=Integer.parseInt(strs[0]);
k=Integer.parseInt(strs[1]);
Long sum=0L;
Queue<Integer>x=new LinkedList<Integer>();
while(n>0){
n--;
line = br.readLine();
strs = line.trim().split("\\s+");
int y=Integer.parseInt(strs[0]);
if(y==1){
y=Integer.parseInt(strs[1]);
x.add(y);
sum += y;
if (x.size() > k) {
sum -= x.peek();
x.remove();
}
}
else{
System.out.print(sum+"\n");
}
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer Q representing the number of queries and an integer K. There are two types of queries:
(i) 1 x : Add the number x to the stream
(ii) 2 : Print the sum of last K numbers of the stream. If there are less than K numbers in the stream, print the sum of current stream.
Process all the queries.First line contains two integers Q and K.
Next Q lines contains the queries.
Constraints
1 <= Q <= 10^5
1 <= x <= 10^5
1 <= K <= Q
There is atleast one query of 2nd type. For each query of type 2, print the sum of last K numbers of the stream on a new line.Sample Input 1:
5 2
1 4
2
1 1
1 3
2
Output
4
4
Explanation:
Initial Stream = {}
Add 4. Stream = {4}
Sum of last two elements = 4
Add 1. Stream = {4, 1}
Add 3. Stream = {4, 1, 3}
Sum of last two elements = 4
Sample Input 2:
3 1
1 1
2
2
Output
1
1
Explanation
Initial Stream = {}
Add 1. Stream = {1}
Sum of last element = 1
Sum of last element = 1, I have written this Solution Code: /**
* author: tourist1256
* created: 2022-06-14 14:26:47
**/
#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() {
int Q, K;
cin >> Q >> K;
deque<int> st;
int sum = 0;
while (Q--) {
int x;
cin >> x;
if (x == 1) {
int y;
cin >> y;
if (st.size() == K) {
sum -= st.back();
st.pop_back();
st.push_front(y);
sum += y;
} else {
st.push_front(y);
sum += y;
}
} else {
cout << sum << "\n";
}
}
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given an array and Q queries. Your task is to perform these operations:-
enqueue: this operation will add an element to your current queue.
dequeue: this operation will delete the element from the starting of the queue
displayfront: this operation will print the element presented at the frontUser task:
Since this will be a functional problem, you don't have to take input. You just have to complete the functions:
<b>enqueue()</b>:- that takes the integer to be added and the maximum size of array as parameter.
<b>dequeue()</b>:- that takes the queue as parameter.
<b>displayfront()</b> :- that takes the queue as parameter.
Constraints:
1 <= Q(Number of queries) <= 10<sup>3</sup>
<b> Custom Input:</b>
First line of input should contains two integer number of queries Q and the size of the array N. Next Q lines contains any of the given three operations:-
enqueue x
dequeue
displayfrontDuring a dequeue operation if queue is empty you need to print "Queue is empty", during enqueue operation if the maximum size of array is reached you need to print "Queue is full" and during displayfront operation you need to print the element which is at the front and if the queue is empty you need to print "Queue is empty".
Note:-Each msg or element is to be printed on a new line
Sample Input:-
8 2
displayfront
enqueue 2
displayfront
enqueue 4
displayfront
dequeue
displayfront
enqueue 5
Sample Output:-
Queue is empty
2
2
4
Queue is full
Explanation:-here size of given array is 2 so when last enqueue operation perfomed the array was already full so we display the msg "Queue is full".
Sample input:
5 5
enqueue 4
enqueue 5
displayfront
dequeue
displayfront
Sample output:-
4
5, I have written this Solution Code: public static void enqueue(int x,int k)
{
if (rear >= k) {
System.out.println("Queue is full");
}
else {
a[rear] = x;
rear++;
}
}
public static void dequeue()
{
if (rear <= front) {
System.out.println("Queue is empty");
}
else {
front++;
}
}
public static void displayfront()
{
if (rear<=front) {
System.out.println("Queue is empty");
}
else {
int x = a[front];
System.out.println(x);
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Jenny was walking along the beach on a Sunday evening. She started from a point 0 and wants to reach a point N. What is the probability that she reaches exactly on point N, if she can only take 2 steps or 3 steps? Probability for step length 2 is P and probability for step length 3 is 1 – P.The first line of the input contains n and p.
<b>Constraints</b>
1<= n <= 1e5
0 <= p <= 1Print the final probability upto 2 decimal places.Sample Input
5 0.2
Sample Output
0.32, I have written this Solution Code: /**
* Author : tourist1256
* Time : 2022-02-18 00:02:21
**/
#include <bits/stdc++.h>
using namespace std;
#ifdef LOCAL
#define debug(...) cerr << "[" << #__VA_ARGS__ << "]:", debug_out(__VA_ARGS__)
#else
#define debug(...) 2351
#endif
float find_prob(int N, float P) {
double dp[N + 1];
dp[0] = 1;
dp[1] = 0;
dp[2] = P;
dp[3] = 1 - P;
for (int i = 4; i <= N; ++i)
dp[i] = (P)*dp[i - 2] + (1 - P) * dp[i - 3];
return dp[N];
}
int main() {
int N;
float p;
cin >> N >> p;
printf("%0.2f", find_prob(N, p));
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Jenny was walking along the beach on a Sunday evening. She started from a point 0 and wants to reach a point N. What is the probability that she reaches exactly on point N, if she can only take 2 steps or 3 steps? Probability for step length 2 is P and probability for step length 3 is 1 – P.The first line of the input contains n and p.
<b>Constraints</b>
1<= n <= 1e5
0 <= p <= 1Print the final probability upto 2 decimal places.Sample Input
5 0.2
Sample Output
0.32, 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().trim().split(" ");
int n = Integer.parseInt(str[0]);
float m = Float.parseFloat(str[1]);
double[] arr = new double[n+1];
arr[0] = 1;
arr[1] = 0;
arr[2] = m;
arr[3] = 1-m;
for(int i = 4; i <= n; ++i) {
arr[i] = m*arr[i-2]+(1-m)*arr[i-3];
}
System.out.printf("%.2f",(float)arr[n]);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array, A of N integers. Find the product of maximum values for every subarray of size K. Print the answer modulo 10<sup>9</sup>+7.
A subarray is any contiguous sequence of elements in an array.The first line contains two integers N and K, denoting the size of the array and the size of the subarray respectively.
The next line contains N integers denoting the elements of the array.
<b>Constarints</b>
1 <= K <= N <= 1000000
1 <= A[i] <= 1000000Print a single integer denoting the product of maximums for every subarray of size K modulo 1000000007Sample Input 1:
6 4
1 5 2 3 6 4
Sample Output 1:
180
<b>Explanation:</b>
For subarray [1, 5, 2, 3], maximum = 5
For subarray [5, 2, 3, 6], maximum = 6
For subarray [2, 3, 6, 4], maximum = 6
Therefore, ans = 5*6*6 = 180, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args)throws IOException {
Reader sc=new Reader();
int n=sc.nextInt();
int b=sc.nextInt();
int[] a=new int[n];
for(int i=0;i<n;i++){
a[i]=sc.nextInt();
}
int j, max;
long mul=1;
Deque<Integer> dq= new LinkedList<Integer>();
for (int i = 0;i<b;i++)
{
while (!dq.isEmpty() && a[i] >=a[dq.peekLast()])
dq.removeLast();
dq.addLast(i);
}
mul=((mul%1000000007)*(a[dq.peek()]%1000000007))%1000000007;
for (int i=b; i < n;i++)
{
while ((!dq.isEmpty()) && dq.peek() <=i-b)
dq.removeFirst();
while ((!dq.isEmpty()) && a[i]>=a[dq.peekLast()])
dq.removeLast();
dq.addLast(i);
mul=((mul%1000000007)*(a[dq.peek()]%1000000007))%1000000007;
}
System.out.println(mul%1000000007);
}
static class Reader
{
final private int BUFFER_SIZE = 1 << 16;
private DataInputStream din;
private byte[] buffer;
private int bufferPointer, bytesRead;
public Reader()
{
din = new DataInputStream(System.in);
buffer = new byte[BUFFER_SIZE];
bufferPointer = bytesRead = 0;
}
public Reader(String file_name) throws IOException
{
din = new DataInputStream(new FileInputStream(file_name));
buffer = new byte[BUFFER_SIZE];
bufferPointer = bytesRead = 0;
}
public String readLine() throws IOException
{
byte[] buf = new byte[64];
int cnt = 0, c;
while ((c = read()) != -1)
{
if (c == '\n')
break;
buf[cnt++] = (byte) c;
}
return new String(buf, 0, cnt);
}
public int nextInt() throws IOException
{
int ret = 0;
byte c = read();
while (c <= ' ')
c = read();
boolean neg = (c == '-');
if (neg)
c = read();
do
{
ret = ret * 10 + c - '0';
} while ((c = read()) >= '0' && c <= '9');
if (neg)
return -ret;
return ret;
}
public long nextLong() throws IOException
{
long ret = 0;
byte c = read();
while (c <= ' ')
c = read();
boolean neg = (c == '-');
if (neg)
c = read();
do {
ret = ret * 10 + c - '0';
}
while ((c = read()) >= '0' && c <= '9');
if (neg)
return -ret;
return ret;
}
public double nextDouble() throws IOException
{
double ret = 0, div = 1;
byte c = read();
while (c <= ' ')
c = read();
boolean neg = (c == '-');
if (neg)
c = read();
do {
ret = ret * 10 + c - '0';
}
while ((c = read()) >= '0' && c <= '9');
if (c == '.')
{
while ((c = read()) >= '0' && c <= '9')
{
ret += (c - '0') / (div *= 10);
}
}
if (neg)
return -ret;
return ret;
}
private void fillBuffer() throws IOException
{
bytesRead = din.read(buffer, bufferPointer = 0, BUFFER_SIZE);
if (bytesRead == -1)
buffer[0] = -1;
}
private byte read() throws IOException
{
if (bufferPointer == bytesRead)
fillBuffer();
return buffer[bufferPointer++];
}
public void close() throws IOException
{
if (din == null)
return;
din.close();
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array, A of N integers. Find the product of maximum values for every subarray of size K. Print the answer modulo 10<sup>9</sup>+7.
A subarray is any contiguous sequence of elements in an array.The first line contains two integers N and K, denoting the size of the array and the size of the subarray respectively.
The next line contains N integers denoting the elements of the array.
<b>Constarints</b>
1 <= K <= N <= 1000000
1 <= A[i] <= 1000000Print a single integer denoting the product of maximums for every subarray of size K modulo 1000000007Sample Input 1:
6 4
1 5 2 3 6 4
Sample Output 1:
180
<b>Explanation:</b>
For subarray [1, 5, 2, 3], maximum = 5
For subarray [5, 2, 3, 6], maximum = 6
For subarray [2, 3, 6, 4], maximum = 6
Therefore, ans = 5*6*6 = 180, I have written this Solution Code: from collections import deque
deq=deque()
n,k=list(map(int,input().split()))
array=list(map(int,input().split()))
for i in range(k):
while(deq and array[deq[-1]]<=array[i]):
deq.pop()
deq.append(i)
ans=1
for j in range(k,n):
ans=ans*array[deq[0]]
ans=(ans)%1000000007
while(deq and deq[0]<=j-k):
deq.popleft()
while(deq and array[deq[-1]]<=array[j]):
deq.pop()
deq.append(j)
ans=ans*array[deq[0]]
ans=(ans)%1000000007
print(ans), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array, A of N integers. Find the product of maximum values for every subarray of size K. Print the answer modulo 10<sup>9</sup>+7.
A subarray is any contiguous sequence of elements in an array.The first line contains two integers N and K, denoting the size of the array and the size of the subarray respectively.
The next line contains N integers denoting the elements of the array.
<b>Constarints</b>
1 <= K <= N <= 1000000
1 <= A[i] <= 1000000Print a single integer denoting the product of maximums for every subarray of size K modulo 1000000007Sample Input 1:
6 4
1 5 2 3 6 4
Sample Output 1:
180
<b>Explanation:</b>
For subarray [1, 5, 2, 3], maximum = 5
For subarray [5, 2, 3, 6], maximum = 6
For subarray [2, 3, 6, 4], maximum = 6
Therefore, ans = 5*6*6 = 180, I have written this Solution Code: #include<bits/stdc++.h>
#define int long long
#define ll long long
#define pb push_back
#define endl '\n'
#define pii pair<int,int>
#define vi vector<int>
#define all(a) (a).begin(),(a).end()
#define F first
#define S second
#define sz(x) (int)x.size()
#define hell 1000000007
#define rep(i,a,b) for(int i=a;i<b;i++)
#define dep(i,a,b) for(int i=a;i>=b;i--)
#define lbnd lower_bound
#define ubnd upper_bound
#define bs binary_search
#define mp make_pair
using namespace std;
const int N = 1e6 + 5;
const int mod = 1e9 + 7;
const int inf = 1e9 + 9;
int a[N];
void solve(){
int n, k;
cin >> n >> k;
for(int i = 1; i <= n; i++)
cin >> a[i];
deque<int> q;
for(int i = 1; i <= k; i++){
while(!q.empty() && a[q.back()] <= a[i])
q.pop_back();
q.push_back(i);
}
int ans = a[q.front()];
for(int i = k+1; i <= n; i++){
if(q.front() == i-k)
q.pop_front();
while(!q.empty() && a[q.back()] <= a[i])
q.pop_back();
q.push_back(i);
ans = (ans*a[q.front()]) % mod;
}
cout << ans;
}
void testcases(){
int tt = 1;
//cin >> tt;
while(tt--){
solve();
}
}
signed main()
{
ios_base::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
clock_t start = clock();
testcases();
cerr << (double)(clock() - start)*1000/CLOCKS_PER_SEC << " ms: ";
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: We are given a string. Your task is to compress the consecutive letters of the string
For example, given string is "AAACCCBBD", thus here
A's occurrence 3 times
C's occurrence 3 times
B's occurrence 2 times
D's occurrence 1 time
So after compressing string becomes "A3C3B2D1".The first line of input contains an integer T denoting the number of test cases. Each test case will have a string provided in the new line.
Constraints:
1 <= T <= 10
1 <= sizeof(String) <= 10^6
All characters of String are upper case letters. (A-Z)
Sum of size of Strings over all testcases is <= 10^6For each testcase, in a new line, print the compressed string for each test case in a new line. Input:
2
AAACCCBBD
ABCD
Output:
A3C3B2D1
A1B1C1D1, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void compress(String str, int l){
for (int i = 0; i < l; i++) {
int count = 1;
while (i < l - 1 && str.charAt(i) == str.charAt(i + 1)) {
count++;
i++;
}
System.out.print(str.charAt(i));
System.out.print(count);
}
System.out.println();
}
public static void main (String[] args) throws IOException{
BufferedReader rd = new BufferedReader(new InputStreamReader(System.in));
int test = Integer.parseInt(rd.readLine());
while(test-->0){
String s = rd.readLine();
int len = s.length();
compress(s,len);
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: We are given a string. Your task is to compress the consecutive letters of the string
For example, given string is "AAACCCBBD", thus here
A's occurrence 3 times
C's occurrence 3 times
B's occurrence 2 times
D's occurrence 1 time
So after compressing string becomes "A3C3B2D1".The first line of input contains an integer T denoting the number of test cases. Each test case will have a string provided in the new line.
Constraints:
1 <= T <= 10
1 <= sizeof(String) <= 10^6
All characters of String are upper case letters. (A-Z)
Sum of size of Strings over all testcases is <= 10^6For each testcase, in a new line, print the compressed string for each test case in a new line. Input:
2
AAACCCBBD
ABCD
Output:
A3C3B2D1
A1B1C1D1, I have written this Solution Code:
def compress(st):
n = len(st)
i = 0
while i < n:
count = 1
while (i < n-1 and st[i] == st[i + 1]):
count += 1
i += 1
i += 1
print(st[i-1] +str(count),end="")
t=int(input())
for i in range(t):
s=input()
compress(s)
print(), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: We are given a string. Your task is to compress the consecutive letters of the string
For example, given string is "AAACCCBBD", thus here
A's occurrence 3 times
C's occurrence 3 times
B's occurrence 2 times
D's occurrence 1 time
So after compressing string becomes "A3C3B2D1".The first line of input contains an integer T denoting the number of test cases. Each test case will have a string provided in the new line.
Constraints:
1 <= T <= 10
1 <= sizeof(String) <= 10^6
All characters of String are upper case letters. (A-Z)
Sum of size of Strings over all testcases is <= 10^6For each testcase, in a new line, print the compressed string for each test case in a new line. Input:
2
AAACCCBBD
ABCD
Output:
A3C3B2D1
A1B1C1D1, 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 = 1e5 + 5;
const int mod = 1e9 + 7;
const int inf = 1e9 + 9;
void solve(){
string s; cin >> s;
int c = 1;
char p = 0;
int n = s.length();
for(int i = 1; i < n; i++){
if(s[i] != s[i-1]){
cout << s[i-1] << c;
c = 1;
}
else
c++;
}
cout << s[n-1] << c << 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 two sorted linked list of size s1 and s2(sizes may or may not be same), your task is to merge them such that resultant list is also sorted.<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>Merge()</b> that takes the head node of both the linked list as the parameter.
Use <b>insert()</b> function for inserting nodes in the linked list.
<b>Constraints:</b>
1 < = s1, s2 < = 1000
1 < = val < = 10000Return the head of the merged linked list, printing will be done by the driver codeSample Input:
5 6
1 2 3 4 5
3 4 6 8 9 10
Sample Output:
1 2 3 3 4 4 5 6 8 9 10, I have written this Solution Code: public static Node Merge (Node head1, Node head2){
Node head =null;
while(head1!=null && head2!=null){
if(head1.val<head2.val){
head=insert(head,head1.val);
head1=head1.next;
}
else{
head=insert(head,head2.val);
head2=head2.next;
}
}
while(head1!=null){
head=insert(head,head1.val);
head1=head1.next;
}
while(head2!=null){
head=insert(head,head2.val);
head2=head2.next;
}
return head;
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You will be given a number N that represents where the minute hand currently is on a clock. Your program
should return the angle that is formed by the minute hand and the 12 o'clock mark on the clock.Number between 1 to 60 inclusivePrint the angle as a numberSample input 1:-
1
Sample output 1:-
6
Sample Input 2:
30
Sample Output2:
180
, I have written this Solution Code: function simpleClockAngle(num) {
console.log(6*num);
}, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You will be given a number N that represents where the minute hand currently is on a clock. Your program
should return the angle that is formed by the minute hand and the 12 o'clock mark on the clock.Number between 1 to 60 inclusivePrint the angle as a numberSample input 1:-
1
Sample output 1:-
6
Sample Input 2:
30
Sample Output2:
180
, I have written this Solution Code: num=int(input())
print(num*6), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You will be given a number N that represents where the minute hand currently is on a clock. Your program
should return the angle that is formed by the minute hand and the 12 o'clock mark on the clock.Number between 1 to 60 inclusivePrint the angle as a numberSample input 1:-
1
Sample output 1:-
6
Sample Input 2:
30
Sample Output2:
180
, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) {
Scanner obj = new Scanner(System.in);
int N = obj.nextInt();
System.out.println(N*6);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number N and a number P, your task is to calculate N<sup>P</sup>. Since the ans can be very long calculate your ans as N<sup>P</sup>%M where M = 10<sup>9</sup>+7Input contains only two integers N and P separated by spaces.
Constraints:-
1 < = N, P <= 1000000000Print N<sup>P</sup>%M.Sample Input:-
2 5
Sample Output:-
32
Sample Input:-
2 100
Sample Output:-
976371285, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args)throws IOException{
BufferedReader rd=new BufferedReader(new InputStreamReader(System.in));
String y[]=rd.readLine().split(" ");
long n=Long.parseLong(y[0]);
long p=Long.parseLong(y[1]);
long v=1;
while(p>0){
if((p&1L)==1L)
v=(v*n)%1000000007;
p/=2;
n=(n*n)%1000000007;
}
System.out.print(v);
}
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number N and a number P, your task is to calculate N<sup>P</sup>. Since the ans can be very long calculate your ans as N<sup>P</sup>%M where M = 10<sup>9</sup>+7Input contains only two integers N and P separated by spaces.
Constraints:-
1 < = N, P <= 1000000000Print N<sup>P</sup>%M.Sample Input:-
2 5
Sample Output:-
32
Sample Input:-
2 100
Sample Output:-
976371285, I have written this Solution Code: n, p =input().split()
n, p =int(n), int(p)
def FastModularExponentiation(b, k, m):
res = 1
b = b % m
while (k > 0):
if ((k & 1) == 1):
res = (res * b) % m
k = k >> 1
b = (b * b) % m
return res
m=pow(10,9)+7
print(FastModularExponentiation(n, p, m)), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number N and a number P, your task is to calculate N<sup>P</sup>. Since the ans can be very long calculate your ans as N<sup>P</sup>%M where M = 10<sup>9</sup>+7Input contains only two integers N and P separated by spaces.
Constraints:-
1 < = N, P <= 1000000000Print N<sup>P</sup>%M.Sample Input:-
2 5
Sample Output:-
32
Sample Input:-
2 100
Sample Output:-
976371285, I have written this Solution Code:
// author-Shivam gupta
#include <bits/stdc++.h>
using namespace std;
#define MEM(a, b) memset(a, (b), sizeof(a))
#define FOREACH(it, l) for (auto it = l.begin(); it != l.end(); it++)
#define IN(A, B, C) assert( B <= A && A <= C)
#define MP make_pair
#define FOR(i,a) for(int i=0;i<a;i++)
#define FOR1(i,j,a) for(int i=j;i<a;i++)
#define EB emplace_back
#define INF (int)1e9
#define EPS 1e-9
#define PI 3.1415926535897932384626433832795
#define MOD 1000000007
#define read(type) readInt<type>()
#define max1 100001
#define out(x) cout<<x<<'\n'
#define out1(x) cout<<x<<" "
#define END cout<<'\n'
#define int long long
void fast(){
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
}
int power(int x, unsigned int y, int p)
{
int res = 1; // Initialize result
x = x % p; // Update x if it is more than or
// equal to p
if (x == 0) return 0; // In case x is divisible by 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;
}
// Driver code
signed main()
{
int x ;
int y;
cin>>x>>y;
int p = 1e9+7;
cout<< power(x, y, p);
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 numbers printed on them. The numbers are in the range of 1 to 6, like any <b>ordinary dice</b>. You will be provided with a face of this cube, your task is to find the number on the opposite face of the cube.
<b>Note</b>:- The sum of numbers on all opposite faces of the die is constant<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>DiceProblem()</b> that takes the integer N(face) as parameter.
<b>Constraints:</b>
1 <= N <= 6Return the number on the opposite side.Sample Input:-
2
Sample Output:-
5
Sample Input:-
1
Sample Output:-
6, I have written this Solution Code: def DiceProblem(N):
return (7-N)
, 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 numbers printed on them. The numbers are in the range of 1 to 6, like any <b>ordinary dice</b>. You will be provided with a face of this cube, your task is to find the number on the opposite face of the cube.
<b>Note</b>:- The sum of numbers on all opposite faces of the die is constant<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>DiceProblem()</b> that takes the integer N(face) as parameter.
<b>Constraints:</b>
1 <= N <= 6Return the number on the opposite side.Sample Input:-
2
Sample Output:-
5
Sample Input:-
1
Sample Output:-
6, I have written this Solution Code:
int diceProblem(int N){
return (7-N);
}
, 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 numbers printed on them. The numbers are in the range of 1 to 6, like any <b>ordinary dice</b>. You will be provided with a face of this cube, your task is to find the number on the opposite face of the cube.
<b>Note</b>:- The sum of numbers on all opposite faces of the die is constant<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>DiceProblem()</b> that takes the integer N(face) as parameter.
<b>Constraints:</b>
1 <= N <= 6Return the number on the opposite side.Sample Input:-
2
Sample Output:-
5
Sample Input:-
1
Sample Output:-
6, I have written this Solution Code:
static int diceProblem(int N){
return (7-N);
}
, In this Programming Language: Java, 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 numbers printed on them. The numbers are in the range of 1 to 6, like any <b>ordinary dice</b>. You will be provided with a face of this cube, your task is to find the number on the opposite face of the cube.
<b>Note</b>:- The sum of numbers on all opposite faces of the die is constant<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>DiceProblem()</b> that takes the integer N(face) as parameter.
<b>Constraints:</b>
1 <= N <= 6Return the number on the opposite side.Sample Input:-
2
Sample Output:-
5
Sample Input:-
1
Sample Output:-
6, I have written this Solution Code:
int diceProblem(int N){
return (7-N);
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a binary tree containing with N nodes and an integer X. Your task is to complete the function countSubtreesWithSumX() that returns the count of the number of subtress having total node’s data sum equal to a value X.
Example: A tree given below<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>countSubtreesWithSumX()</b> that takes "root" node and the integer x as parameter.
<b>Constraints:</b>
1 <= T <= 100
1 <= N <= 10^3
1 <= node values <= 10^4
<b>Sum of "N" over all testcases does not exceed 10^5</b>Return the number of subtrees with sum X. The driver code will take care of printing it.Sample Input:
1
3 5
1 2 3
Sum=5
Tree:-
1
/ \
2 3
Sample Output:
0
Explanation:
No subtree has a sum equal to 5.
Sample Input:-
1
5 5
2 1 3 4 5
Sum=5
Tree:-
2
/ \
1 3
/ \
4 5
Sample Output:-
1, I have written this Solution Code: static int c = 0;
static int countSubtreesWithSumXUtil(Node root,int x)
{
// if tree is empty
if (root==null)return 0;
// sum of nodes in the left subtree
int ls = countSubtreesWithSumXUtil(root.left,x);
// sum of nodes in the right subtree
int rs = countSubtreesWithSumXUtil(root.right, x);
int sum = ls + rs + root.data;
// if tree's nodes sum == x
if (sum == x)c++;
return sum;
}
static int countSubtreesWithSumX(Node root, int x)
{
c = 0;
// if tree is empty
if (root==null)return 0;
// sum of nodes in the left subtree
int ls = countSubtreesWithSumXUtil(root.left, x);
// sum of nodes in the right subtree
int rs = countSubtreesWithSumXUtil(root.right, x);
// check if above sum is equal to x
if ((ls + rs + root.data) == x)c++;
return c;
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a one-dimensional sorted array A containing N integers.
You are also given an integer target, find if there exists a pair of elements in the array whose difference is equal to the target.
Approach this problem in O(n).The first line contains a single integer N.
The second line contains N space- separated integer A[i].
The third line contains an integer target.
Constraints
1<=N<=10^5
0<=A[i]<=10^9
0<=target<=10^9Print Yes if pair with given difference exists in our array otherwise print No.Sample Input 1:
5
1 2 7 9 11
5
Sample Output 1:
Yes
Sample Input 2:
5
1 1 8 8 25
0
Sample Output 2:
Yes, 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);
try{
int n = sc.nextInt();
int arr[] = new int[n];
HashMap<Integer, Integer> map =new HashMap<>();
for(int i=0; i<n; i++){
arr[i]=sc.nextInt();
map.put(arr[i], 1);
}
int target = sc.nextInt();
boolean found = false;
for(int i =0; i<n; i++){
if(map.containsKey(arr[i]+target)){
found=true;
break;
}
}
if(found)
System.out.println("Yes");
else
System.out.println("No");
}
catch(Exception e){
System.out.println("No");
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a one-dimensional sorted array A containing N integers.
You are also given an integer target, find if there exists a pair of elements in the array whose difference is equal to the target.
Approach this problem in O(n).The first line contains a single integer N.
The second line contains N space- separated integer A[i].
The third line contains an integer target.
Constraints
1<=N<=10^5
0<=A[i]<=10^9
0<=target<=10^9Print Yes if pair with given difference exists in our array otherwise print No.Sample Input 1:
5
1 2 7 9 11
5
Sample Output 1:
Yes
Sample Input 2:
5
1 1 8 8 25
0
Sample Output 2:
Yes, I have written this Solution Code: #include<bits/stdc++.h>
using namespace std;
bool check(vector<int> A,int target,int n) {
int i=0,j=1,x=0;
while(i<=n && j<=n)
{
x=A[j]-A[i];
if(x==target && i!=j)return 1;
else if(x<target)j++;
else i++;
}
return 0;
}
int main() {
int n,target;
cin>>n;
vector<int>arr(n);
for(int i=0;i<n;i++)cin>>arr[i];
cin>>target;
cout<<(check(arr,target,n)==1 ? "Yes" : "No");
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
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