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For this Question: Given a matrix of size M*N, your task is to find the column having the maximum sum and print the sum.The first line of input contains two space-separated integers M and N. The following M lines of input contain N space-separated integers each depicting the values of the matrix.
Constraints:-
1 <= M, N <= 100
1 <= Matrix[][] <= 100000Print the maximum sum.Sample Input:-
3 3
1 2 3
4 5 6
7 8 9
Sample Output:-
18
Explanation:-
1 + 4 + 7 = 12
2 + 5 + 8 = 15
3 + 6 + 9 = 18
maximum = 18
Sample Input:-
3 2
1 4
9 6
9 1
Sample Output:-
19, 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 1001
#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,m;
cin>>n>>m;
int a[m];
for(int i=0;i<m;i++){
a[i]=0;
}
int x;
int sum=0;
FOR(i,n){
FOR(j,m){
cin>>x;
a[j]+=x;
sum=max(sum,a[j]);
}
}
out(sum);
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a matrix of size M*N, your task is to find the column having the maximum sum and print the sum.The first line of input contains two space-separated integers M and N. The following M lines of input contain N space-separated integers each depicting the values of the matrix.
Constraints:-
1 <= M, N <= 100
1 <= Matrix[][] <= 100000Print the maximum sum.Sample Input:-
3 3
1 2 3
4 5 6
7 8 9
Sample Output:-
18
Explanation:-
1 + 4 + 7 = 12
2 + 5 + 8 = 15
3 + 6 + 9 = 18
maximum = 18
Sample Input:-
3 2
1 4
9 6
9 1
Sample Output:-
19, 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 m = sc.nextInt();
int n = sc.nextInt();
int a[][] = new int[m][n];
for(int i=0;i<m;i++){
for(int j=0;j<n;j++){
a[i][j]=sc.nextInt();
}
}
int sum=0;
int ans=0;
for(int i=0;i<n;i++){
sum=0;
for(int j=0;j<m;j++){
sum+=a[j][i];
}
if(sum>ans){ans=sum;}
}
System.out.print(ans);
}
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a 2D screen, you are given the location of a pixel in the screen i.e. (x, y) and a color(K), your task is to replace color of the given pixel at (x, y) and all pixels connected to it with the given color K.
Two pixels are connected if these conditions satisfy:
1. They both must have same color
2. They share an edge or they are connected to a common pixel
Example:
{{1, 1, 1, 1, 1, 1, 1, 1},
{1, 1, 1, 1, 1, 1, 0, 0},
{1, 0, 0, 1, 1, 0, 1, 1},
{1, 2, 2, 2, 2, 0, 1, 0},
{1, 1, 1, 2, 2, 0, 1, 0},
{1, 1, 1, 2, 2, 2, 2, 0},
{1, 1, 1, 1, 1, 2, 1, 1},
{1, 1, 1, 1, 1, 2, 2, 1},
};
x=4, y=4, color=3
{{1, 1, 1, 1, 1, 1, 1, 1},
{1, 1, 1, 1, 1, 1, 0, 0},
{1, 0, 0, 1, 1, 0, 1, 1},
{1, 3, 3, 3, 3, 0, 1, 0},
{1, 1, 1, 3, 3, 0, 1, 0},
{1, 1, 1, 3, 3, 3, 3, 0},
{1, 1, 1, 1, 1, 3, 1, 1},
{1, 1, 1, 1, 1, 3, 3, 1}, };
Note: Use zero- based indexing.The first line of each test case contains two integers N and M denoting the size of the matrix. Then in the next line are N*M space-separated values of the matrix. Then in the next line are three values x, y and K.
Constraints:
1 <= N, M <= 300
0 <= A[][] <= 100For each test case print the space- separated values of the new matrix.Input:
3 4
0 1 1 0 1 1 1 1 0 1 2 3
0 1 5
Output
0 5 5 0 5 5 5 5 0 5 2 3
Input
2 2
1 1 1 1
0 1 8
Output
8 8 8 8, I have written this Solution Code: import java.io.*;
import java.util.*;
class Point {
public int x;
public int y;
public Point(int xx, int yy) {
x=xx;
y=yy;
}
}
class Main {
public static void main (String[] args) {
Scanner in = new Scanner(System.in);
int n = in.nextInt(), m = in.nextInt();
int[][] grid = new int[n][m];
for(int i = 0; i < n; i++) {
for(int j = 0; j < m; j++) {
grid[i][j] = in.nextInt();
}
}
int startx = in.nextInt(), starty = in.nextInt();
int targetColor = in.nextInt();
int expectedColor = grid[startx][starty];
Queue<Point> q = new LinkedList<Point>();
q.add(new Point(startx, starty));
while(!q.isEmpty()) {
Point front = q.poll();
int x = front.x, y = front.y;
grid[x][y] = targetColor;
if(x > 0 && grid[x-1][y] == expectedColor) {
q.add(new Point(x-1, y));
}
if(y > 0 && grid[x][y-1] == expectedColor) {
q.add(new Point(x, y-1));
}
if(x < n-1 && grid[x+1][y] == expectedColor) {
q.add(new Point(x+1, y));
}
if(y < m-1 && grid[x][y+1] == expectedColor) {
q.add(new Point(x, y+1));
}
}
for(int i = 0; i < n; i++) {
for(int j = 0; j < m; j++) {
System.out.print(grid[i][j] + " ");
}
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a 2D screen, you are given the location of a pixel in the screen i.e. (x, y) and a color(K), your task is to replace color of the given pixel at (x, y) and all pixels connected to it with the given color K.
Two pixels are connected if these conditions satisfy:
1. They both must have same color
2. They share an edge or they are connected to a common pixel
Example:
{{1, 1, 1, 1, 1, 1, 1, 1},
{1, 1, 1, 1, 1, 1, 0, 0},
{1, 0, 0, 1, 1, 0, 1, 1},
{1, 2, 2, 2, 2, 0, 1, 0},
{1, 1, 1, 2, 2, 0, 1, 0},
{1, 1, 1, 2, 2, 2, 2, 0},
{1, 1, 1, 1, 1, 2, 1, 1},
{1, 1, 1, 1, 1, 2, 2, 1},
};
x=4, y=4, color=3
{{1, 1, 1, 1, 1, 1, 1, 1},
{1, 1, 1, 1, 1, 1, 0, 0},
{1, 0, 0, 1, 1, 0, 1, 1},
{1, 3, 3, 3, 3, 0, 1, 0},
{1, 1, 1, 3, 3, 0, 1, 0},
{1, 1, 1, 3, 3, 3, 3, 0},
{1, 1, 1, 1, 1, 3, 1, 1},
{1, 1, 1, 1, 1, 3, 3, 1}, };
Note: Use zero- based indexing.The first line of each test case contains two integers N and M denoting the size of the matrix. Then in the next line are N*M space-separated values of the matrix. Then in the next line are three values x, y and K.
Constraints:
1 <= N, M <= 300
0 <= A[][] <= 100For each test case print the space- separated values of the new matrix.Input:
3 4
0 1 1 0 1 1 1 1 0 1 2 3
0 1 5
Output
0 5 5 0 5 5 5 5 0 5 2 3
Input
2 2
1 1 1 1
0 1 8
Output
8 8 8 8, I have written this Solution Code: #include<bits/stdc++.h>
using namespace std;
#define pu push_back
#define fi first
#define se second
#define mp make_pair
#define int long long
#define pii pair<int,int>
#define mm (s+e)/2
#define all(x) x.begin(), x.end()
#define For(i, st, en) for(int i=st; i<en; i++)
#define tr(x) for(auto it=x.begin(); it!=x.end(); it++)
#define fast std::ios::sync_with_stdio(false);cin.tie(NULL);
#define sz 200000
int A[2000][2000];
int vis[2000][2000];
int dist[2000][2000];
int xa[8]={0,0,1,-1,1,1,-1,-1};
int ya[8]={1,-1,0,0,1,-1,1,-1};
int n,m;
int val,hh;
int check(int x,int y)
{
if (x>=0 && y>=0 && x<n && y<m && A[x][y]==val)return 1;
else return 0;
}
#define qw1 freopen("input1.txt", "r", stdin); freopen("output1.txt", "w", stdout);
#define qw2 freopen("input2.txt", "r", stdin); freopen("output2.txt", "w", stdout);
#define qw3 freopen("input3.txt", "r", stdin); freopen("output3.txt", "w", stdout);
#define qw4 freopen("input4.txt", "r", stdin); freopen("output4.txt", "w", stdout);
#define qw5 freopen("input5.txt", "r", stdin); freopen("output5.txt", "w", stdout);
#define qw6 freopen("input6.txt", "r", stdin); freopen("output6.txt", "w", stdout);
#define qw freopen("input.txt", "r", stdin); freopen("output.txt", "w", stdout);
int dfs(int x,int y)
{
vis[x][y]=1;
A[x][y]=hh;
for(int i=0;i<4;i++)
{
int a=x+xa[i];
int b=y+ya[i];
if(check(a,b)==1 && vis[a][b]==0) dfs(a,b);
}
}
signed main()
{
cin>>n>>m;
for(int i=0;i<n;i++)
{
for(int j=0;j<m;j++)
{
cin>>A[i][j];
}
}
int xx,yy,gg;
cin>>xx>>yy>>gg;
val=A[xx][yy];hh=gg;
dfs(xx,yy);
for(int i=0;i<n;i++)
{
for(int j=0;j<m;j++)
{
cout<<A[i][j]<<" ";
}
}
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a binary tree with root 1. You have to make some other node as the root such that all conditions of a binary tree is satisfied. Print all nodes other than 1 that can be made as the root of the tree in sorted order.
Note - You cannot add/remove a node. You cannot break any edge.First line contains the integer N, denoting the number of nodes in the binary tree.
Next N lines contains two integers denoting the left and right child of the i'th node respectively.
If the node doesn't have a left or right child, it is denoted by '-1'
1 <= N <= 100000Print in a single line all nodes other than 1 that can be made as the root. You must print in increasing order.Sample Input 1:
5
2 4
5 -1
-1 -1
-1 3
-1 -1
Sample output 1:
2 3 4 5
Explanation: Given binary tree
1
/ \
2 4
/ \
5 3
One possible configuration of tree when 4 is made as the root
4
/ \
1 3
/
2
/
5 , I have written this Solution Code: import java.util.*;
import java.io.*;
class Node
{
int data;
Node left;
Node right;
Node(int data)
{
this.data=data;
left = right = null;
}
}
class Main
{
static void printSorted(Node root, ArrayList<Integer> ar)
{
if (root == null) {
return;
}
Queue<Node> queue = new LinkedList<Node>();
queue.add(root);
while(!queue.isEmpty())
{
int curr=queue.size();
for(int i=0;i<curr;i++)
{
Node temp=queue.poll();
boolean m=(temp.left!=null && temp.right!=null);
if(m==false)
{
if(temp.data!=1)
{
ar.add(temp.data);
}
}
if(temp.left!=null)
{
queue.add(temp.left);
}
if(temp.right!=null)
{
queue.add(temp.right);
}
}
}
Collections.sort(ar);
for(int k=0;k<ar.size();k++)
{
System.out.print(ar.get(k)+" ");
}
}
public static void main(String args[]) throws Exception
{
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
int n=Integer.parseInt(br.readLine());
Node head=new Node(1);
Node arr1[]=new Node[n+1];
arr1[1]=head;
for(int i=1;i<n+1;i++)
{
String str[]=br.readLine().trim().split(" ");
int a=Integer.parseInt(str[0]);
int b=Integer.parseInt(str[1]);
if(a!=-1){
Node newNode = new Node(a);
arr1[i].left = newNode;
arr1[a] = newNode;
}
if(b!=-1){
Node newNode = new Node(b);
arr1[i].right = newNode;
arr1[b] = newNode;
}
}
ArrayList<Integer> arr = new ArrayList<>();
printSorted(head,arr);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a binary tree with root 1. You have to make some other node as the root such that all conditions of a binary tree is satisfied. Print all nodes other than 1 that can be made as the root of the tree in sorted order.
Note - You cannot add/remove a node. You cannot break any edge.First line contains the integer N, denoting the number of nodes in the binary tree.
Next N lines contains two integers denoting the left and right child of the i'th node respectively.
If the node doesn't have a left or right child, it is denoted by '-1'
1 <= N <= 100000Print in a single line all nodes other than 1 that can be made as the root. You must print in increasing order.Sample Input 1:
5
2 4
5 -1
-1 -1
-1 3
-1 -1
Sample output 1:
2 3 4 5
Explanation: Given binary tree
1
/ \
2 4
/ \
5 3
One possible configuration of tree when 4 is made as the root
4
/ \
1 3
/
2
/
5 , I have written this Solution Code:
n=int(input())
A=[]
for i in range(1,n+1):
u,v=[int(x) for x in input().split()]
if(u==-1 or v== -1):
A.append(i)
A.sort()
for x in A:
print(x,end=" "), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a binary tree with root 1. You have to make some other node as the root such that all conditions of a binary tree is satisfied. Print all nodes other than 1 that can be made as the root of the tree in sorted order.
Note - You cannot add/remove a node. You cannot break any edge.First line contains the integer N, denoting the number of nodes in the binary tree.
Next N lines contains two integers denoting the left and right child of the i'th node respectively.
If the node doesn't have a left or right child, it is denoted by '-1'
1 <= N <= 100000Print in a single line all nodes other than 1 that can be made as the root. You must print in increasing order.Sample Input 1:
5
2 4
5 -1
-1 -1
-1 3
-1 -1
Sample output 1:
2 3 4 5
Explanation: Given binary tree
1
/ \
2 4
/ \
5 3
One possible configuration of tree when 4 is made as the root
4
/ \
1 3
/
2
/
5 , I have written this Solution Code: #include "bits/stdc++.h"
using namespace std;
#define int long long int
#define ld long double
#define pi pair<int, int>
#define pb push_back
#define fi first
#define se second
#define IOS ios::sync_with_stdio(false); cin.tie(0); cout.tie(0)
#ifndef LOCAL
#define endl '\n'
#endif
const int N = 2e5 + 5;
const int mod = 1e9 + 7;
const int inf = 1e9 + 9;
int d[N];
void solve(){
int n; cin >> n;
for(int i = 1; i <= n; i++){
int x, y;
cin >> x >> y;
if(x != -1) d[i]++, d[x]++;
if(y != -1) d[i]++, d[y]++;
}
for(int i = 2; i <= n; i++){
if(d[i] <= 2)
cout << i << " ";
}
}
signed main() {
IOS;
clock_t start = clock();
int Test = 1;
// cin >> Test;
for(int tt = 0; tt < Test; tt++){
solve();
}
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: There are n people in a circle, numbered from 1 to n, each of whom always tells the truth or always lies.
Each person i makes a claim of the form: “the number of truth-tellers in this circle is between a<sub>i</sub> and b<sub>i</sub>, inclusive.”
Compute the maximum number of people who could be telling the truth.The first line contains a single integer n (1 ≤ n ≤ 10^3).
Each of the next n lines contains two space-separated integers a<sub>i</sub> and b<sub>i</sub>
(0 ≤ a<sub>i</sub> ≤ b<sub>i</sub> ≤ n).Print, on a single line, the maximum number of people who could be telling the truth. If the given
set of statements are inconsistent, print -1 instead.Sample input
3
1 1
2 3
2 2
Sample output
2
Sample input
8
0 1
1 7
4 8
3 7
1 2
4 5
3 7
1 8
Sample output
-1, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws IOException{
BufferedReader read=new BufferedReader(new InputStreamReader(System.in));
int n=Integer.parseInt(read.readLine());
int count=0;
int cnt[]=new int[n+2];
for(int i=0;i<n;i++){
StringTokenizer st=new StringTokenizer(read.readLine()," ");
int ai=Integer.parseInt(st.nextToken());
int bi=Integer.parseInt(st.nextToken());
for(int j=ai;j<=bi;j++)
{
cnt[j]++;
}
}
int ans=-1;
for(int i=0;i<=n;i++){
if(cnt[i]==i)
{
ans=i;
}
}
System.out.println(ans);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: There are n people in a circle, numbered from 1 to n, each of whom always tells the truth or always lies.
Each person i makes a claim of the form: “the number of truth-tellers in this circle is between a<sub>i</sub> and b<sub>i</sub>, inclusive.”
Compute the maximum number of people who could be telling the truth.The first line contains a single integer n (1 ≤ n ≤ 10^3).
Each of the next n lines contains two space-separated integers a<sub>i</sub> and b<sub>i</sub>
(0 ≤ a<sub>i</sub> ≤ b<sub>i</sub> ≤ n).Print, on a single line, the maximum number of people who could be telling the truth. If the given
set of statements are inconsistent, print -1 instead.Sample input
3
1 1
2 3
2 2
Sample output
2
Sample input
8
0 1
1 7
4 8
3 7
1 2
4 5
3 7
1 8
Sample output
-1, I have written this Solution Code: n=int(input())
a=[0]*(n+1)
m=0
for i in range(n):
b=[int(k) for k in input().split()]
for j in range(b[0],b[1]+1):
a[j]+=1;
for i in range(n,0,-1):
if a[i]==i:
print(i)
exit()
print(-1), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: There are n people in a circle, numbered from 1 to n, each of whom always tells the truth or always lies.
Each person i makes a claim of the form: “the number of truth-tellers in this circle is between a<sub>i</sub> and b<sub>i</sub>, inclusive.”
Compute the maximum number of people who could be telling the truth.The first line contains a single integer n (1 ≤ n ≤ 10^3).
Each of the next n lines contains two space-separated integers a<sub>i</sub> and b<sub>i</sub>
(0 ≤ a<sub>i</sub> ≤ b<sub>i</sub> ≤ n).Print, on a single line, the maximum number of people who could be telling the truth. If the given
set of statements are inconsistent, print -1 instead.Sample input
3
1 1
2 3
2 2
Sample output
2
Sample input
8
0 1
1 7
4 8
3 7
1 2
4 5
3 7
1 8
Sample output
-1, I have written this Solution Code: #include<stdio.h>
#define maxn 1100
struct node{
int l,r;
}a[maxn];
int main(){
int n,i,cnt,j,ans=-1;
scanf("%d",&n);
for (i=1;i<=n;i++) scanf("%d%d",&a[i].l,&a[i].r);
for (i=n;i>=0;i--) {
cnt=0;
for (j=1;j<=n;j++) if(a[j].l<=i&&i<=a[j].r) cnt++;
if (cnt==i) {ans=i;break;}
}
printf("%d\n",ans);
return 0;
}, In this Programming Language: C, Now tell me if this Code is compilable or not? | Compilable |
For this Question: For an integer N, your task is to calculate sum of first N natural numbers.<b>User Task:</b>
Since this will be a functional problem, you don't have to worry about input. You just have to complete the function <b>sum()</b> which takes the integer N as a parameter.
Constraints:
1 <= N < = 100000000Print the sum of first N natural numbers.Sample Input:-
5
Sample Output:-
15
Sample Input:-
3
Sample Output:-
6, I have written this Solution Code:
static void sum(int N){
long x=N;
x=x*(x+1);
x=x/2;
System.out.print(x);
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given three integers A, B and C. Your task is to calculate the number of integers which are factors of atleast two given numbers.<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>CommonFactors()</b> that takes the integers A, B and C as parameters.
<b>Constraints:</b>
1 <= A, B, C <= 100000Return the count of common factors.Sample Input:-
3 6 12
Sample Output:-
4
Explanation:- 1, 2, 3, 6 are the required factors
Sample Input:-
1 2 3
Sample Output:-
1, I have written this Solution Code:
int commonFactors(int A,int B,int C){
int cnt=0;
for(int i=1;i<=A;i++){
if(A%i==0){
if(B%i==0 || C%i==0){cnt++;}
}
}
for(int i=1;i<=B;i++){
if(B%i==0){
if(A%i!=0 && C%i==0){cnt++;}
}
}
return cnt;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given three integers A, B and C. Your task is to calculate the number of integers which are factors of atleast two given numbers.<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>CommonFactors()</b> that takes the integers A, B and C as parameters.
<b>Constraints:</b>
1 <= A, B, C <= 100000Return the count of common factors.Sample Input:-
3 6 12
Sample Output:-
4
Explanation:- 1, 2, 3, 6 are the required factors
Sample Input:-
1 2 3
Sample Output:-
1, I have written this Solution Code:
static int commonFactors(int A,int B,int C){
int cnt=0;
for(int i=1;i<=A;i++){
if(A%i==0){
if(B%i==0 || C%i==0){cnt++;}
}
}
for(int i=1;i<=B;i++){
if(B%i==0){
if(A%i!=0 && C%i==0){cnt++;}
}
}
return cnt;
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given three integers A, B and C. Your task is to calculate the number of integers which are factors of atleast two given numbers.<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>CommonFactors()</b> that takes the integers A, B and C as parameters.
<b>Constraints:</b>
1 <= A, B, C <= 100000Return the count of common factors.Sample Input:-
3 6 12
Sample Output:-
4
Explanation:- 1, 2, 3, 6 are the required factors
Sample Input:-
1 2 3
Sample Output:-
1, I have written this Solution Code:
int commonFactors(int A,int B,int C){
int cnt=0;
for(int i=1;i<=A;i++){
if(A%i==0){
if(B%i==0 || C%i==0){cnt++;}
}
}
for(int i=1;i<=B;i++){
if(B%i==0){
if(A%i!=0 && C%i==0){cnt++;}
}
}
return cnt;
}
, In this Programming Language: C, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given three integers A, B and C. Your task is to calculate the number of integers which are factors of atleast two given numbers.<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>CommonFactors()</b> that takes the integers A, B and C as parameters.
<b>Constraints:</b>
1 <= A, B, C <= 100000Return the count of common factors.Sample Input:-
3 6 12
Sample Output:-
4
Explanation:- 1, 2, 3, 6 are the required factors
Sample Input:-
1 2 3
Sample Output:-
1, I have written this Solution Code: def CommonFactors(A,B,C):
cnt=0
for i in range (1,A+1):
if(A%i==0):
if(B%i==0) or (C%i==0):
cnt=cnt+1
for i in range (1,B+1):
if(B%i==0):
if(A%i!=0) and (C%i==0):
cnt = cnt +1
return cnt
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Ready for a super simple challenge!
You are given two integers A and B. You will create a line of boys and girls in the following manner:
First A girls, then B boys, then A girls, then B boys, and so on.
You will also be given an integer N. Find the number of girls in the first N people of the row.The first and the only line of input contains 3 integers N, A, B.
Constraints
1 < N, A, B <= 10^18 (1000000000000000000)
A+B <= 10^18Output a singer integer, the number of girls in the first N people in the queue.Sample Input
8 3 4
Sample Output
4
Explanation
The arrangement will look like GGGBBBBG, where G represents girl and B represents boy., 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[] nab = rd.readLine().split(" ");
long n = Long.parseLong(nab[0]);
long girl = Long.parseLong(nab[1]);
long boy = Long.parseLong(nab[2]);
long total = 0;
if(n>1 && girl>1 && boy>1){
long temp = n/(girl+boy);
total = (girl*temp);
if(n%(girl+boy)<girl)
total+=n%(girl+boy);
else
total+=girl;
System.out.print(total);
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Ready for a super simple challenge!
You are given two integers A and B. You will create a line of boys and girls in the following manner:
First A girls, then B boys, then A girls, then B boys, and so on.
You will also be given an integer N. Find the number of girls in the first N people of the row.The first and the only line of input contains 3 integers N, A, B.
Constraints
1 < N, A, B <= 10^18 (1000000000000000000)
A+B <= 10^18Output a singer integer, the number of girls in the first N people in the queue.Sample Input
8 3 4
Sample Output
4
Explanation
The arrangement will look like GGGBBBBG, where G represents girl and B represents boy., I have written this Solution Code: n,a,b=map(int,input().split())
if n%(a+b)== 0:
print(n//(a+b)*a)
else:
k=n-(n//(a+b))*(a+b)
if k >= a:
print((n//(a+b))*a+a)
else:
print((n//(a+b))*a+k), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Ready for a super simple challenge!
You are given two integers A and B. You will create a line of boys and girls in the following manner:
First A girls, then B boys, then A girls, then B boys, and so on.
You will also be given an integer N. Find the number of girls in the first N people of the row.The first and the only line of input contains 3 integers N, A, B.
Constraints
1 < N, A, B <= 10^18 (1000000000000000000)
A+B <= 10^18Output a singer integer, the number of girls in the first N people in the queue.Sample Input
8 3 4
Sample Output
4
Explanation
The arrangement will look like GGGBBBBG, where G represents girl and B represents boy., 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
int begtime = clock();
#define end_routine() cout << "\n\nTime elapsed: " << (clock() - begtime)*1000/CLOCKS_PER_SEC << " ms\n\n";
#else
#define endl '\n'
#define end_routine()
#endif
signed main()
{
fast
int n, a, b; cin>>n>>a>>b;
int x = a+b;
int y = (n/x);
n -= (y*x);
int ans = y*a;
if(n > a)
ans += a;
else
ans += n;
cout<<ans;
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array arr[] of size N and a number K, the task is to find the smallest number M such that the sum of the array becomes lesser than or equal to the number K when every element of that array is divided by the number M.
Note: Each result of the division is rounded to the nearest integer greater than or equal to that element. For example: 10/3 = 4 and 6/2 = 3The first line contains two integers N and K.
Next line contains N integers denoting the elements of arr[]
<b>Constraints:</b>
1 <= N <= 100000
1 <= arr[i] <= 100000
1 <= K <= 100000000Print a single integer the value of smallest number MSample Input:
4 6
2 3 4 9
Sample Output:
4
Explanation:
When every element is divided by 4 - 2/4 + 3/4 + 4/4 + 9/4 = 1 + 1 + 1 + 3 = 6
When every element is divided by 3 - 2/3 + 3/3 + 4/3 + 9/3 = 1 + 1 + 2 + 3 = 7 which is greater than K.
Hence the smallest integer which makes the sum less than or equal to K = 6 is 4., I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws IOException{
InputStreamReader ir = new InputStreamReader(System.in);
BufferedReader br = new BufferedReader(ir);
String s[] = br.readLine().split(" ");
int n = Integer.parseInt(s[0]);
int k = Integer.parseInt(s[1]);
String str[] = br.readLine().split(" ");
int[] arr = new int[n];
for(int i=0;i<n;i++){
arr[i] = Integer.parseInt(str[i]);
}
System.out.print(minDivisor(arr,n,k));
}
static int minDivisor(int arr[],int N, int limit) {
int low = 0, high = 1000000000;
while (low < high)
{
int mid = (low + high) / 2;
int sum = 0;
for(int i = 0; i < N; i++)
{
sum += Math.ceil((double) arr[i] / (double) mid);
}
if(sum <= limit){
high = mid;
}
else{
low = mid + 1;
}
}
return low;
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array arr[] of size N and a number K, the task is to find the smallest number M such that the sum of the array becomes lesser than or equal to the number K when every element of that array is divided by the number M.
Note: Each result of the division is rounded to the nearest integer greater than or equal to that element. For example: 10/3 = 4 and 6/2 = 3The first line contains two integers N and K.
Next line contains N integers denoting the elements of arr[]
<b>Constraints:</b>
1 <= N <= 100000
1 <= arr[i] <= 100000
1 <= K <= 100000000Print a single integer the value of smallest number MSample Input:
4 6
2 3 4 9
Sample Output:
4
Explanation:
When every element is divided by 4 - 2/4 + 3/4 + 4/4 + 9/4 = 1 + 1 + 1 + 3 = 6
When every element is divided by 3 - 2/3 + 3/3 + 4/3 + 9/3 = 1 + 1 + 2 + 3 = 7 which is greater than K.
Hence the smallest integer which makes the sum less than or equal to K = 6 is 4., I have written this Solution Code: from math import ceil
def kSum(arr, n, k):
low = 0
high = 10 ** 9
while (low < high):
mid = (low + high) // 2
sum = 0
for i in range(int(n)):
sum += ceil( int(arr[i]) / mid)
if (sum <= int(k)):
high = mid
else:
low = mid + 1
return low
n, k =input().split()
arr = input().split()
print( kSum(arr, n, k) ), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array arr[] of size N and a number K, the task is to find the smallest number M such that the sum of the array becomes lesser than or equal to the number K when every element of that array is divided by the number M.
Note: Each result of the division is rounded to the nearest integer greater than or equal to that element. For example: 10/3 = 4 and 6/2 = 3The first line contains two integers N and K.
Next line contains N integers denoting the elements of arr[]
<b>Constraints:</b>
1 <= N <= 100000
1 <= arr[i] <= 100000
1 <= K <= 100000000Print a single integer the value of smallest number MSample Input:
4 6
2 3 4 9
Sample Output:
4
Explanation:
When every element is divided by 4 - 2/4 + 3/4 + 4/4 + 9/4 = 1 + 1 + 1 + 3 = 6
When every element is divided by 3 - 2/3 + 3/3 + 4/3 + 9/3 = 1 + 1 + 2 + 3 = 7 which is greater than K.
Hence the smallest integer which makes the sum less than or equal to K = 6 is 4., I have written this Solution Code: #include "bits/stdc++.h"
using namespace std;
#define ll long long int
#define ld long double
#define pi pair<int, int>
#define pb push_back
#define fi first
#define se second
#define IOS ios::sync_with_stdio(false); cin.tie(0); cout.tie(0)
#ifndef LOCAL
#define endl '\n'
#endif
const int N = 1e5 + 5;
const int mod = 1e9 + 7;
const int inf = 1e9 + 9;
int a[N], n, k;
bool f(int x){
int sum = 0;
for(int i = 1; i <= n; i++)
sum += (a[i]-1)/x + 1;
return (sum <= k);
}
void solve(){
cin >> n >> k;
for(int i = 1; i <= n; i++)
cin >> a[i];
int l = 0, h = N;
while(l+1 < h){
int m = (l + h) >> 1;
if(f(m))
h = m;
else
l = m;
}
cout << h;
}
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 a sorted array arr[] of N integers and a number K is given. The task is to check if the element K is present in the array or not.
<b>Note:</b> Use binary search to solve the problemThe first line of input contains a number of test cases T. For each test case, the first line of input contains a number of elements in the array, and the number K is separated by space. The next line contains N elements.
<b>Constraints:</b>
1 ≤ T ≤ 10
1 ≤ N ≤ 10<sup>5</sup>
1 ≤ K ≤ 10<sup>9</sup>
1 ≤ arr[i] ≤ 10<sup>9</sup>
<b>Sum of N over all test cases doesn't exceed 10<sup>6</sup></b>If the element is present in the array print "1" else print "-1".Sample Input:
2
5 6
1 2 3 4 6
5 2
1 3 4 5 6
Sample Output:
1
-1, I have written this Solution Code: static int isPresent(long arr[], int n, long k)
{
int left = 0;
int right = n-1;
int res = -1;
while(left<=right){
int mid = (left+right)/2;
if(arr[mid] == k){
res = 1;
break;
}else if(arr[mid] < k){
left = mid + 1;
}else{
right = mid - 1;
}
}
return res;
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a sorted array arr[] of N integers and a number K is given. The task is to check if the element K is present in the array or not.
<b>Note:</b> Use binary search to solve the problemThe first line of input contains a number of test cases T. For each test case, the first line of input contains a number of elements in the array, and the number K is separated by space. The next line contains N elements.
<b>Constraints:</b>
1 ≤ T ≤ 10
1 ≤ N ≤ 10<sup>5</sup>
1 ≤ K ≤ 10<sup>9</sup>
1 ≤ arr[i] ≤ 10<sup>9</sup>
<b>Sum of N over all test cases doesn't exceed 10<sup>6</sup></b>If the element is present in the array print "1" else print "-1".Sample Input:
2
5 6
1 2 3 4 6
5 2
1 3 4 5 6
Sample Output:
1
-1, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main(){
int t;
cin>>t;
while(t--){
int n;
cin>>n;
unordered_map<long long,int> m;
long k;
cin>>k;
long long a;
for(int i=0;i<n;i++){
cin>>a;
m[a]++;
}
if(m.find(k)!=m.end()){
cout<<1<<endl;
}
else{
cout<<-1<<endl;
}
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a sorted array arr[] of N integers and a number K is given. The task is to check if the element K is present in the array or not.
<b>Note:</b> Use binary search to solve the problemThe first line of input contains a number of test cases T. For each test case, the first line of input contains a number of elements in the array, and the number K is separated by space. The next line contains N elements.
<b>Constraints:</b>
1 ≤ T ≤ 10
1 ≤ N ≤ 10<sup>5</sup>
1 ≤ K ≤ 10<sup>9</sup>
1 ≤ arr[i] ≤ 10<sup>9</sup>
<b>Sum of N over all test cases doesn't exceed 10<sup>6</sup></b>If the element is present in the array print "1" else print "-1".Sample Input:
2
5 6
1 2 3 4 6
5 2
1 3 4 5 6
Sample Output:
1
-1, I have written this Solution Code: def binary_search(arr, low, high, x):
if high >= low:
mid = (high + low) // 2
if arr[mid] == x:
return 1
elif arr[mid] > x:
return binary_search(arr, low, mid - 1, x)
else:
return binary_search(arr, mid + 1, high, x)
else:
return -1
def position(n,arr,x):
return binary_search(arr,0,n-1,x)
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a sorted array arr[] of N integers and a number K is given. The task is to check if the element K is present in the array or not.
<b>Note:</b> Use binary search to solve the problemThe first line of input contains a number of test cases T. For each test case, the first line of input contains a number of elements in the array, and the number K is separated by space. The next line contains N elements.
<b>Constraints:</b>
1 ≤ T ≤ 10
1 ≤ N ≤ 10<sup>5</sup>
1 ≤ K ≤ 10<sup>9</sup>
1 ≤ arr[i] ≤ 10<sup>9</sup>
<b>Sum of N over all test cases doesn't exceed 10<sup>6</sup></b>If the element is present in the array print "1" else print "-1".Sample Input:
2
5 6
1 2 3 4 6
5 2
1 3 4 5 6
Sample Output:
1
-1, I have written this Solution Code: // arr is they array to search from
// x is target
function binSearch(arr, x) {
// write code here
// do not console.log
// return the 1 or -1
let l = 0;
let r = arr.length - 1;
let mid;
while (r >= l) {
mid = l + Math.floor((r - l) / 2);
// If the element is present at the middle
// itself
if (arr[mid] == x)
return 1;
// If element is smaller than mid, then
// it can only be present in left subarray
if (arr[mid] > x)
r = mid - 1;
// Else the element can only be present
// in right subarray
else
l = mid + 1;
}
// We reach here when element is not
// present in array
return -1;
}, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given two strings Str1 and Str2 your task is to print the concatenation of the given two strings.<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>StringConcatenation()</b> that takes the string Str1 and Str2 as input.
<b>Note</b>:- String will contain uppercase and lowercase English letters.Return the concatenation of both the strings.Sample Input:-
Newton
School
Sample Output:-
NewtonSchool
Sample Input:-
Women
InTech
Sample Output:-
WomenInTech, I have written this Solution Code: def StringConcatenation(Str1,Str2):
return Str1+Str2
Str1 = input()
Str2 = input()
result = StringConcatenation(Str1,Str2)
print(result), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given two strings Str1 and Str2 your task is to print the concatenation of the given two strings.<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>StringConcatenation()</b> that takes the string Str1 and Str2 as input.
<b>Note</b>:- String will contain uppercase and lowercase English letters.Return the concatenation of both the strings.Sample Input:-
Newton
School
Sample Output:-
NewtonSchool
Sample Input:-
Women
InTech
Sample Output:-
WomenInTech, I have written this Solution Code: static String StringConcatenation(String Str1, String Str2){
String P = Str1+Str2;
return P;
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given two strings Str1 and Str2 your task is to print the concatenation of the given two strings.<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>StringConcatenation()</b> that takes the string Str1 and Str2 as input.
<b>Note</b>:- String will contain uppercase and lowercase English letters.Return the concatenation of both the strings.Sample Input:-
Newton
School
Sample Output:-
NewtonSchool
Sample Input:-
Women
InTech
Sample Output:-
WomenInTech, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main() {
string a,b;
cin>>a>>b;
a+=b;
cout<<a;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a matrix of size N*N, your task is to find the sum of the primary and secondary diagonal of the matrix.
For Matrix:-
M<sub>00</sub> M<sub>01</sub> M<sub>02</sub>
M<sub>10</sub> M<sub>11</sub> M<sub>12</sub>
M<sub>20</sub> M<sub>21</sub> M<sub>22</sub>
Primary diagonal:- M<sub>00</sub> M<sub>11</sub> M<sub>22</sub>
Secondary diagonal:- M<sub>02</sub> M<sub>11</sub> M<sub>20</sub>The first line of input contains a single integer N, The next N lines of input contains N space-separated integers depicting the values of the matrix.
Constraints:-
1 <= N <= 500
1 <= Matrix[][] <= 100000Print the sum of primary and secondary diagonal separated by a space.Sample Input:-
2
1 4
2 6
Sample Output:-
7 6
Sample Input:-
3
1 4 2
1 5 7
3 8 1
Sample Output:-
7 10, 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
#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[n][n];
FOR(i,n){
FOR(j,n){
cin>>a[i][j];}}
int sum=0,sum1=0;;
FOR(i,n){
sum+=a[i][i];
sum1+=a[n-i-1][i];
}
out1(sum);out(sum1);
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a matrix of size N*N, your task is to find the sum of the primary and secondary diagonal of the matrix.
For Matrix:-
M<sub>00</sub> M<sub>01</sub> M<sub>02</sub>
M<sub>10</sub> M<sub>11</sub> M<sub>12</sub>
M<sub>20</sub> M<sub>21</sub> M<sub>22</sub>
Primary diagonal:- M<sub>00</sub> M<sub>11</sub> M<sub>22</sub>
Secondary diagonal:- M<sub>02</sub> M<sub>11</sub> M<sub>20</sub>The first line of input contains a single integer N, The next N lines of input contains N space-separated integers depicting the values of the matrix.
Constraints:-
1 <= N <= 500
1 <= Matrix[][] <= 100000Print the sum of primary and secondary diagonal separated by a space.Sample Input:-
2
1 4
2 6
Sample Output:-
7 6
Sample Input:-
3
1 4 2
1 5 7
3 8 1
Sample Output:-
7 10, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String args[])throws Exception {
InputStreamReader inr= new InputStreamReader(System.in);
BufferedReader br= new BufferedReader(inr);
String str=br.readLine();
int row = Integer.parseInt(str);
int col=row;
int [][] arr=new int [row][col];
for(int i=0;i<row;i++){
String line =br.readLine();
String[] elements = line.split(" ");
for(int j=0;j<col;j++){
arr[i][j]= Integer.parseInt(elements[j]);
}
}
int sumPrimary=0;
int sumSecondary=0;
for(int i=0;i<row;i++){
sumPrimary=sumPrimary + arr[i][i];
sumSecondary= sumSecondary + arr[i][row-1-i];
}
System.out.println(sumPrimary+ " " +sumSecondary);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a matrix of size N*N, your task is to find the sum of the primary and secondary diagonal of the matrix.
For Matrix:-
M<sub>00</sub> M<sub>01</sub> M<sub>02</sub>
M<sub>10</sub> M<sub>11</sub> M<sub>12</sub>
M<sub>20</sub> M<sub>21</sub> M<sub>22</sub>
Primary diagonal:- M<sub>00</sub> M<sub>11</sub> M<sub>22</sub>
Secondary diagonal:- M<sub>02</sub> M<sub>11</sub> M<sub>20</sub>The first line of input contains a single integer N, The next N lines of input contains N space-separated integers depicting the values of the matrix.
Constraints:-
1 <= N <= 500
1 <= Matrix[][] <= 100000Print the sum of primary and secondary diagonal separated by a space.Sample Input:-
2
1 4
2 6
Sample Output:-
7 6
Sample Input:-
3
1 4 2
1 5 7
3 8 1
Sample Output:-
7 10, I have written this Solution Code: // mat is the matrix/ 2d array
// the dimensions of array are n * n
function diagonalSum(mat, n) {
// write code here
// console.log the answer as in example
let principal = 0, secondary = 0;
for (let i = 0; i < n; i++) {
principal += mat[i][i];
secondary += mat[i][n - i - 1];
}
console.log(`${principal} ${secondary}`);
}
, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a matrix of size N*N, your task is to find the sum of the primary and secondary diagonal of the matrix.
For Matrix:-
M<sub>00</sub> M<sub>01</sub> M<sub>02</sub>
M<sub>10</sub> M<sub>11</sub> M<sub>12</sub>
M<sub>20</sub> M<sub>21</sub> M<sub>22</sub>
Primary diagonal:- M<sub>00</sub> M<sub>11</sub> M<sub>22</sub>
Secondary diagonal:- M<sub>02</sub> M<sub>11</sub> M<sub>20</sub>The first line of input contains a single integer N, The next N lines of input contains N space-separated integers depicting the values of the matrix.
Constraints:-
1 <= N <= 500
1 <= Matrix[][] <= 100000Print the sum of primary and secondary diagonal separated by a space.Sample Input:-
2
1 4
2 6
Sample Output:-
7 6
Sample Input:-
3
1 4 2
1 5 7
3 8 1
Sample Output:-
7 10, I have written this Solution Code: n = int(input())
sum1 = 0
sum2 = 0
for i in range(n):
a = [int(j) for j in input().split()]
sum1 = sum1+a[i]
sum2 = sum2+a[n-1-i]
print(sum1,sum2), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Alice has Q queries for you. If you can solve all her queries, she will invite you to participate in a fireworks display with her.
In each query, she gives you a positive integer N, and you have to find the number of positive integers M such that M × d(M) = N.
Here, d(M) refers to the sum of all digits of M. For example, d(1023) = 6.The first line consists of a single integer Q – the number of queries.
Then Q lines follow, each line containing a single integer N denoting a query.
<b> Constraints: </b>
1 ≤ Q ≤ 1000
1 ≤ N ≤ 10<sup>18</sup>Print Q lines, the i<sup>th</sup> line containing the answer to the i<sup>th</sup> query.Sample Input 1:
3
1
6
36
Sample Output 1:
1
0
2
Sample Explanation 1:
For the first query, the only possibility is 1.
For the third query, the only possibilities are 6 and 12., I have written this Solution Code: import java.io.*;
import java.util.*;
public class Main {
private static final int START_TEST_CASE = 1;
public static void solveCase(FastIO io, int testCase) {
final long N = io.nextLong();
int count = 0;
for (int i = 1; i <= 200; ++i) {
if (N % i == 0) {
long j = N / i;
if (digitSum(j) == i) {
++count;
}
}
}
io.println(count);
}
private static long digitSum(long x) {
long s = 0;
while (x > 0) {
s += x % 10;
x /= 10;
}
return s;
}
public static void solve(FastIO io) {
final int T = io.nextInt();
for (int t = 0; t < T; ++t) {
solveCase(io, START_TEST_CASE + t);
}
}
public static class FastIO {
private InputStream reader;
private PrintWriter writer;
private byte[] buf = new byte[1024];
private int curChar;
private int numChars;
public FastIO(InputStream r, OutputStream w) {
reader = r;
writer = new PrintWriter(new BufferedWriter(new OutputStreamWriter(w)));
}
public int read() {
if (numChars == -1)
throw new InputMismatchException();
if (curChar >= numChars) {
curChar = 0;
try {
numChars = reader.read(buf);
} catch (IOException e) {
throw new InputMismatchException();
}
if (numChars <= 0)
return -1;
}
return buf[curChar++];
}
public String nextLine() {
int c = read();
while (isSpaceChar(c))
c = read();
StringBuilder res = new StringBuilder();
do {
res.appendCodePoint(c);
c = read();
} while (!isEndOfLine(c));
return res.toString();
}
public String nextString() {
int c = read();
while (isSpaceChar(c))
c = read();
StringBuilder res = new StringBuilder();
do {
res.appendCodePoint(c);
c = read();
} while (!isSpaceChar(c));
return res.toString();
}
public long nextLong() {
int c = read();
while (isSpaceChar(c))
c = read();
int sgn = 1;
if (c == '-') {
sgn = -1;
c = read();
}
long res = 0;
do {
if (c < '0' || c > '9')
throw new InputMismatchException();
res *= 10;
res += c - '0';
c = read();
} while (!isSpaceChar(c));
return res * sgn;
}
public int nextInt() {
int c = read();
while (isSpaceChar(c))
c = read();
int sgn = 1;
if (c == '-') {
sgn = -1;
c = read();
}
int res = 0;
do {
if (c < '0' || c > '9')
throw new InputMismatchException();
res *= 10;
res += c - '0';
c = read();
} while (!isSpaceChar(c));
return res * sgn;
}
public double nextDouble() {
return Double.parseDouble(nextString());
}
public int[] nextIntArray(int n) {
return nextIntArray(n, 0);
}
public int[] nextIntArray(int n, int off) {
int[] arr = new int[n + off];
for (int i = 0; i < n; i++) {
arr[i + off] = nextInt();
}
return arr;
}
public long[] nextLongArray(int n) {
return nextLongArray(n, 0);
}
public long[] nextLongArray(int n, int off) {
long[] arr = new long[n + off];
for (int i = 0; i < n; i++) {
arr[i + off] = nextLong();
}
return arr;
}
private boolean isSpaceChar(int c) {
return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
}
private boolean isEndOfLine(int c) {
return c == '\n' || c == '\r' || c == -1;
}
public void print(Object... objects) {
for (int i = 0; i < objects.length; i++) {
if (i != 0) {
writer.print(' ');
}
writer.print(objects[i]);
}
}
public void println(Object... objects) {
print(objects);
writer.println();
}
public void printArray(int[] arr) {
for (int i = 0; i < arr.length; i++) {
if (i != 0) {
writer.print(' ');
}
writer.print(arr[i]);
}
}
public void printArray(long[] arr) {
for (int i = 0; i < arr.length; i++) {
if (i != 0) {
writer.print(' ');
}
writer.print(arr[i]);
}
}
public void printlnArray(int[] arr) {
printArray(arr);
writer.println();
}
public void printlnArray(long[] arr) {
printArray(arr);
writer.println();
}
public void printf(String format, Object... args) {
print(String.format(format, args));
}
public void flush() {
writer.flush();
}
}
private static class Solution implements Runnable {
@Override
public void run() {
FastIO io = new FastIO(System.in, System.out);
solve(io);
io.flush();
}
}
public static void main(String[] args) throws InterruptedException {
Thread t = new Thread(null, new Solution(), "Solution", 1 << 30);
t.start();
t.join();
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Alice has Q queries for you. If you can solve all her queries, she will invite you to participate in a fireworks display with her.
In each query, she gives you a positive integer N, and you have to find the number of positive integers M such that M × d(M) = N.
Here, d(M) refers to the sum of all digits of M. For example, d(1023) = 6.The first line consists of a single integer Q – the number of queries.
Then Q lines follow, each line containing a single integer N denoting a query.
<b> Constraints: </b>
1 ≤ Q ≤ 1000
1 ≤ N ≤ 10<sup>18</sup>Print Q lines, the i<sup>th</sup> line containing the answer to the i<sup>th</sup> query.Sample Input 1:
3
1
6
36
Sample Output 1:
1
0
2
Sample Explanation 1:
For the first query, the only possibility is 1.
For the third query, the only possibilities are 6 and 12., I have written this Solution Code:
q = int(input())
for _ in range(q):
n = int(input())
count = 0
for i in range(1,9*len(str(n))):
if not n % i:
dig = n//i
if sum(map(int,str(dig))) == i:
count += 1
print(count), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Alice has Q queries for you. If you can solve all her queries, she will invite you to participate in a fireworks display with her.
In each query, she gives you a positive integer N, and you have to find the number of positive integers M such that M × d(M) = N.
Here, d(M) refers to the sum of all digits of M. For example, d(1023) = 6.The first line consists of a single integer Q – the number of queries.
Then Q lines follow, each line containing a single integer N denoting a query.
<b> Constraints: </b>
1 ≤ Q ≤ 1000
1 ≤ N ≤ 10<sup>18</sup>Print Q lines, the i<sup>th</sup> line containing the answer to the i<sup>th</sup> query.Sample Input 1:
3
1
6
36
Sample Output 1:
1
0
2
Sample Explanation 1:
For the first query, the only possibility is 1.
For the third query, the only possibilities are 6 and 12., I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define int long long
// #pragma gcc optimize("ofast")
// #pragma gcc target("avx,avx2,fma")
#define all(x) (x).begin(), (x).end()
#define pb push_back
#define endl '\n'
#define fi first
#define se second
// const int mod = 1e9 + 7;
const int mod=998'244'353;
const long long INF = 2e18 + 10;
// const int INF=4e9+10;
#define readv(x, n) \
vector<int> x(n); \
for (auto &i : x) \
cin >> i;
template <typename t>
using v = vector<t>;
template <typename t>
using vv = vector<vector<t>>;
template <typename t>
using vvv = vector<vector<vector<t>>>;
typedef vector<int> vi;
typedef vector<double> vd;
typedef vector<vector<int>> vvi;
typedef vector<vector<vector<int>>> vvvi;
typedef vector<vector<vector<vector<int>>>> vvvvi;
typedef vector<vector<double>> vvd;
typedef pair<int, int> pii;
int multiply(int a, int b, int in_mod) { return (int)(1ll * a * b % in_mod); }
int mult_identity(int a) { return 1; }
const double pi = acosl(-1);
vector<vector<int> > multiply(vector<vector<int>> a, vector<vector<int>> b, int in_mod)
{
int n = a.size();
int l = b.size();
int m = b[0].size();
vector<vector<int> > result(n,vector<int>(n));
for(int i=0;i<n;i++)
{
for(int j=0;j<m;j++)
{
for(int k=0;k<l;k++)
{
result[i][j] = (result[i][j] + a[i][k]*b[k][j])%in_mod;
}
}
}
return result;
}
vector<vector<int>> operator%(vector<vector<int>> a, int in_mod)
{
for(auto &i:a)
for(auto &j:i)
j%=in_mod;
return a;
}
vector<vector<int>> mult_identity(vector<vector<int>> a)
{
int n=a.size();
vector<vector<int>> output(n, vector<int> (n));
for(int i=0;i<n;i++)
output[i][i]=1;
return output;
}
vector<int> mat_vector_product(vector<vector<int>> a, vector<int> b, int in_mod)
{
int n =a.size();
vector<int> output(n);
for(int i=0;i<n;i++)
{
for(int j=0;j<n;j++)
{
output[i]+=a[i][j]*b[j];
output[i]%=in_mod;
}
}
return output;
}
auto power(auto a, auto b, const int in_mod)
{
auto prod = mult_identity(a);
auto mult = a % in_mod;
while (b != 0)
{
if (b % 2)
{
prod = multiply(prod, mult, in_mod);
}
if(b/2)
mult = multiply(mult, mult, in_mod);
b /= 2;
}
return prod;
}
auto mod_inv(auto q, const int in_mod)
{
return power(q, in_mod - 2, in_mod);
}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
#define stp cout << fixed << setprecision(20);
int digit_sum(int x){
int sm = 0;
while(x){
sm += x%10;
x/= 10;
}
return sm;
}
void solv(){
int n;
cin>>n;
int cnt = 0;
for(int sm = 1;sm<= 200;sm++){
if(n %sm == 0){
if( digit_sum(n/sm) == sm){
cnt++;
}
}
}
cout<<cnt<<endl;
}
void solve()
{
int t = 1;
cin>>t;
for(int T=1;T<=t;T++)
{
// cout<<"Case #"<<T<<": ";
solv();
}
}
signed main()
{
// freopen("input.txt", "r", stdin);
// freopen("output.txt", "w", stdout);
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cerr.tie(NULL);
auto clk = clock();
// -------------------------------------Code starts here---------------------------------------------------------------------
signed t = 1;
// cin >> t;
for (signed test = 1; test <= t; test++)
{
// cout<<"Case #"<<test<<": ";
solve();
}
// -------------------------------------Code ends here------------------------------------------------------------------
clk = clock() - clk;
#ifndef ONLINE_JUDGE
cerr << fixed << setprecision(6) << "\nTime: " << ((float)clk) / CLOCKS_PER_SEC << "\n";
#endif
return 0;
}
/*
000100
1000100
1 0 -1 -2 -1 -2 -3
*/
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a binary sorted non-increasing array arr of size <b>N</b>. You need to print the count of <b>1's</b> in the binary array.
Try to solve the problem using binary searchThe input line contains T, denotes the number of testcases.
Each test case contains two lines. The first line contains N (size of binary array). The second line contains N elements of binary array separated by space.
<b>Constraints:</b>
1 <= T <= 100
1 <= N <= 10^6
arr[i] = 0,1
<b>Sum of N over all testcases does not exceed 10^6</b>For each testcase in new line, print the count 1's in binary array.Input:
2
8
1 1 1 1 1 0 0 0
8
1 1 0 0 0 0 0 0
Output:
5
2
Explanation:
Testcase 1: Number of 1's in given binary array : 1 1 1 1 1 0 0 0 is 5.
Testcase 2: Number of 1's in given binary array : 1 1 0 0 0 0 0 0 is 2., 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')
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();
}
}
public static void main (String[] args) throws IOException {
Reader sc=new Reader();
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 count=search(a,0,n-1);
System.out.println(count);
}
}
public static int search(int[] a,int l,int h){
while(l<=h){
int mid=l+(h-l)/2;
if ((mid==h||a[mid+1]==0)&&(a[mid]==1))
return mid+1;
if (a[mid]==1)
l=mid+1;
else h=mid-1;
}
return 0;
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a binary sorted non-increasing array arr of size <b>N</b>. You need to print the count of <b>1's</b> in the binary array.
Try to solve the problem using binary searchThe input line contains T, denotes the number of testcases.
Each test case contains two lines. The first line contains N (size of binary array). The second line contains N elements of binary array separated by space.
<b>Constraints:</b>
1 <= T <= 100
1 <= N <= 10^6
arr[i] = 0,1
<b>Sum of N over all testcases does not exceed 10^6</b>For each testcase in new line, print the count 1's in binary array.Input:
2
8
1 1 1 1 1 0 0 0
8
1 1 0 0 0 0 0 0
Output:
5
2
Explanation:
Testcase 1: Number of 1's in given binary array : 1 1 1 1 1 0 0 0 is 5.
Testcase 2: Number of 1's in given binary array : 1 1 0 0 0 0 0 0 is 2., I have written this Solution Code: #include "bits/stdc++.h"
#pragma GCC optimize "03"
using namespace std;
#define int long long int
#define ld long double
#define pi pair<int, int>
#define pb push_back
#define fi first
#define se second
#define IOS ios::sync_with_stdio(false); cin.tie(0); cout.tie(0)
#ifndef LOCAL
#define endl '\n'
#endif
const int N = 1e6 + 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];
int l = 0, h = n+1;
while(l+1 < h){
int m = (l + h) >> 1;
if(a[m] == 1)
l = m;
else
h = m;
}
cout << l << endl;
}
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a binary sorted non-increasing array arr of size <b>N</b>. You need to print the count of <b>1's</b> in the binary array.
Try to solve the problem using binary searchThe input line contains T, denotes the number of testcases.
Each test case contains two lines. The first line contains N (size of binary array). The second line contains N elements of binary array separated by space.
<b>Constraints:</b>
1 <= T <= 100
1 <= N <= 10^6
arr[i] = 0,1
<b>Sum of N over all testcases does not exceed 10^6</b>For each testcase in new line, print the count 1's in binary array.Input:
2
8
1 1 1 1 1 0 0 0
8
1 1 0 0 0 0 0 0
Output:
5
2
Explanation:
Testcase 1: Number of 1's in given binary array : 1 1 1 1 1 0 0 0 is 5.
Testcase 2: Number of 1's in given binary array : 1 1 0 0 0 0 0 0 is 2., I have written this Solution Code: c=int(input())
for x in range(c):
size=int(input())
s=input()
print(s.count('1')), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Alice's power is currently known to be an integer X. It is also known that her power doubles every second. For example, if Alice's power was currently 20, then after 2 seconds her power would have grown to 80.
Your task is to find out Alice's power after N seconds.The input consists of a single line containing two space-separated integers X and N.
<b>Constraints:</b>
1 ≤ X ≤ 1000
1 ≤ N ≤ 10Print a single integer – the power of Alice after N seconds.Sample Input 1:
5 2
Sample Output 1:
20
Sample Explanation 1:
Alice's power after 1 second will be 5*2 = 10. After 2 seconds it will be 10*2 = 20.
Sample Input 2:
4 3
Sample Output 2:
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 s = br.readLine();
if(s==null){
System.exit(0);
}
StringTokenizer st = new StringTokenizer(s, " ");
int power = Integer.parseInt(st.nextToken());
int multiple = Integer.parseInt(st.nextToken());
int res = power;
for(int i = 1;i<=multiple;i++){
res = res*2;
}
System.out.println(res);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Alice's power is currently known to be an integer X. It is also known that her power doubles every second. For example, if Alice's power was currently 20, then after 2 seconds her power would have grown to 80.
Your task is to find out Alice's power after N seconds.The input consists of a single line containing two space-separated integers X and N.
<b>Constraints:</b>
1 ≤ X ≤ 1000
1 ≤ N ≤ 10Print a single integer – the power of Alice after N seconds.Sample Input 1:
5 2
Sample Output 1:
20
Sample Explanation 1:
Alice's power after 1 second will be 5*2 = 10. After 2 seconds it will be 10*2 = 20.
Sample Input 2:
4 3
Sample Output 2:
32, I have written this Solution Code: #include <iostream>
using namespace std;
int main()
{
int x, n;
cin >> x >> n;
cout << x*(1 << n);
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Alice's power is currently known to be an integer X. It is also known that her power doubles every second. For example, if Alice's power was currently 20, then after 2 seconds her power would have grown to 80.
Your task is to find out Alice's power after N seconds.The input consists of a single line containing two space-separated integers X and N.
<b>Constraints:</b>
1 ≤ X ≤ 1000
1 ≤ N ≤ 10Print a single integer – the power of Alice after N seconds.Sample Input 1:
5 2
Sample Output 1:
20
Sample Explanation 1:
Alice's power after 1 second will be 5*2 = 10. After 2 seconds it will be 10*2 = 20.
Sample Input 2:
4 3
Sample Output 2:
32, I have written this Solution Code: x,n = map(int,input().split())
print(x*(2**n)), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given two arrays - value and frequency both containing N elements.
There is also a third array C which is currently empty. Then you perform N insertion operation in the array. For ith operation you insert value[i] to the end of the array frequency[i] number of times.
Finally you have to tell the kth smallest element in the array C.First line of input contains N.
Second line contains N integers denoting array - value
Third line contains N integers denoting array - frequency
Fourth line contains single integer K.
Constraints
1 <= N, value[i], frequency[i] <= 100000
1 <= k <= frequency[1] + frequency[2] +frequency[3] +........ + frequency[N]
Output a single integer which is the kth smallest element of the array C.Sample input 1
5
1 2 3 4 5
1 1 1 2 2
3
Sample output 1
3
Explanation 1:
Array C constructed is 1 2 3 4 4 5 5
Third smallest element is 3
Sample input 2
3
2 1 3
3 3 2
2
sample output 2
1
Explanation 2:
Array C constructed is 2 2 2 1 1 1 3 3
Second smallest element is 1, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args)throws Exception {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int n = Integer.parseInt(br.readLine());
StringTokenizer st = new StringTokenizer(br.readLine());
int[] val = new int[n];
for(int i=0; i<n; i++){
val[i] = Integer.parseInt(st.nextToken());
}
st = new StringTokenizer(br.readLine());
int[] freq = new int[n];
for(int i=0; i<n; i++){
freq[i] = Integer.parseInt(st.nextToken());
}
int k = Integer.parseInt(br.readLine());
for (int i=0; i<n; i++) {
for (int j=i+1; j<n; j++) {
if (val[j] < val[i]) {
int temp = val[i];
val[i] = val[j];
val[j] = temp;
int temp1 = freq[i];
freq[i] = freq[j];
freq[j] = temp1;
}
}
}
int element=0;
for(int i=0; i<n; i++){
for(int j=0; j<freq[i]; j++){
element++;
int value = val[i];
if(element==k){
System.out.print(value);
break;
}
}
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given two arrays - value and frequency both containing N elements.
There is also a third array C which is currently empty. Then you perform N insertion operation in the array. For ith operation you insert value[i] to the end of the array frequency[i] number of times.
Finally you have to tell the kth smallest element in the array C.First line of input contains N.
Second line contains N integers denoting array - value
Third line contains N integers denoting array - frequency
Fourth line contains single integer K.
Constraints
1 <= N, value[i], frequency[i] <= 100000
1 <= k <= frequency[1] + frequency[2] +frequency[3] +........ + frequency[N]
Output a single integer which is the kth smallest element of the array C.Sample input 1
5
1 2 3 4 5
1 1 1 2 2
3
Sample output 1
3
Explanation 1:
Array C constructed is 1 2 3 4 4 5 5
Third smallest element is 3
Sample input 2
3
2 1 3
3 3 2
2
sample output 2
1
Explanation 2:
Array C constructed is 2 2 2 1 1 1 3 3
Second smallest element is 1, I have written this Solution Code: def myFun():
n = int(input())
arr1 = list(map(int,input().strip().split()))
arr2 = list(map(int,input().strip().split()))
k = int(input())
arr = []
for i in range(n):
arr.append((arr1[i], arr2[i]))
arr.sort()
c = 0
for i in arr:
k -= i[1]
if k <= 0:
print(i[0])
return
myFun()
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given two arrays - value and frequency both containing N elements.
There is also a third array C which is currently empty. Then you perform N insertion operation in the array. For ith operation you insert value[i] to the end of the array frequency[i] number of times.
Finally you have to tell the kth smallest element in the array C.First line of input contains N.
Second line contains N integers denoting array - value
Third line contains N integers denoting array - frequency
Fourth line contains single integer K.
Constraints
1 <= N, value[i], frequency[i] <= 100000
1 <= k <= frequency[1] + frequency[2] +frequency[3] +........ + frequency[N]
Output a single integer which is the kth smallest element of the array C.Sample input 1
5
1 2 3 4 5
1 1 1 2 2
3
Sample output 1
3
Explanation 1:
Array C constructed is 1 2 3 4 4 5 5
Third smallest element is 3
Sample input 2
3
2 1 3
3 3 2
2
sample output 2
1
Explanation 2:
Array C constructed is 2 2 2 1 1 1 3 3
Second smallest element is 1, I have written this Solution Code: #include<bits/stdc++.h>
using namespace std;
#define ll long long
#define VV vector
#define pb push_back
#define bitc __builtin_popcountll
#define m_p make_pair
#define inf 1e18+1
#define eps 0.000000000001
#define fastio ios_base::sync_with_stdio(false);cin.tie(NULL);
string char_to_str(char c){string tem(1,c);return tem;}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
template<class T>//usage rand<long long>()
T rand() {
return uniform_int_distribution<T>()(rng);
}
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
typedef tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> oset;
// string to integer stoi()
// string to long long stoll()
// string.substr(position,length);
// integer to string to_string();
//////////////
auto clk=clock();
#define all(x) x.begin(),x.end()
#define S second
#define F first
#define sz(x) ((long long)x.size())
#define int long long
#define f80 __float128
#define pii pair<int,int>
/////////////
signed main()
{
fastio;
#ifdef ANIKET_GOYAL
freopen("inputf.in","r",stdin);
freopen("outputf.in","w",stdout);
#endif
int N;ll K;
cin>>N;
int c=0;
pair<int, ll> A[N];
for(int i=0;i<N;++i){
cin >> A[i].first ;
}
for(int i=0;i<N;++i){
cin >> A[i].second ;
}
cin>>K;
sort(A, A+N);
for(int i=0;i<N;++i){
K -= A[i].second;
if(K <= 0){
cout << A[i].first << endl;;
break;
}
}
#ifdef ANIKET_GOYAL
// cout<<endl<<endl<<endl<<endl<<"Time elapsed: "<<(double)(clock()-clk)/CLOCKS_PER_SEC<<endl;
#endif
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Let's define P[i] as the ith Prime Number. Therefore, P[1]=2, P[2]=3, P[3]=5, so on.
Given two integers L, R (L<=R), find the value of P[L]+P[L+1]+P[L+2]...+P[R].The first line of the input contains an integer T denoting the number of test cases.
The next T lines contain two integers L and R.
Constraints
1 <= T <= 50000
1 <= L <= R <= 50000For each test case, print one line corresponding to the required valueSample Input
4
1 3
2 4
5 5
1 5
Sample Output
10
15
11
28, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args)throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int t= Integer.parseInt(br.readLine());
int max = 1000001;
boolean isNotPrime[] = new boolean[max];
ArrayList<Integer> arr = new ArrayList<Integer>();
isNotPrime[0] = true; isNotPrime[1] = true;
for (int i=2; i*i <max; i++) {
if (!isNotPrime[i]) {
for (int j=i*i; j<max; j+= i) {
isNotPrime[j] = true;
}
}
}
for(int i=2; i<max; i++) {
if(!isNotPrime[i]) {
arr.add(i);
}
}
while(t-- > 0) {
String str[] = br.readLine().trim().split(" ");
int l = Integer.parseInt(str[0]);
int r = Integer.parseInt(str[1]);
System.out.println(primeRangeSum(l,r,arr));
}
}
static long primeRangeSum(int l , int r, ArrayList<Integer> arr) {
long sum = 0;
for(int i=l; i<=r;i++) {
sum += arr.get(i-1);
}
return sum;
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Let's define P[i] as the ith Prime Number. Therefore, P[1]=2, P[2]=3, P[3]=5, so on.
Given two integers L, R (L<=R), find the value of P[L]+P[L+1]+P[L+2]...+P[R].The first line of the input contains an integer T denoting the number of test cases.
The next T lines contain two integers L and R.
Constraints
1 <= T <= 50000
1 <= L <= R <= 50000For each test case, print one line corresponding to the required valueSample Input
4
1 3
2 4
5 5
1 5
Sample Output
10
15
11
28, I have written this Solution Code: def SieveOfEratosthenes(n):
prime = [True for i in range(n+1)]
pri = []
p = 2
while (p * p <= n):
if (prime[p] == True):
for i in range(p * p, n+1, p):
prime[i] = False
p += 1
for p in range(2, n+1):
if prime[p]:
pri.append(p)
return pri
N = int(input())
X = []
prim = SieveOfEratosthenes(1000000)
for i in range(1,len(prim)):
prim[i] = prim[i]+prim[i-1]
for i in range(N):
nnn = input()
X.append((int(nnn.split()[0]),int(nnn.split()[1])))
for xx,yy in X:
if xx==1:
print(prim[yy-1])
else:
print(prim[yy-1]-prim[xx-2])
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Let's define P[i] as the ith Prime Number. Therefore, P[1]=2, P[2]=3, P[3]=5, so on.
Given two integers L, R (L<=R), find the value of P[L]+P[L+1]+P[L+2]...+P[R].The first line of the input contains an integer T denoting the number of test cases.
The next T lines contain two integers L and R.
Constraints
1 <= T <= 50000
1 <= L <= R <= 50000For each test case, print one line corresponding to the required valueSample Input
4
1 3
2 4
5 5
1 5
Sample Output
10
15
11
28, 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 = 1e6 + 5;
const int mod = 1e9 + 7;
const int inf = 1e9 + 9;
int a[N];
signed main() {
IOS;
vector<int> v;
v.push_back(0);
for(int i = 2; i < N; i++){
if(a[i]) continue;
v.push_back(i);
for(int j = i*i; j < N; j += i)
a[j] = 1;
}
int p = 0;
for(auto &i: v){
i += p;
p = i;
}
int t; cin >> t;
while(t--){
int l, r;
cin >> l >> r;
cout << v[r] - v[l-1] << endl;
}
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer n, For each i from 1 to N if i is even print "<b>even</b>" else print "<b>odd</b>".<b>User task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the functions <b>For_Loop()</b> that takes the integer n as parameter.
</b>Constraints:</b>
1 <= n <= 100
<b>Note:</b>
<i>But there is a catch here given user function has already code in it which may or may not be correct, now you need to figure out these and correct if it is required</i>Print even or odd for each i separated by white spaces.Sample Input:
5
Sample Output:
odd even odd even odd
Sample Input:
2
Sample Output:
odd even, I have written this Solution Code: public static void For_Loop(int n){
for(int i=0;i<n;i++){
if(i%2==1){System.out.print("even ");}
else{
System.out.print("odd ");
}
}
}, 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. <b>Value</b> of an element A<sub>i</sub> is defined as the sum of the absolute difference of all elements of the array with A<sub>i</sub>. More formally, the value of an element at index i is the sum of |A<sub>i</sub> - A<sub>j</sub>| over all j (1 <= j <= N). Find the maximum such value over all i (1 <= i <= N) in the array.
<b>Note</b>: Given array is 1-based indexThe first line of the input contains a single integer N.
The second line of the input contains N space-separated integers.
<b>Constraints:</b>
1 ≤ N ≤ 10<sup>5</sup>
1 ≤ A<sub>i</sub> ≤ 10<sup>9</sup>Print the maximum such value over all i (1 <= i <= N) in the array.Sample Input:
6
1 1 5 5 8 9
Sample Output:
25
<b>Explanation:</b>
For, i = 6,
|A<sub>1</sub> - A<sub>6</sub>| = 8
|A<sub>2</sub> - A<sub>6</sub>| = 8
|A<sub>3</sub> - A<sub>6</sub>| = 4
|A<sub>4</sub> - A<sub>6</sub>| = 4
|A<sub>5</sub> - A<sub>6</sub>| = 1
Value = 8 + 8 + 4 + 4 + 1 = 25
For all other i, give values less than 25., I have written this Solution Code: #include<bits/stdc++.h>
using namespace std;
#define fast ios_base::sync_with_stdio(false); cin.tie(NULL);
#define int long long
#define pb push_back
#define ff first
#define ss second
#define endl '\n'
#define all(a) a.begin(), a.end()
#define rall(a) a.rbegin(), a.rend()
using T = pair<int, int>;
typedef long double ld;
const int mod = 1e9 + 7;
const int INF = 2e9;
void solve(){
int n;
cin >> n;
vector<int> a(n + 1);
for(int i = 1; i <= n; i++) cin >> a[i];
sort(a.begin(), a.end());
vector<int> pre(n + 1);
for(int i = 1; i <= n; i++){
pre[i] = pre[i - 1] + a[i];
}
int ans = 0;
for(int i = 1; i <= n; i++){
ans = max(ans, (pre[n] - pre[i]) - ((n - i) * (a[i])) + ((i-1)* a[i]) - pre[i - 1]);
}
cout << ans;
}
signed main(){
fast
int t = 1;
// cin >> t;
for(int i = 1; i <= t; i++){
solve();
if(i != t) cout << endl;
}
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a matrix of size M*N, your task is to find the column having the maximum sum and print the sum.The first line of input contains two space-separated integers M and N. The following M lines of input contain N space-separated integers each depicting the values of the matrix.
Constraints:-
1 <= M, N <= 100
1 <= Matrix[][] <= 100000Print the maximum sum.Sample Input:-
3 3
1 2 3
4 5 6
7 8 9
Sample Output:-
18
Explanation:-
1 + 4 + 7 = 12
2 + 5 + 8 = 15
3 + 6 + 9 = 18
maximum = 18
Sample Input:-
3 2
1 4
9 6
9 1
Sample Output:-
19, I have written this Solution Code: m,n=map(int ,input().split())
matrix=[]
for i in range(m):
l1=[eval(x) for x in input().split()]
matrix.append(l1)
l2=[]
for coloumn in range(n):
sum1=0
for row in range(m):
sum1+= matrix[row][coloumn]
l2.append(sum1)
print(max(l2))
'''for row in range(n):
sum2=0
for col in range(m):
sum2 += matrix[row][col]
print(sum2)''', In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a matrix of size M*N, your task is to find the column having the maximum sum and print the sum.The first line of input contains two space-separated integers M and N. The following M lines of input contain N space-separated integers each depicting the values of the matrix.
Constraints:-
1 <= M, N <= 100
1 <= Matrix[][] <= 100000Print the maximum sum.Sample Input:-
3 3
1 2 3
4 5 6
7 8 9
Sample Output:-
18
Explanation:-
1 + 4 + 7 = 12
2 + 5 + 8 = 15
3 + 6 + 9 = 18
maximum = 18
Sample Input:-
3 2
1 4
9 6
9 1
Sample Output:-
19, I have written this Solution Code: // mat is the matrix/ 2d array
// the dimensions of array are:- a rows, b columns
function colMaxSum(mat,a,b) {
// write code here
// do not console.log
// return the answer as a number
let idx = -1;
// Variable to store max sum
let maxSum = Number.MIN_VALUE;
// Traverse matrix column wise
for (let i = 0; i < b; i++) {
let sum = 0;
// calculate sum of column
for (let j = 0; j < a; j++) {
sum += mat[j][i];
}
// Update maxSum if it is
// less than current sum
if (sum > maxSum) {
maxSum = sum;
// store index
idx = i;
}
}
let res;
res = [idx, maxSum];
// return result
return maxSum;
}
, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a matrix of size M*N, your task is to find the column having the maximum sum and print the sum.The first line of input contains two space-separated integers M and N. The following M lines of input contain N space-separated integers each depicting the values of the matrix.
Constraints:-
1 <= M, N <= 100
1 <= Matrix[][] <= 100000Print the maximum sum.Sample Input:-
3 3
1 2 3
4 5 6
7 8 9
Sample Output:-
18
Explanation:-
1 + 4 + 7 = 12
2 + 5 + 8 = 15
3 + 6 + 9 = 18
maximum = 18
Sample Input:-
3 2
1 4
9 6
9 1
Sample Output:-
19, 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 1001
#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,m;
cin>>n>>m;
int a[m];
for(int i=0;i<m;i++){
a[i]=0;
}
int x;
int sum=0;
FOR(i,n){
FOR(j,m){
cin>>x;
a[j]+=x;
sum=max(sum,a[j]);
}
}
out(sum);
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a matrix of size M*N, your task is to find the column having the maximum sum and print the sum.The first line of input contains two space-separated integers M and N. The following M lines of input contain N space-separated integers each depicting the values of the matrix.
Constraints:-
1 <= M, N <= 100
1 <= Matrix[][] <= 100000Print the maximum sum.Sample Input:-
3 3
1 2 3
4 5 6
7 8 9
Sample Output:-
18
Explanation:-
1 + 4 + 7 = 12
2 + 5 + 8 = 15
3 + 6 + 9 = 18
maximum = 18
Sample Input:-
3 2
1 4
9 6
9 1
Sample Output:-
19, 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 m = sc.nextInt();
int n = sc.nextInt();
int a[][] = new int[m][n];
for(int i=0;i<m;i++){
for(int j=0;j<n;j++){
a[i][j]=sc.nextInt();
}
}
int sum=0;
int ans=0;
for(int i=0;i<n;i++){
sum=0;
for(int j=0;j<m;j++){
sum+=a[j][i];
}
if(sum>ans){ans=sum;}
}
System.out.print(ans);
}
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a string S you have to remove all pairs of identical letters, which are consecutive. If after the removal there appear new pairs, remove them as well. Repeat this untill no adjacent letter in the string is same.
Note that deleting of the consecutive identical letters can be done in any order, as any order leads to the same result.The input data consists of a single string S.
Constraints:
1 <= |S| <= 100000
S contains lowercase english letters only.Print the given string after it is processed. It is guaranteed that the result will contain at least one character.Sample Input
hhoowaaaareyyoouu
Sample Output
wre
Explanation:
First we remove "hh" then "oo" then "aa" then "yy" then "oo" then "uu" and we are left with "wre".
Now we cannot remove anything.
Sample Input:-
abcde
Sample Output:-
abcde
Sample Input:-
abcddcb
Sample Output:-
a, 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));
StringBuilder s = new StringBuilder();
String text=null;
while ((text = in.readLine ()) != null)
{
s.append(text);
}
int len=s.length();
for(int i=0;i<len-1;i++){
if(s.charAt(i)==s.charAt(i+1)){
int flag=0;
s.delete(i,i+2);
int left=i-1;
len=len-2;
i=i-2;
if(i<0){
i=-1;
}
}
}
System.out.println(s);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a string S you have to remove all pairs of identical letters, which are consecutive. If after the removal there appear new pairs, remove them as well. Repeat this untill no adjacent letter in the string is same.
Note that deleting of the consecutive identical letters can be done in any order, as any order leads to the same result.The input data consists of a single string S.
Constraints:
1 <= |S| <= 100000
S contains lowercase english letters only.Print the given string after it is processed. It is guaranteed that the result will contain at least one character.Sample Input
hhoowaaaareyyoouu
Sample Output
wre
Explanation:
First we remove "hh" then "oo" then "aa" then "yy" then "oo" then "uu" and we are left with "wre".
Now we cannot remove anything.
Sample Input:-
abcde
Sample Output:-
abcde
Sample Input:-
abcddcb
Sample Output:-
a, I have written this Solution Code: s=input()
l=["aa","bb","cc","dd","ee","ff","gg","hh","ii","jj","kk","ll","mm","nn","oo","pp","qq","rr","ss","tt","uu","vv","ww","xx","yy","zz"]
while True:
do=False
for i in range(len(l)):
if l[i] in s:
do=True
while l[i] in s:
s=s.replace(l[i],"")
if do==False:
break
print(s), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a string S you have to remove all pairs of identical letters, which are consecutive. If after the removal there appear new pairs, remove them as well. Repeat this untill no adjacent letter in the string is same.
Note that deleting of the consecutive identical letters can be done in any order, as any order leads to the same result.The input data consists of a single string S.
Constraints:
1 <= |S| <= 100000
S contains lowercase english letters only.Print the given string after it is processed. It is guaranteed that the result will contain at least one character.Sample Input
hhoowaaaareyyoouu
Sample Output
wre
Explanation:
First we remove "hh" then "oo" then "aa" then "yy" then "oo" then "uu" and we are left with "wre".
Now we cannot remove anything.
Sample Input:-
abcde
Sample Output:-
abcde
Sample Input:-
abcddcb
Sample Output:-
a, 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
string s;
cin>>s;
int len=s.length();
char stk[410000];
int k = 0;
for (int i = 0; i < len; i++)
{
stk[k++] = s[i];
while (k > 1 && stk[k - 1] == stk[k - 2])
k -= 2;
}
for (int i = 0; i < k; i++)
cout << stk[i];
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Find factorial of a given number where n! = n * n-1 * n-2 .....* 1First line consists of a single integer denoting n
Constraints:-
0 <= n <= 20Output is a single line containing factorial(n)Sample Input
5
Sample Output
120
Explanation:-
5!= 5 * 4 * 3 * 2 * 1 = 120
Sample Input
10
Sample Output
3628800, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args)throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int n=Integer.parseInt(br.readLine());
long fact=1;
for(int i=1; i<=n;i++){
fact*=i;
}
System.out.print(fact);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Find factorial of a given number where n! = n * n-1 * n-2 .....* 1First line consists of a single integer denoting n
Constraints:-
0 <= n <= 20Output is a single line containing factorial(n)Sample Input
5
Sample Output
120
Explanation:-
5!= 5 * 4 * 3 * 2 * 1 = 120
Sample Input
10
Sample Output
3628800, I have written this Solution Code: def fact(n):
if( n==0 or n==1):
return 1
return n*fact(n-1);
n=int(input())
print(fact(n)), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Find factorial of a given number where n! = n * n-1 * n-2 .....* 1First line consists of a single integer denoting n
Constraints:-
0 <= n <= 20Output is a single line containing factorial(n)Sample Input
5
Sample Output
120
Explanation:-
5!= 5 * 4 * 3 * 2 * 1 = 120
Sample Input
10
Sample Output
3628800, I have written this Solution Code: #include <bits/stdc++.h>
// #define ll long long
using namespace std;
int main(){
int t;
t=1;
while(t--){
int n;
cin>>n;
unsigned long long sum=1;
for(int i=1;i<=n;i++){
sum*=i;
}
cout<<sum<<endl;
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Chef has a garden of size N x N. The garden is divided into N<sup>2</sup> squares of size 1 x 1 each. Chef has plants in some unit squares of the garden, because of which, that part looks green. Formally, for the i<sup>th</sup> row (0 ≤ i < N), Chef has plants in all the columns from A<sub>i</sub>
to B<sub>i</sub> (both inclusive) where 0 ≤ A<sub>i</sub> ≤ B<sub>i</sub> < N. Help Chef find the length of the side of the largest square that is green.
The first line contains an integer N, the size of the garden. The next N lines contain two space separated integers A<sub>i</sub> and B<sub>i</sub>, representing that plants are present in all the columns from A<sub>i</sub> to B<sub>i</sub> (both inclusive) of the i<sup>th</sup> row.
<b>Constraints </b>
1 ≤ N ≤ 10<sup>6</sup>
0 ≤ A<sub>i</sub> ≤ B<sub>i</sub> < NOutput a single integer denoting the length of the side of the largest square that is green. In other words, output the length of the side of the square with maximum size, inside which all unit squares have plants.Sample Input :
3
1 1
0 2
1 2
Sample Output :
2, I have written this Solution Code: #include<bits/stdc++.h>
namespace {
using namespace std;
// https://github.com/atcoder/ac-library/
#define rep(i,n) for(int i = 0; i < (int)(n); i++)
#define rrep(i,n) for(int i = (int)(n) - 1; i >= 0; i--)
#define all(x) begin(x), end(x)
#define rall(x) rbegin(x), rend(x)
template<class T> bool chmax(T& a, const T& b) { if (a < b) { a = b; return true; } else return false; }
template<class T> bool chmin(T& a, const T& b) { if (b < a) { a = b; return true; } else return false; }
using ll = long long;
using P = pair<int,int>;
using VI = vector<int>;
using VVI = vector<VI>;
using VL = vector<ll>;
using VVL = vector<VL>;
template <class S, S (*op)(S, S)>
struct DisjointSparseTable {
const int n;
const vector<unsigned char> msb;
const vector<vector<S>> d;
DisjointSparseTable(vector<S> a) : n(a.size()), msb(build_msb_table(n)), d(build_table(move(a))) {}
vector<unsigned char> build_msb_table(int n) {
if (n <= 1) return {};
unsigned char k_max = 32 - __builtin_clz(n - 1);
vector<unsigned char> res(1 << k_max);
unsigned char* p = res.data();
for (unsigned char k = 0; k < k_max; k++) {
memset(p + (1U << k), k, 1U << k);
}
return res;
}
vector<vector<S>> build_table(vector<S> a) {
const int n = a.size();
vector<vector<S>> res(1);
if (n >= 2) {
const int i_max = n - 1, k_max = 32 - __builtin_clz(i_max);
res.resize(k_max);
for (int k = 1; k < k_max; k++) {
vector<S> t(i_max >> k & 1 ? n : i_max & ~0U << k);
for (int c = 1 << k; c < n; c += 1 << (k + 1)) {
int l = c - (1 << k);
int r = min(n, c + (1 << k));
t[c - 1] = a[c - 1];
for (int i = c - 2; i >= l; i--) t[i] = op(a[i], t[i + 1]);
t[c] = a[c];
for (int i = c + 1; i < r; i++) t[i] = op(t[i - 1], a[i]);
}
res[k] = move(t);
}
}
res[0] = move(a);
return res;
}
S query(int l, int r) { return query_closed(l, r - 1); }
S query_closed(int l, int r) {
assert(0 <= l && l <= r && r < n);
if (l == r) return d[0][l];
auto k = msb[l ^ r];
return op(d[k][l], d[k][r]);
}
};
int op_min(int x, int y) { return min(x, y); }
int op_max(int x, int y) { return max(x, y); }
} int main() {
ios::sync_with_stdio(false);
cin.tie(0);
int n;
cin >> n;
VI a(n), b(n);
rep(i, n) cin >> a[i] >> b[i];
DisjointSparseTable<int, op_max> da(move(a));
DisjointSparseTable<int, op_min> db(move(b));
int l = 1, r = n + 1;
while(r - l > 1) {
int c = (l + r) / 2;
bool ok = false;
rep(i, n - c + 1) {
int lb = da.query(i, i + c);
int ub = db.query(i, i + c);
if (ub - lb + 1 >= c) {
ok = true;
break;
}
}
(ok ? l : r) = c;
}
cout << l << '\n';
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer n holding some random value, your task is to assign value 10 to the given integer.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>Assignment_Operator()</b>, which takes no parameter.
<b>Constraints:</b>
1 <= n <= 100You don't need to print anything you just need to assign value 10 to the integer n.Sample Input:-
48
Sample output:-
10
Sample Input:-
24
Sample Output:-
10, I have written this Solution Code: static void Assignment_Operator(){
n=10;
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a Binary Search Tree and 2 node value <b>n1</b> and <b>n2</b>, your task is to find the <b>lowest common ancestor(LCA)</b> of the two nodes given.
It may happen that values n1 and n2 may or may be not present.
<b>Note:</b> Duplicates are not inserted in the BST.<b>User Task:</b>
This is a function problem. You don't have to take any input. You are required to complete the function LCA() that takes root node, n1, n2 as parameters and returns the node that is LCA of n1 and n2.
<b>Constraints:</b>
1 <= T <= 100
1 <= N <= 10^4
1 <= node values <= 10^4
<b>Sum of "N" over all testcases does not exceed 10^5</b>
For <b>Custom Input:</b>
First line of input should contains the number of test cases T. For each test case, there will be two lines of input.
First line will be a string representing the tree as described below:
The values in the string are in the order of level order traversal of the tree where, numbers denote node values, and a character “N” denotes NULL child.
Second line will contain the values of two nodes
<b>Note:</b> If a node has been declared Null using 'N', no information about its children will be given further in the array.
Second line will contain the values of two nodesFor each testcase, you need to return the node containing LCA of n1 and n2. The driver code will print the data. If any of the node values (n1, n2) is not present then return null driver code will print -1 for that.Sample Input
2
5 4 6 3 N N 7 N N N 8
7 8
2 1 3
1 3
Sample Output
7
2
Explanation:
Testcase1:
The BST in above test case will look like
5
/ \
4 6
/ \
3 7
\
8
Here the LCA of 7 and 8 is 7., I have written this Solution Code:
static Node LCA(Node node, int n1, int n2) {
if (node == null) {
return null;
}
// If both n1 and n2 are smaller than root, then LCA lies in left
if (node.data > n1 && node.data > n2) {
return LCA(node.left, n1, n2);
}
// If both n1 and n2 are greater than root, then LCA lies in right
if (node.data < n1 && node.data < n2) {
return LCA(node.right, n1, n2);
}
return node;
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Newton is forced to solve a puzzle given by his friend Ramanujan. In the puzzle Newton is given N non-negative integers T<sub>1</sub>, T<sub>2</sub>, ... T<sub>N</sub>, and another positive integer X. Now, Newton has to choose an integer Z between 0 and X, such that the function f(Z) = (Z xor T<sub>1</sub>) + (Z xor T<sub>2</sub>) + ... + (Z xor T<sub>N</sub>) returns maximum value. Help Newton calculate the maximum value that is possible under the given constraints.The first line contains 2 integers, N and X.
The second line contains N integers T<sub>1</sub>, T<sub>2</sub>, ... T<sub>N</sub>
<b>Constraints:</b>
1 <= N <= 2 x 10<sup>5</sup>
0 <= X <= 10<sup>12</sup>
0 <= T<sub>i</sub> <= 10<sup>12</sup>Find and print the maximum value.<b>Sample Input 1:</b>
4 10
1 2 3 4
<b>Sample Output 1:</b>
42
<b>Explanation:</b>
Maximum value can be achieved by choosing Z as 9.
(9 xor 1) + (9 xor 2) + (9 xor 3) + (9 xor 4) = 42
<b>Sample Input 2:</b>
4 10
11 9 6 98
<b>Sample Output 2:</b>
132, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define all(p) p.begin(),p.end()
#define rep(i,a) for(int i=0;i<a;i++)
using ll=long long;
template<class T> bool chmax(T &a,T const b){if(a<b){a=b;return true;}return false;}
int main() {
ll N,K;
cin>>N>>K;
K++;
int D=42;
vector<ll> p(D);
rep(i,N){
ll a;
cin>>a;
rep(j,D){
if((a&(1ll<<j))==0) p[j]++;
}
}
ll ans=-(1ll<<60),tmp=0;
for(int i=D-1;i>=0;i--){
if(K&(1ll<<i)){
ans+=(1ll<<i)*max(p[i],N-p[i]);
chmax(ans,tmp+(1ll<<i)*(N-p[i]));
tmp+=(1ll<<i)*p[i];
}else{
tmp+=(1ll<<i)*(N-p[i]);
ans+=(1ll<<i)*max(p[i],N-p[i]);
}
}
cout<<ans<<"\n";
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: We have N+M balls, each of which has an integer written on it.
It is known that:
1) The numbers written on the N of the balls are even.
2) The numbers written on the M of the balls are odd.
Find the number of ways to choose two of the N+M balls (disregarding order) so that the sum of the numbers written on them is even.The first and the only line of the input contains 2 space separated integers, N and M
<b>Constraints:</b>
1) 0 ≤ N, M ≤ 100
2) 2 ≤ N + MPrint the answer<b>Sample Input 1:</b>
2 1
<b>Sample Output 1:</b>
1
<b>Sample Input 2:</b>
4 3
<b>Sample Output 2:</b>
9
<b>Sample Input 3:</b>
13 3
<b>Sample Output 3:</b>
81, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
using ll = long long int;
using pii = pair<int,int>;
using vi = vector<int>;
using vll = vector<ll>;
using vvi = vector<vector<int>>;
using ss = string;
using db = double;
const int dx[4] = {1,0,-1,0};
const int dy[4] = {0,1,0,-1};
#define pi 3.14159265358979
#define rep(i,s,n) for(int i=(s);i<(int)(n);i++)
#define all(v) v.begin(),v.end()
#define ci(x) cin >> (x)
#define cii(x) int (x);cin >> (x)
#define cci(x,y) int (x),(y);cin >> (x) >>(y)
#define co(x) cout << (x) << endl
int main() {
cci(n, m);
co((n*(n-1))/2 + (m*(m-1))/2);
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Alice's power is currently known to be an integer X. It is also known that her power doubles every second. For example, if Alice's power was currently 20, then after 2 seconds her power would have grown to 80.
Your task is to find out Alice's power after N seconds.The input consists of a single line containing two space-separated integers X and N.
<b>Constraints:</b>
1 ≤ X ≤ 1000
1 ≤ N ≤ 10Print a single integer – the power of Alice after N seconds.Sample Input 1:
5 2
Sample Output 1:
20
Sample Explanation 1:
Alice's power after 1 second will be 5*2 = 10. After 2 seconds it will be 10*2 = 20.
Sample Input 2:
4 3
Sample Output 2:
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 s = br.readLine();
if(s==null){
System.exit(0);
}
StringTokenizer st = new StringTokenizer(s, " ");
int power = Integer.parseInt(st.nextToken());
int multiple = Integer.parseInt(st.nextToken());
int res = power;
for(int i = 1;i<=multiple;i++){
res = res*2;
}
System.out.println(res);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Alice's power is currently known to be an integer X. It is also known that her power doubles every second. For example, if Alice's power was currently 20, then after 2 seconds her power would have grown to 80.
Your task is to find out Alice's power after N seconds.The input consists of a single line containing two space-separated integers X and N.
<b>Constraints:</b>
1 ≤ X ≤ 1000
1 ≤ N ≤ 10Print a single integer – the power of Alice after N seconds.Sample Input 1:
5 2
Sample Output 1:
20
Sample Explanation 1:
Alice's power after 1 second will be 5*2 = 10. After 2 seconds it will be 10*2 = 20.
Sample Input 2:
4 3
Sample Output 2:
32, I have written this Solution Code: #include <iostream>
using namespace std;
int main()
{
int x, n;
cin >> x >> n;
cout << x*(1 << n);
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Alice's power is currently known to be an integer X. It is also known that her power doubles every second. For example, if Alice's power was currently 20, then after 2 seconds her power would have grown to 80.
Your task is to find out Alice's power after N seconds.The input consists of a single line containing two space-separated integers X and N.
<b>Constraints:</b>
1 ≤ X ≤ 1000
1 ≤ N ≤ 10Print a single integer – the power of Alice after N seconds.Sample Input 1:
5 2
Sample Output 1:
20
Sample Explanation 1:
Alice's power after 1 second will be 5*2 = 10. After 2 seconds it will be 10*2 = 20.
Sample Input 2:
4 3
Sample Output 2:
32, I have written this Solution Code: x,n = map(int,input().split())
print(x*(2**n)), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Naruto and Sasuke are on a S- Rank mission. Now They got bored and thought of competing in a race against each other in a horizontal plane. They decides a common spot C at which both of them will try to reach. Whoever reaches first wins the race, both of them run at the same speed.
Given initial positions of Naruto and Sasuke as A and B recpectively.
you need to tell which of them will win the race. If Naruto wins print "N" ( without the quotes ), if Sasuke wins print "S" ( without the quotes ).
if both of them reach the common spot at the same time, print "D" (for draw, without the quotes ).<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function
<b>Race</b> that takes the integer A (initial position of Naruto) and B (initial position of Sasuke) and C (position of common spot) as parameter.
Constraints
1 <= A, B, C <= 100Return the character according to the given scenario.Sample Input
1 2 3
Sample Output
S
Sample Input
1 3 2
Sample Output
D, I have written this Solution Code:
char Race(int A, int B, int C){
if(abs(C-A)==abs(C-B)){return 'D';}
if(abs(C-A)>abs(C-B)){return 'S';}
else{
return 'N';}
}
, In this Programming Language: C, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Naruto and Sasuke are on a S- Rank mission. Now They got bored and thought of competing in a race against each other in a horizontal plane. They decides a common spot C at which both of them will try to reach. Whoever reaches first wins the race, both of them run at the same speed.
Given initial positions of Naruto and Sasuke as A and B recpectively.
you need to tell which of them will win the race. If Naruto wins print "N" ( without the quotes ), if Sasuke wins print "S" ( without the quotes ).
if both of them reach the common spot at the same time, print "D" (for draw, without the quotes ).<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function
<b>Race</b> that takes the integer A (initial position of Naruto) and B (initial position of Sasuke) and C (position of common spot) as parameter.
Constraints
1 <= A, B, C <= 100Return the character according to the given scenario.Sample Input
1 2 3
Sample Output
S
Sample Input
1 3 2
Sample Output
D, I have written this Solution Code: def Race(A,B,C):
if abs(C-A) ==abs(C-B):
return 'D'
if abs(C-A)>abs(C-B):
return 'S'
return 'N'
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Naruto and Sasuke are on a S- Rank mission. Now They got bored and thought of competing in a race against each other in a horizontal plane. They decides a common spot C at which both of them will try to reach. Whoever reaches first wins the race, both of them run at the same speed.
Given initial positions of Naruto and Sasuke as A and B recpectively.
you need to tell which of them will win the race. If Naruto wins print "N" ( without the quotes ), if Sasuke wins print "S" ( without the quotes ).
if both of them reach the common spot at the same time, print "D" (for draw, without the quotes ).<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function
<b>Race</b> that takes the integer A (initial position of Naruto) and B (initial position of Sasuke) and C (position of common spot) as parameter.
Constraints
1 <= A, B, C <= 100Return the character according to the given scenario.Sample Input
1 2 3
Sample Output
S
Sample Input
1 3 2
Sample Output
D, I have written this Solution Code:
char Race(int A, int B, int C){
if(abs(C-A)==abs(C-B)){return 'D';}
if(abs(C-A)>abs(C-B)){return 'S';}
else{
return 'N';}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Naruto and Sasuke are on a S- Rank mission. Now They got bored and thought of competing in a race against each other in a horizontal plane. They decides a common spot C at which both of them will try to reach. Whoever reaches first wins the race, both of them run at the same speed.
Given initial positions of Naruto and Sasuke as A and B recpectively.
you need to tell which of them will win the race. If Naruto wins print "N" ( without the quotes ), if Sasuke wins print "S" ( without the quotes ).
if both of them reach the common spot at the same time, print "D" (for draw, without the quotes ).<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function
<b>Race</b> that takes the integer A (initial position of Naruto) and B (initial position of Sasuke) and C (position of common spot) as parameter.
Constraints
1 <= A, B, C <= 100Return the character according to the given scenario.Sample Input
1 2 3
Sample Output
S
Sample Input
1 3 2
Sample Output
D, I have written this Solution Code: static char Race(int A,int B,int C){
if(Math.abs(C-A)==Math.abs(C-B)){return 'D';}
if(Math.abs(C-A)>Math.abs(C-B)){return 'S';}
else{
return 'N';}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an unsorted integer array having distinct integers, find a pair with the given sum in it.
First element of pair > second element of pairFirst line of input contains a single integer N, next line contains N space separated integers depicting the values of array and third line consist target sum.
Constraints:
2<=N<=5*(10^5)
1<=A[i], target<=2*(10^9)
TargetPrint the pair of integers which sum is target.
Sample Input1:-
6
8 7 4 5 3 1
10
Sample Output:-
Pair found (7, 3)
Sample Input2:
6
5 2 6 8 1 9
12
Sample Output:
Pair not found, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main{
public static void swap(int[] arr,int i, int j){
int temp=arr[i];
arr[i]=arr[j];
arr[j]=temp;
}
public static int partition(int[] arr,int l,int r){
int pivot=arr[r];
int i=l-1;
for(int j=l;j<r;j++){
if(arr[j]>pivot){
i++;
swap(arr,i,j);
}
}
swap(arr,i+1,r);
return i+1;
}
public static void quickSort(int[] arr,int l,int r){
if(l<r){
int pivot=partition(arr,l,r);
quickSort(arr,l,pivot-1);
quickSort(arr,pivot+1,r);
}
}
public static void main (String[] args) {
Scanner sc=new Scanner(System.in);
int n=sc.nextInt();
int arr[]=new int[n];
for(int i=0;i<n;i++){
arr[i]=sc.nextInt();
}
int target=sc.nextInt();
quickSort(arr,0,n-1);
int i=0,j=n-1;
while(i<j){
if((arr[i]+arr[j])==target){
System.out.print("Pair found ("+arr[i]+", "+arr[j]+")");
return;
}
else if((arr[i]+arr[j])<target){
j--;
}
else{
i++;
}
}
System.out.print("Pair not found");
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an unsorted integer array having distinct integers, find a pair with the given sum in it.
First element of pair > second element of pairFirst line of input contains a single integer N, next line contains N space separated integers depicting the values of array and third line consist target sum.
Constraints:
2<=N<=5*(10^5)
1<=A[i], target<=2*(10^9)
TargetPrint the pair of integers which sum is target.
Sample Input1:-
6
8 7 4 5 3 1
10
Sample Output:-
Pair found (7, 3)
Sample Input2:
6
5 2 6 8 1 9
12
Sample Output:
Pair not found, I have written this Solution Code:
N=int(input())
arr=list(map(int,input().split()))
target=int(input())
arr.sort(reverse=True)
'''
ans='Pair not found'
for i in range(N):
for j in range(i+1,N):
if arr[i]+arr[j]==target:
ans='Pair found ({}, {})'.format(arr[i],arr[j])
print(ans)
'''
i=0
j=N-1
ans='Pair not found'
while i<j:
if arr[i]+arr[j]<target:
j-=1
elif arr[i]+arr[j]>target:
i+=1
j+=1
else:
ans='Pair found ({}, {})'.format(arr[i],arr[j])
break
print(ans), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an unsorted integer array having distinct integers, find a pair with the given sum in it.
First element of pair > second element of pairFirst line of input contains a single integer N, next line contains N space separated integers depicting the values of array and third line consist target sum.
Constraints:
2<=N<=5*(10^5)
1<=A[i], target<=2*(10^9)
TargetPrint the pair of integers which sum is target.
Sample Input1:-
6
8 7 4 5 3 1
10
Sample Output:-
Pair found (7, 3)
Sample Input2:
6
5 2 6 8 1 9
12
Sample Output:
Pair not found, I have written this Solution Code: #include <iostream>
#include <algorithm>
#include <bits/stdc++.h>
using namespace std;
// Function to find a pair in an array with a given sum using sorting
void findPair(vector<int> &nums, int n, int target)
{
// sort the array in ascending order
sort(nums.begin(), nums.end());
// maintain two indices pointing to endpoints of the array
int low = 0;
int high = n - 1;
// reduce the search space `nums[low…high]` at each iteration of the loop
// loop till the search space is exhausted
while (low < high)
{
// sum found
if (nums[low] + nums[high] == target)
{
if (nums[low] < nums[high])
swap(nums[low], nums[high]);
cout << "Pair found (" << nums[low] << ", " << nums[high] << ")\n";
return;
}
// increment `low` index if the total is less than the desired sum;
// decrement `high` index if the total is more than the desired sum
(nums[low] + nums[high] < target) ? low++ : high--;
}
// we reach here if the pair is not found
cout << "Pair not found";
}
int main()
{
int n;
cin >> n;
vector<int> nums(n);
for (int i = 0; i < n; i++)
cin >> nums[i];
int target;
cin >> target;
findPair(nums, n, target);
return 0;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara wants to be the best programmer in the world but today she is stuck on an easy problem. Help her to solve it.
<b>Problem description:- </b>
Choose an integer N and subtract the sum of digits of the number N from it i. e. if N is 245 then subtract 2+4+5 from it making it 245 - 11 = 234.
Keep on doing this operation until the number becomes 0 (for eg. 25 requires 3 operations to reduce to 0
25 - > 18 - > 9 - > 0).
Given a number N, your task is to print the number of operations required to make the number 0.The first line of input contains a single integer containing the number of test cases T. Next T lines contain a single integer N.
Constraints:-
1 <= T <= 10000
1 <= N <= 1000000For each test case print the number of operations required to make the number 0.Sample Input:-
4
25
8
17
842
Sample Output:-
3
1
2
72
Explanation:-
25 - > 18 - > 9 - > 0
8 - > 0
17 - > 9 - > 0, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
static int[] dp = new int[1000002];
public static void main (String[] args) throws IOException {
int x = 1000000;
int sum = 0;
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
for(int i = 1; i <= x; ++i) {
int j = i;
sum = total(j);
dp[i] = dp[i-sum]+1;
}
int t = Integer.parseInt(br.readLine());
while(t-- > 0) {
int count = 0;
int n = Integer.parseInt(br.readLine());
System.out.println(dp[n]);
}
}
static int total(int n) {
int temp = 0;
while(n > 0) {
temp += n%10;
n = n/10;
}
return temp;
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara wants to be the best programmer in the world but today she is stuck on an easy problem. Help her to solve it.
<b>Problem description:- </b>
Choose an integer N and subtract the sum of digits of the number N from it i. e. if N is 245 then subtract 2+4+5 from it making it 245 - 11 = 234.
Keep on doing this operation until the number becomes 0 (for eg. 25 requires 3 operations to reduce to 0
25 - > 18 - > 9 - > 0).
Given a number N, your task is to print the number of operations required to make the number 0.The first line of input contains a single integer containing the number of test cases T. Next T lines contain a single integer N.
Constraints:-
1 <= T <= 10000
1 <= N <= 1000000For each test case print the number of operations required to make the number 0.Sample Input:-
4
25
8
17
842
Sample Output:-
3
1
2
72
Explanation:-
25 - > 18 - > 9 - > 0
8 - > 0
17 - > 9 - > 0, I have written this Solution Code: import sys
sys.setrecursionlimit(40000)
dic = {}
def sumi(n):
x = 0
while(n>0):
x += n%10
n //= 10
return x
def Num_of_times(n):
if (n==0):
return 0
if dic.get(n):
count = dic[n]
return count
else:
x = sumi(n)
count = 1+Num_of_times(n-x)
dic[n] = count
return count
arr = []
maxi = 0
for i in range(int(input())):
x = int(input())
maxi = max(maxi,x)
arr.append(x)
maxi_ans = Num_of_times(maxi)
for i in arr:
if i == maxi:
print(maxi_ans)
else:
print(Num_of_times(i)), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara wants to be the best programmer in the world but today she is stuck on an easy problem. Help her to solve it.
<b>Problem description:- </b>
Choose an integer N and subtract the sum of digits of the number N from it i. e. if N is 245 then subtract 2+4+5 from it making it 245 - 11 = 234.
Keep on doing this operation until the number becomes 0 (for eg. 25 requires 3 operations to reduce to 0
25 - > 18 - > 9 - > 0).
Given a number N, your task is to print the number of operations required to make the number 0.The first line of input contains a single integer containing the number of test cases T. Next T lines contain a single integer N.
Constraints:-
1 <= T <= 10000
1 <= N <= 1000000For each test case print the number of operations required to make the number 0.Sample Input:-
4
25
8
17
842
Sample Output:-
3
1
2
72
Explanation:-
25 - > 18 - > 9 - > 0
8 - > 0
17 - > 9 - > 0, I have written this Solution Code: #include<bits/stdc++.h>
using namespace std;
#define LL long long
int dp[1000001];
void solve() {
int n; cin >> n;
cout << dp[n] << '\n';
}
int main() {
ios::sync_with_stdio(0), cin.tie(0);
for(int i = 1; i <= 1000000; i++) {
int s = 0, j = i;
while(j > 0) {
s += j % 10;
j /= 10;
}
dp[i] = dp[i - s] + 1;
}
int tt; cin >> tt;
while(tt--) solve();
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Rick has got a huge candy bag. The bag contains a total of O orange candies, A apple candies, and U candies with an unknown flavor (can be orange or apple as well).
Now Morty wants to know the maximum number of candies that can be drawn from the bag blindfolded such that no more than X orange and Y apple candies are drawn out.
Can you solve the problem for Morty?The first line of the input contains three integers O, A, and U, representing the number of orange, apple, and unknown flavored candies respectively.
The second line of the input contains two integers X and Y representing the maximum number of orange and apple candies that can be drawn of the bag.
Constraints
0 <= O, A, U <= 10<sup>9</sup>
0 <= X, Y <= 10<sup>9</sup>Output a single integer, the maximum number of candies that can be drawn out of Rick's bag.Sample Input 1
4 2 7
2 7
Sample Output 1
2
Explanation: If we draw out more than 2 candies, there is a possible chance for 3 candies to be orange.
Sample Input 2
1 2 3
8 4
Sample Output 2
4, I have written this Solution Code: [o,a,u]=[int(i) for i in input().split()]
s=o+a+u
[x,y]=[int(i) for i in input().split()]
o+=u
a+=u
if o<=x and a<=y:
print(s)
elif o<=x:
print(y)
elif a<=y:
print(x)
else:
print(min(x,y)), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Rick has got a huge candy bag. The bag contains a total of O orange candies, A apple candies, and U candies with an unknown flavor (can be orange or apple as well).
Now Morty wants to know the maximum number of candies that can be drawn from the bag blindfolded such that no more than X orange and Y apple candies are drawn out.
Can you solve the problem for Morty?The first line of the input contains three integers O, A, and U, representing the number of orange, apple, and unknown flavored candies respectively.
The second line of the input contains two integers X and Y representing the maximum number of orange and apple candies that can be drawn of the bag.
Constraints
0 <= O, A, U <= 10<sup>9</sup>
0 <= X, Y <= 10<sup>9</sup>Output a single integer, the maximum number of candies that can be drawn out of Rick's bag.Sample Input 1
4 2 7
2 7
Sample Output 1
2
Explanation: If we draw out more than 2 candies, there is a possible chance for 3 candies to be orange.
Sample Input 2
1 2 3
8 4
Sample Output 2
4, 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(){
long a=in.nl(),b=in.nl(),c= in.nl();
long x=in.nl(),y=in.nl();
long ans =0;
if(x>y){
long t = x;
x=y;
y=t;
t = a;
a=b;
b=t;
}
if(x-a-c <0){
ans = x;
}else if(y-b-c<0) {
ans = y;
}else {
ans = a+b+c;
}
out.printLn(ans);
}
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 has got a huge candy bag. The bag contains a total of O orange candies, A apple candies, and U candies with an unknown flavor (can be orange or apple as well).
Now Morty wants to know the maximum number of candies that can be drawn from the bag blindfolded such that no more than X orange and Y apple candies are drawn out.
Can you solve the problem for Morty?The first line of the input contains three integers O, A, and U, representing the number of orange, apple, and unknown flavored candies respectively.
The second line of the input contains two integers X and Y representing the maximum number of orange and apple candies that can be drawn of the bag.
Constraints
0 <= O, A, U <= 10<sup>9</sup>
0 <= X, Y <= 10<sup>9</sup>Output a single integer, the maximum number of candies that can be drawn out of Rick's bag.Sample Input 1
4 2 7
2 7
Sample Output 1
2
Explanation: If we draw out more than 2 candies, there is a possible chance for 3 candies to be orange.
Sample Input 2
1 2 3
8 4
Sample Output 2
4, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define sd(x) scanf("%d", &x)
#define sz(v) (int) v.size()
#define pr(v) For(i, 0, sz(v)) {cout<<v[i]<<" ";} cout<<endl;
#define slld(x) scanf("%lld", &x)
#define all(x) x.begin(), x.end()
#define For(i, st, en) for(ll i=st; i<en; i++)
#define tr(x) for(auto it=x.begin(); it!=x.end(); it++)
#define fast std::ios::sync_with_stdio(false);cin.tie(NULL);
#define pb push_back
#define ll long long
#define ld long double
#define int long long
#define double long double
#define mp make_pair
#define F first
#define S second
typedef pair<int, int> pii;
typedef vector<int> vi;
#define pi 3.141592653589793238
const int MOD = 1e9+7;
const int INF = 1LL<<60;
const int N = 2e5+5;
// it's swapnil07 ;)
#ifdef SWAPNIL07
#define trace(...) __f(#__VA_ARGS__, __VA_ARGS__)
template <typename Arg1>
void __f(const char* name, Arg1&& arg1){
cout << name << " : " << arg1 << endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args){
const char* comma = strchr(names + 1, ',');cout.write(names, comma - names) << " : " << arg1<<" | ";__f(comma+1, args...);
}
int begtime = clock();
#define end_routine() cout << "\n\nTime elapsed: " << (clock() - begtime)*1000/CLOCKS_PER_SEC << " ms\n\n";
#else
#define endl '\n'
#define trace(...)
#define end_routine()
#endif
void solve(){
int a, b, c;
cin >> a >> b >> c;
int n, m;
cin >> n >> m;
int ans = a+b+c;
if (n < a + c) ans = min(n, ans);
if (m < b + c) ans = min(m, ans);
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: Sky (the blue ranger) wants to replace Jack (the red ranger) from his position. For this, he needs to conquer the entire Base.
The Base can be represented as an M*N grid, and Sky needs to conquer it cell by cell. Sky starts conquering the Base from the cell (1, 1). In each move, he conquers the cell, and moves to an adjacent non- conquered cell (he cannot move if there is no adjacent non- conquered cell). Now, there is a catch, the last cell he needs to conquer is (M, N) so as to complete the quest for the red ranger tag!
Please let us know if Sky can replace Jack by conquering all the cells in the Base.
Note: The diagonal cells are not considered as adjacent cells.The first and the only line of input contains two integers M and N.
Constraints
1 <= M, N <= 1000Output "YES" (without quotes) if Sky can conquer the entire Base to replace Jack, else output "NO" (without quotes).Sample Input
2 2
Sample Output
NO
Explanation
The possible journeys of Sky ending at (2, 2) can be:
(1, 1) - > (1, 2) - > (2, 2)
(1, 1) - > (2, 1) - > (2, 2)
Since, in each of the path that Sky takes, the total cells covered are not 4, hence Sky cannot conquer the entire base.
Sample Input
3 3
Sample Output
YES, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws IOException{
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
String s[]=br.readLine().split(" ");
int m=Integer.parseInt(s[0]);
int n=Integer.parseInt(s[1]);
if(m%2==0 && n%2==0)
System.out.println("NO");
else
System.out.println("YES");
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sky (the blue ranger) wants to replace Jack (the red ranger) from his position. For this, he needs to conquer the entire Base.
The Base can be represented as an M*N grid, and Sky needs to conquer it cell by cell. Sky starts conquering the Base from the cell (1, 1). In each move, he conquers the cell, and moves to an adjacent non- conquered cell (he cannot move if there is no adjacent non- conquered cell). Now, there is a catch, the last cell he needs to conquer is (M, N) so as to complete the quest for the red ranger tag!
Please let us know if Sky can replace Jack by conquering all the cells in the Base.
Note: The diagonal cells are not considered as adjacent cells.The first and the only line of input contains two integers M and N.
Constraints
1 <= M, N <= 1000Output "YES" (without quotes) if Sky can conquer the entire Base to replace Jack, else output "NO" (without quotes).Sample Input
2 2
Sample Output
NO
Explanation
The possible journeys of Sky ending at (2, 2) can be:
(1, 1) - > (1, 2) - > (2, 2)
(1, 1) - > (2, 1) - > (2, 2)
Since, in each of the path that Sky takes, the total cells covered are not 4, hence Sky cannot conquer the entire base.
Sample Input
3 3
Sample Output
YES, I have written this Solution Code: m,n=map(int, input().split())
if(m%2 or n%2):
print("YES")
else:
print("NO"), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sky (the blue ranger) wants to replace Jack (the red ranger) from his position. For this, he needs to conquer the entire Base.
The Base can be represented as an M*N grid, and Sky needs to conquer it cell by cell. Sky starts conquering the Base from the cell (1, 1). In each move, he conquers the cell, and moves to an adjacent non- conquered cell (he cannot move if there is no adjacent non- conquered cell). Now, there is a catch, the last cell he needs to conquer is (M, N) so as to complete the quest for the red ranger tag!
Please let us know if Sky can replace Jack by conquering all the cells in the Base.
Note: The diagonal cells are not considered as adjacent cells.The first and the only line of input contains two integers M and N.
Constraints
1 <= M, N <= 1000Output "YES" (without quotes) if Sky can conquer the entire Base to replace Jack, else output "NO" (without quotes).Sample Input
2 2
Sample Output
NO
Explanation
The possible journeys of Sky ending at (2, 2) can be:
(1, 1) - > (1, 2) - > (2, 2)
(1, 1) - > (2, 1) - > (2, 2)
Since, in each of the path that Sky takes, the total cells covered are not 4, hence Sky cannot conquer the entire base.
Sample Input
3 3
Sample Output
YES, 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, m; cin>>n>>m;
if(n%2 || m%2){
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: N+1 towns are going to be attacked by monsters. The i-th town will be attacked by Ai monsters. Saloni has got N of her friends for rescue. The i-th friend has the capability to kill at most Bi monsters. The only constraint is that the i-th friend can kill monsters from town i and i+1 only.
Find the maximum number of monsters the friends can kill.The first line of input contains an integer N.
The next line contains N+1 integers Ai.
The next line contains N integers Bi.
Constraints
1 <= N <= 100000
1 <= Ai, Bi <= 1000000000Output a single integer, the maximum number of monsters that can be killed.Sample Input
2
3 5 2
4 5
Sample Output
9
Explanation
The first friend will kill 2 monsters from town 1 and 2 monsters from town 2.
The second friend will kill 3 monsters from town 2 and 2 monsters from town 3.
Sample Input:
2
100 1 1
1 100
Sample Output:
3, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args)throws IOException {
BufferedReader buf=new BufferedReader(new InputStreamReader(System.in));
int t=Integer.parseInt(buf.readLine());
String []inp=buf.readLine().split(" ");
String []inp1=buf.readLine().split(" ");
int []a=new int[inp.length];
int []b=new int[inp.length];
for(int i=0;i<inp.length;i++)a[i]=Integer.parseInt(inp[i]);
for(int i=0;i<inp1.length;i++)b[i]=Integer.parseInt(inp1[i]);
long ans=0;
int last=0;
for(int i=0;i<b.length;i++)
{
int r=a[i];
int s=b[i];
r-=Math.min(r,last);
int k=Math.min(r,s);
r-=k;
last=s-k;
ans+=a[i]-r;
}
ans+=Math.min(a[inp.length-1],last);
System.out.println(ans);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: N+1 towns are going to be attacked by monsters. The i-th town will be attacked by Ai monsters. Saloni has got N of her friends for rescue. The i-th friend has the capability to kill at most Bi monsters. The only constraint is that the i-th friend can kill monsters from town i and i+1 only.
Find the maximum number of monsters the friends can kill.The first line of input contains an integer N.
The next line contains N+1 integers Ai.
The next line contains N integers Bi.
Constraints
1 <= N <= 100000
1 <= Ai, Bi <= 1000000000Output a single integer, the maximum number of monsters that can be killed.Sample Input
2
3 5 2
4 5
Sample Output
9
Explanation
The first friend will kill 2 monsters from town 1 and 2 monsters from town 2.
The second friend will kill 3 monsters from town 2 and 2 monsters from town 3.
Sample Input:
2
100 1 1
1 100
Sample Output:
3, I have written this Solution Code: n = int(input())
a= list(map(int,input().strip().split()))
b= list(map(int,input().strip().split()))
b.insert(0,0)
ans=0
for i in range (1,n+1):
cnt=b[i]
if i>0:
mn = min(cnt,a[i-1])
cnt-=mn
ans+=mn
mn = min(cnt,a[i])
ans += mn;
a[i] -= mn;
print(ans), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: N+1 towns are going to be attacked by monsters. The i-th town will be attacked by Ai monsters. Saloni has got N of her friends for rescue. The i-th friend has the capability to kill at most Bi monsters. The only constraint is that the i-th friend can kill monsters from town i and i+1 only.
Find the maximum number of monsters the friends can kill.The first line of input contains an integer N.
The next line contains N+1 integers Ai.
The next line contains N integers Bi.
Constraints
1 <= N <= 100000
1 <= Ai, Bi <= 1000000000Output a single integer, the maximum number of monsters that can be killed.Sample Input
2
3 5 2
4 5
Sample Output
9
Explanation
The first friend will kill 2 monsters from town 1 and 2 monsters from town 2.
The second friend will kill 3 monsters from town 2 and 2 monsters from town 3.
Sample Input:
2
100 1 1
1 100
Sample Output:
3, I have written this Solution Code: #include <bits/stdc++.h> // header file includes every Standard library
using namespace std;
int main() {
int n;
cin>>n;
vector<int> v(n+1);
for(int i=0;i<=n;i++)cin>>v[i];
vector<int> vv(n+1);
for(int i=1;i<=n;i++)cin>>vv[i];
int ans=0;
for(int i=1;i<=n;i++){
int cnt=vv[i];
if(i>0){
int mn=min(cnt,v[i-1]);
cnt-=mn;
ans+=mn;
}
int mn=min(cnt,v[i]);
ans+=mn;
v[i]-=mn;
}
cout<<ans<<endl;
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Complete the function <code>solve</code> which takes <code>obj</code> as a parameter which is an <code>object</code>.
Iterate <code>obj</code> using a <code>for of</code> and a <code>for in</code> loop. The <code>for...of</code> loop should print all the <code>values</code> of the <code>obj</code> in the console and the <code>for...in</code> loop should print the <code>key</code> and <code>values</code> of <code>obj</code> in the format <code>{key}: {value}</code>.
See the example for more clarity.
Note: Generate Expected Output section will not work for this questionInput will have the object which is passed as a parameter to the function solve
You need not worry about the same, it is handled properly by the hidden pre-function code.
Example:
{"name": "John","age": "32","location": "New York"}The <code>for of</code> loop should print all the <code>values</code> of the <code>obj</code> object in the console
The <code>for in</code> loop should print the <code>key</code> and <code>values</code> of <code>obj</code> object in the format <code>{key}: {value}</code>const obj = {"name": "John","age": "32","location": "New York"}
solve(obj)
/*
The console output should be
John
32
New York
name: John
age: 32
location: New York
*/, I have written this Solution Code: function solve(obj){
for (const value of Object.values(obj)) {
console.log(value);
}
for (const key in obj) {
console.log(`${key}: ${obj[key]}`);
}
}, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Siddharth, a great entrepreneur, chooses a person as an intern if he/she good at maths. Thus Sid gives a simple maths problem to the candidate based on which he decides to who is gonna get the offer. Below is the description of the problem asked by Sid
Given arr <b>A[]</b> containing <b>N</b> positive integers. The task is to check whether it is possible to divide the given array into an <b>odd number of non-empty subarrays each of odd length such that each sub-array starts and ends with an odd number</b>. If it is possible to divide the array then print "<b>YES</b>" otherwise "<b>NO</b>".
<b>Note:</b> Don't forget that the complete array is also considered as a subarrayThe first line of each test case contains N size of array. Second-line contains N space-separated elements.
<b>Constraints:</b>
1 <= N <= 999999
0 <= A[i] <= 999999999Print "YES", if it is possible to divide the array. Otherwise, print "NO". Without quotes.Sample Input 1:
5
9 0 3 1 5
Sample Output 1:
YES
Sample Input 2:
3
1 2 0
Sample Output 2:
NO
Explanation:
Testcase 1: Array {9, 0, 3, 1, 5} can be divided as {9, 0, 3}, {1}, {5}., 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')
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();
}
}
public static void main (String[] args) throws IOException {
Reader sc=new Reader();
int n=sc.nextInt();
int[] a=new int[n];
int counter=0;
for(int i=0;i<n;i++){
a[i]=sc.nextInt();
if(a[i]%2!=0) counter++;
}
String ans="NO";
if((a[0]%2!=0)&&(a[n-1]%2!=0)&&(n%2!=0))
ans="YES";
System.out.println(ans);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Siddharth, a great entrepreneur, chooses a person as an intern if he/she good at maths. Thus Sid gives a simple maths problem to the candidate based on which he decides to who is gonna get the offer. Below is the description of the problem asked by Sid
Given arr <b>A[]</b> containing <b>N</b> positive integers. The task is to check whether it is possible to divide the given array into an <b>odd number of non-empty subarrays each of odd length such that each sub-array starts and ends with an odd number</b>. If it is possible to divide the array then print "<b>YES</b>" otherwise "<b>NO</b>".
<b>Note:</b> Don't forget that the complete array is also considered as a subarrayThe first line of each test case contains N size of array. Second-line contains N space-separated elements.
<b>Constraints:</b>
1 <= N <= 999999
0 <= A[i] <= 999999999Print "YES", if it is possible to divide the array. Otherwise, print "NO". Without quotes.Sample Input 1:
5
9 0 3 1 5
Sample Output 1:
YES
Sample Input 2:
3
1 2 0
Sample Output 2:
NO
Explanation:
Testcase 1: Array {9, 0, 3, 1, 5} can be divided as {9, 0, 3}, {1}, {5}., I have written this Solution Code:
def fun(n,arr):
if n%2==0:
if arr[0]%2==1 and arr[n-1]%2==1:
for i in range(1,n):
if arr[i]%2==1 and arr[i-1]%2==1:
return "YES"
else:
return "NO"
else:
if arr[0]%2==1 and arr[n-1]%2==1:
return "YES"
else:
return "NO"
n=int(input())
arr=[int(i) for i in input().split()]
print(fun(n,arr)), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Siddharth, a great entrepreneur, chooses a person as an intern if he/she good at maths. Thus Sid gives a simple maths problem to the candidate based on which he decides to who is gonna get the offer. Below is the description of the problem asked by Sid
Given arr <b>A[]</b> containing <b>N</b> positive integers. The task is to check whether it is possible to divide the given array into an <b>odd number of non-empty subarrays each of odd length such that each sub-array starts and ends with an odd number</b>. If it is possible to divide the array then print "<b>YES</b>" otherwise "<b>NO</b>".
<b>Note:</b> Don't forget that the complete array is also considered as a subarrayThe first line of each test case contains N size of array. Second-line contains N space-separated elements.
<b>Constraints:</b>
1 <= N <= 999999
0 <= A[i] <= 999999999Print "YES", if it is possible to divide the array. Otherwise, print "NO". Without quotes.Sample Input 1:
5
9 0 3 1 5
Sample Output 1:
YES
Sample Input 2:
3
1 2 0
Sample Output 2:
NO
Explanation:
Testcase 1: Array {9, 0, 3, 1, 5} can be divided as {9, 0, 3}, {1}, {5}., I have written this Solution Code: 'use strict'
function main(input)
{
const inputList = input.split('\n');
const arrSize = Number(inputList[0].split(' ')[0])
const arr = inputList[1].split(' ').
map(x => Number(x))
if(arrSize %2 == 0 || Number(arr[0]) %2 == 0 ||
Number(arr[arrSize-1])%2 == 0)
console.log("NO")
else console.log("YES")
}
main(require("fs").readFileSync("/dev/stdin", "utf8"));, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given N flags, initially set to 0. Now you have to perform two operations on them:
1. Increase(F) by 1: flag F is increased by 1.
2. max_flag: all flags are set to a maximum value of any flag.
A non-empty array arr[] will be given of size M. This array represents consecutive operations:
a) If arr[K] = F, such that 1 <= F <= N then operation K is Increase(F).
b) If arr[K] = N+1 then operation K is max_flag.
The goal is to calculate the value of every flag after all operations.The input line contains T, denoting the number of test cases.
Each test case contains two lines.
The first line contains an integer N, the number of flags, and an integer M, the size of the array 'arr'.The second line contains elements of the array 'arr' separated by space.
<b>Constraints:</b>
1 <= T <= 100
1 <= N, M <= 10^5
1 <= arr[i] <= N+1
Sum of N and M for each test case is less than or equal to 10^5For each testcase you need to print the updated array after all operations in new line.Sample Input:
1
5 7
3 4 4 6 1 4 4
Sample Output:
3 2 2 4 2
<b>Explanation:</b>
Testcase 1:
the values of the flags after each consecutive operation will be:
(0, 0, 1, 0, 0)
(0, 0, 1, 1, 0)
(0, 0, 1, 2, 0)
(2, 2, 2, 2, 2)
(3, 2, 2, 2, 2)
(3, 2, 2, 3, 2)
(3, 2, 2, 4, 2), I have written this Solution Code: t=int(input())
while t>0:
t-=1
n,m=map(int,input().split())
a=map(int,input().split())
b=[0]*(n+1)
for i in a:
if i==n+1:
v=max(b)
for i in range(1,n+1):
b[i]=v
else:b[i]+=1
for i in range(1,n+1):
print(b[i],end=' ')
print(), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given N flags, initially set to 0. Now you have to perform two operations on them:
1. Increase(F) by 1: flag F is increased by 1.
2. max_flag: all flags are set to a maximum value of any flag.
A non-empty array arr[] will be given of size M. This array represents consecutive operations:
a) If arr[K] = F, such that 1 <= F <= N then operation K is Increase(F).
b) If arr[K] = N+1 then operation K is max_flag.
The goal is to calculate the value of every flag after all operations.The input line contains T, denoting the number of test cases.
Each test case contains two lines.
The first line contains an integer N, the number of flags, and an integer M, the size of the array 'arr'.The second line contains elements of the array 'arr' separated by space.
<b>Constraints:</b>
1 <= T <= 100
1 <= N, M <= 10^5
1 <= arr[i] <= N+1
Sum of N and M for each test case is less than or equal to 10^5For each testcase you need to print the updated array after all operations in new line.Sample Input:
1
5 7
3 4 4 6 1 4 4
Sample Output:
3 2 2 4 2
<b>Explanation:</b>
Testcase 1:
the values of the flags after each consecutive operation will be:
(0, 0, 1, 0, 0)
(0, 0, 1, 1, 0)
(0, 0, 1, 2, 0)
(2, 2, 2, 2, 2)
(3, 2, 2, 2, 2)
(3, 2, 2, 3, 2)
(3, 2, 2, 4, 2), I have written this Solution Code: #include "bits/stdc++.h"
#pragma GCC optimize "03"
using namespace std;
#define int long long int
#define ld long double
#define pi pair<int, int>
#define pb push_back
#define fi first
#define se second
#define IOS ios::sync_with_stdio(false); cin.tie(0); cout.tie(0)
#ifndef LOCAL
#define endl '\n'
#endif
const int N = 1e5 + 5;
const int mod = 1e9 + 7;
const int inf = 1e9 + 9;
int a[N];
signed main() {
IOS;
int t; cin >> t;
while(t--){
memset(a, 0, sizeof a);
int n, m;
cin >> n >> m;
int mx = 0, flag = 0;
for(int i = 1; i <= m; i++){
int p; cin >> p;
if(p == n+1){
flag = mx;
}
else{
a[p] = max(a[p], flag) + 1;
mx = max(mx, a[p]);
}
}
for(int i = 1; i <= n; i++){
a[i] = max(a[i], flag);
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: Gruumm has set up N lasers with i<sup>th</sup> laser having attack power Attack[i], all pointing earth. The damage caused by lasers on earth is the XOR of the attack powers of all the lasers. Cruger wants to set up a shield of defensive strength K. As Cruger has budget constraints, K can only be a non-negative integer with value strictly less than the strength of the weakest laser (minimum value of Attack array) set by Gruumm. Due to the defense shield, the attack power of each laser reduces by K. Help Kruger find the number of values of K such that the damage caused by lasers on earth is 0.First line of input contains a single integer, N.
Second line of input contains N integers, denoting the array Attack.
Constraints:
1 <= N <= 10<sup>5</sup>
1 <= Attack[i] <= 10<sup>18</sup>Print a single integer denoting the number of values of K such that the damage on earth is 0.Sample Input 1
3
3 4 5
Sample Output 1
1
Explanation: Only valid value of K is 2. With K = 2, attack powers become [1, 2, 3] and (1 xor 2 xor 3) = 0
Sample Input 2
2
4 4
Sample Output 2
4
Explanation: valid values of K are [0, 1, 2, 3], 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>
/////////////
const int MAXN = 100000;
const int MAXBIT = 62;
int n;
long long a[MAXN];
int cnt_with_bit[MAXBIT], ones_pref[MAXBIT][MAXN+1];
long long d[MAXBIT][MAXN+1];
long long get_d (int bit, int greater) {
if (bit == MAXBIT)
return greater ? 0 : 1;
long long & my = d[bit][greater];
if (my != -1) return my;
my = 0;
for (int x_bit=0; x_bit<2; ++x_bit) {
int ngreater = 0, cur_xor = 0;
for (int tp=0; tp<2; ++tp)
for (int a_bit=0; a_bit<2; ++a_bit) {
int cnt = ones_pref[bit][greater],
total = tp==0 ? greater : n-greater;
if (tp)
cnt = cnt_with_bit[bit] - cnt;
if (!a_bit)
cnt = total - cnt;
if (x_bit > a_bit || x_bit == a_bit && tp == 0)
ngreater += cnt;
int res_bit = (a_bit - x_bit - (1-tp)) & 1;
if (res_bit)
cur_xor ^= cnt & 1;
}
if (!cur_xor)
my += get_d (bit+1, ngreater);
}
return my;
}
void calc_tables() {
for (int bit=0; bit<MAXBIT; ++bit) {
cnt_with_bit[bit] = 0;
for (int i=0; i<n; ++i)
if (a[i] & (1ll << bit))
++cnt_with_bit[bit];
}
vector<int> order (n);
for (int i=0; i<n; ++i)
order[i] = i;
for (int bit=0; bit<MAXBIT; ++bit) {
vector<int> v[2];
for (int i=0; i<n; ++i) {
ones_pref[bit][i] = (int) v[1].size();
v[(a[order[i]] >> bit) & 1].push_back (order[i]);
}
ones_pref[bit][n] = (int) v[1].size();
order = v[0];
order.insert (order.end(), v[1].begin(), v[1].end());
}
}
signed main() {
fastio;
#ifdef ANIKET_GOYAL
freopen("inputf.in","r",stdin);
freopen("outputf.in","w",stdout);
#endif
cin >> n;
for (int i=0; i<n; ++i){
cin >> a[i];
}
calc_tables();
memset (d, -1, sizeof d);
long long ans = get_d (0, 0);
long long mi = * min_element (a, a+n);
long long x = 0;
for (int i=0; i<n; ++i)
x ^= a[i] - mi;
if (!x)
--ans;
cout << ans << endl;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Find factorial of a given number N.
<b>Note: </b>
The Factorial of a number is the product of an integer and all the integers below it;
e.g. factorial four ( 4! ) is equal to 24 (4*3*2*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 <i>Factorial()</i> which contains the given number N.
<b>Constraints:</b>
1 <= N <= 15
Return the factorial of the given number.Sample Input:-
5
Sample Output:-
120
Sample Input:-
3
Sample Output:-
6, I have written this Solution Code: def factorial(n):
if(n == 1):
return 1
return n * factorial(n-1)
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Find factorial of a given number N.
<b>Note: </b>
The Factorial of a number is the product of an integer and all the integers below it;
e.g. factorial four ( 4! ) is equal to 24 (4*3*2*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 <i>Factorial()</i> which contains the given number N.
<b>Constraints:</b>
1 <= N <= 15
Return the factorial of the given number.Sample Input:-
5
Sample Output:-
120
Sample Input:-
3
Sample Output:-
6, I have written this Solution Code: static int Factorial(int N)
{
if(N==0){
return 1;}
return N*Factorial(N-1);
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Find factorial of a given number N.
<b>Note: </b>
The Factorial of a number is the product of an integer and all the integers below it;
e.g. factorial four ( 4! ) is equal to 24 (4*3*2*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 <i>Factorial()</i> which contains the given number N.
<b>Constraints:</b>
1 <= N <= 15
Return the factorial of the given number.Sample Input:-
5
Sample Output:-
120
Sample Input:-
3
Sample Output:-
6, I have written this Solution Code: // n is the input number
function factorial(n) {
// write code here
// do not console.log
// return the answer as a number
if (n == 1 ) return 1;
return n * factorial(n-1)
}, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an unsorted array A of size N of non-negative integers, find a continuous sub-array which adds to a given number S.Each test case consists of two lines. The first line of each test case is N and S, where N is the size of the array and S is the sum. The second line of each test case contains N space-separated integers denoting the array elements.
<b>Constraints:-</b>
1 ≤ N ≤ 10<sup>5</sup>
1 ≤ Ai ≤ 10<sup>5</sup>Print the starting and ending positions (1 indexing) of first such occurring subarray from the left if sum equals to subarray, else print -1.Sample Input
5 12
1 2 3 7 5
Sample Output
2 4
Explanation:
subarray starting from index 2 and ending at index 4 => {2 , 3 , 7}
sum = 2 + 3 + 7 = 12
Sample Input
10 15
1 2 3 4 5 6 7 8 9 10
Sample Output
1 5, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws IOException
{
try{
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
String StrInput[] = br.readLine().trim().split(" ");
int n = Integer.parseInt(StrInput[0]);
int s = Integer.parseInt(StrInput[1]);
int arr[] = new int[n];
String StrInput2[] = br.readLine().trim().split(" ");
for(int i=0;i<n;i++)
{
arr[i] = Integer.parseInt(StrInput2[i]);
}
int sum = arr[0];
int startingindex = 0;
int endingindex = 1;
int j = 0;
int i;
for(i=1;i<=n;i++)
{
if(sum < s && arr[i] != 0)
{
sum += arr[i];
}
while(sum > s && startingindex < i-1)
{
sum -= arr[startingindex];
startingindex++;
}
if(sum == s)
{
endingindex = i+1;
if(arr[0] == 0)
{
System.out.print(startingindex+2 + " " + endingindex);
}
else
{
System.out.print(startingindex+1 + " "+ endingindex);
}
break;
}
if(i == n && sum < s)
{
System.out.print(-1);
break;
}
}
}
catch(Exception e)
{
System.out.print(-1);
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an unsorted array A of size N of non-negative integers, find a continuous sub-array which adds to a given number S.Each test case consists of two lines. The first line of each test case is N and S, where N is the size of the array and S is the sum. The second line of each test case contains N space-separated integers denoting the array elements.
<b>Constraints:-</b>
1 ≤ N ≤ 10<sup>5</sup>
1 ≤ Ai ≤ 10<sup>5</sup>Print the starting and ending positions (1 indexing) of first such occurring subarray from the left if sum equals to subarray, else print -1.Sample Input
5 12
1 2 3 7 5
Sample Output
2 4
Explanation:
subarray starting from index 2 and ending at index 4 => {2 , 3 , 7}
sum = 2 + 3 + 7 = 12
Sample Input
10 15
1 2 3 4 5 6 7 8 9 10
Sample Output
1 5, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main(){
int n,k;
cin>>n>>k;
int a[n];
for(int i=0;i<n;i++)
{
cin>>a[i];
}
int sum=0;
unordered_map<int,int> m;
for(int i=0;i<n;i++){
sum+=a[i];
if(sum==k){cout<<1<<" "<<i+1;return 0;}
if(m.find(sum-k)!=m.end()){
cout<<m[sum-k]+2<<" "<<i+1;
return 0;
}
m[sum]=i;
}
cout<<-1;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an unsorted array A of size N of non-negative integers, find a continuous sub-array which adds to a given number S.Each test case consists of two lines. The first line of each test case is N and S, where N is the size of the array and S is the sum. The second line of each test case contains N space-separated integers denoting the array elements.
<b>Constraints:-</b>
1 ≤ N ≤ 10<sup>5</sup>
1 ≤ Ai ≤ 10<sup>5</sup>Print the starting and ending positions (1 indexing) of first such occurring subarray from the left if sum equals to subarray, else print -1.Sample Input
5 12
1 2 3 7 5
Sample Output
2 4
Explanation:
subarray starting from index 2 and ending at index 4 => {2 , 3 , 7}
sum = 2 + 3 + 7 = 12
Sample Input
10 15
1 2 3 4 5 6 7 8 9 10
Sample Output
1 5, I have written this Solution Code: def sumFinder(N,S,a):
currentSum = a[0]
start = 0
i = 1
while i <= N:
while currentSum > S and start < i-1:
currentSum = currentSum - a[start]
start += 1
if currentSum == S:
return (start+1,i)
if i < N:
currentSum = currentSum + a[i]
i += 1
return(-1)
N, S = [int(x) for x in input().split()]
a = [int(x) for x in input().split()]
ans = sumFinder(N,S,a)
if(ans==-1):
print(ans)
else:
print(ans[0],ans[1]), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Bob was working on a new algorithm to make an array non-decreasing. he came up with the following algorithm :
For an array a of size n
Step 1: Select two indexes i and j, 0 ≤ i < j < n.
Step 2: if a[i] + a[j] is odd then do a[i] = a[j]
if a[i] + a[j] is even then do a[j] = a[i]
Step 3: Repeat steps 1 and 2 until the array is non-decreasing.
Bob is not sure about his algorithm, he asks you if it is possible to make the given array non-decreasing using the mentioned algorithm.The first line contains n.
The next line contains n space-separated integers.
<b>Constraints</b>
1 ≤ n ≤ 10<sup>5</sup>
1 ≤ a[i] ≤ 10<sup>9</sup>Print "YES" of it is possible to make array non- decreasing, otherwise print "NO".Input:
4
3 2 5 3
Output:
YES
Explanation:
(i, j) => (0, 1) => 3+2 = 5 => odd => a[0] = a[1] => a[0] = 2
(i, j) => (2, 3) => 5+3 = 8 => even=> a[3]= a[2] => a[3] = 5
final aray => 2 2 5 5, I have written this Solution Code: import java.io.*;
import java.util.*;
public class Main {
public static void main(String[] args) {
InputStream inputStream = System.in;
OutputStream outputStream = System.out;
InputReader in = new InputReader(inputStream);
OutputWriter out = new OutputWriter(outputStream);
int n=Integer.parseInt(in.next());
Integer a[] = new Integer[n];
for(int t=0;t<n;t++){
a[t] = Integer.parseInt(in.next());
}
out.print("YES");
out.close();
}
static class InputReader {
BufferedReader reader;
StringTokenizer tokenizer;
public InputReader(InputStream stream) {
reader = new BufferedReader(new InputStreamReader(stream), 32768);
tokenizer = null;
}
public String next() {
while (tokenizer == null || !tokenizer.hasMoreTokens()) {
try {
tokenizer = new StringTokenizer(reader.readLine());
} catch (IOException e) {
throw new RuntimeException(e);
}
}
return tokenizer.nextToken();
}
public int nextInt() {
return Integer.parseInt(next());
}
}
static class OutputWriter {
private final PrintWriter writer;
public OutputWriter(OutputStream outputStream) {
writer = new PrintWriter(new BufferedWriter(new OutputStreamWriter(outputStream)));
}
public OutputWriter(Writer writer) {
this.writer = new PrintWriter(writer);
}
public void print(Object... objects) {
for (int i = 0; i < objects.length; i++) {
if (i != 0) {
writer.print(' ');
}
writer.print(objects[i]);
}
}
public void println(Object... objects) {
print(objects);
writer.println();
}
public void close() {
writer.close();
}
public void println(int i) {
writer.println(i);
}
}
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |