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For this Question: Given an array <b>Arr[]</b> of size <b>N</b> as input, your task is to count the number of triplets Arr[i], Arr[j] and Arr[k] such that:-
i < j < k and the difference between every 2 elements of triplets is less than or equal to P i. e |Arr[i] - Arr[j]| <= P, |Arr[i] - Arr[k]| <= P and |Arr[j] - Arr[k]| <= PThe first line of input contains two space- separated integers N and P.
next line contains N space separated integers depicting the values of the Arr[].
Constraints:-
3 <= N <= 10<sup>5</sup>
1 <= Arr[i], P <= 10<sup>9</sup>
0 <= i <= N-1Return the count of triplets that satisfies the above conditions.Sample Input:-
5 4
1 3 2 5 9
Sample Output:-
4
Explanation:-
(1, 3, 2), (1, 3, 5), (1, 2, 5), (2, 3, 5) are the required triplets
Sample Input:-
5 3
1 8 4 2 9
Sample Output:-
1, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
String str1 = br.readLine();
String str2[] = str1.split(" ");
int n = Integer.parseInt(str2[0]);
int k = Integer.parseInt(str2[1]);
String str3 = br.readLine();
String str4[] = str3.split(" ");
long[] arr = new long[n];
for(int i = 0; i < n; ++i) {
arr[i] = Long.parseLong(str4[i]);
}
Arrays.sort(arr);
int i=0,j=2;
long ans=0;
while(j!=n){
if(i==j-1){j++;continue;}
if((arr[j]-arr[i])>k){i++;}
else{
int x = j-i;
ans+=(x*(x-1))/2;
j++;
}
}
System.out.print(ans);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array <b>Arr[]</b> of size <b>N</b> as input, your task is to count the number of triplets Arr[i], Arr[j] and Arr[k] such that:-
i < j < k and the difference between every 2 elements of triplets is less than or equal to P i. e |Arr[i] - Arr[j]| <= P, |Arr[i] - Arr[k]| <= P and |Arr[j] - Arr[k]| <= PThe first line of input contains two space- separated integers N and P.
next line contains N space separated integers depicting the values of the Arr[].
Constraints:-
3 <= N <= 10<sup>5</sup>
1 <= Arr[i], P <= 10<sup>9</sup>
0 <= i <= N-1Return the count of triplets that satisfies the above conditions.Sample Input:-
5 4
1 3 2 5 9
Sample Output:-
4
Explanation:-
(1, 3, 2), (1, 3, 5), (1, 2, 5), (2, 3, 5) are the required triplets
Sample Input:-
5 3
1 8 4 2 9
Sample Output:-
1, I have written this Solution Code: n, p = input().split(' ')
n = int(n)
p = int(p)
arr = input().split(' ')[:n]
for i in range(n):
arr[i] = int(arr[i])
arr.sort()
i = 0
k = 2
count = 0
while k!=n:
if i == k-1:
k = k + 1
continue
if (arr[k] - arr[i]) <= p:
count = count + (int)(((k-i)*(k-i-1))/2)
k = k + 1
else:
i = i + 1
print(count)
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array <b>Arr[]</b> of size <b>N</b> as input, your task is to count the number of triplets Arr[i], Arr[j] and Arr[k] such that:-
i < j < k and the difference between every 2 elements of triplets is less than or equal to P i. e |Arr[i] - Arr[j]| <= P, |Arr[i] - Arr[k]| <= P and |Arr[j] - Arr[k]| <= PThe first line of input contains two space- separated integers N and P.
next line contains N space separated integers depicting the values of the Arr[].
Constraints:-
3 <= N <= 10<sup>5</sup>
1 <= Arr[i], P <= 10<sup>9</sup>
0 <= i <= N-1Return the count of triplets that satisfies the above conditions.Sample Input:-
5 4
1 3 2 5 9
Sample Output:-
4
Explanation:-
(1, 3, 2), (1, 3, 5), (1, 2, 5), (2, 3, 5) are the required triplets
Sample Input:-
5 3
1 8 4 2 9
Sample Output:-
1, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define int long long
signed main(){
int n,p;
cin>>n>>p;
int a[n];
for(int i=0;i<n;i++){
cin>>a[i];
}
sort(a,a+n);
int i=0,j=2;
int ans=0;
while(j!=n){
if(i==j-1){j++;continue;}
if((a[j]-a[i])>p){i++;}
else{
int x = j-i;
ans+=(x*(x-1))/2;
j++;
}
}
cout<<ans;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given two integer N and K, find the minimum number of times N must be multiplied by X to make it divisible by (2<sup>K</sup>).
Where X = Summation (2<sup>(4*i)</sup>) for 1 <= i <= 25.The first and the only line of input contains two space separated integers N and K.
Constraints
1 <= N <= 10^18
1 <= K <= 10^18Print a single integer which is the minimum number of times N must be multiplied by X to make it divisible by (2<sup>K</sup>).Sample Input 1
40 4
Sample Output 1
1
Sample Input
40 3
Sample Output
0, I have written this Solution Code: import java.util.InputMismatchException;
import java.io.ByteArrayInputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Arrays;
public class Main {
InputStream is;
PrintWriter out;
String INPUT = "";
int MAX = (int) 1e5, MOD = (int)1e9+7;
void solve(int TC) {
long n = nl();
long k = nl();
int p = 0;
while(n>0 && n%2==0) {
n/=2;
++p;
}
if(p>=k) {pn(0);return;}
k -= p;
long ans = (k+3L)/4L;
pn(ans);
}
boolean TestCases = false;
public static void main(String[] args) throws Exception { new Main().run(); }
long pow(long a, long b) {
if(b==0 || a==1) return 1;
long o = 1;
for(long p = b; p > 0; p>>=1) {
if((p&1)==1) o = (o*a) % MOD;
a = (a*a) % MOD;
} return o;
}
long inv(long x) {
long o = 1;
for(long p = MOD-2; p > 0; p>>=1) {
if((p&1)==1)o = (o*x)%MOD;
x = (x*x)%MOD;
} return o;
}
long gcd(long a, long b) { return (b==0) ? a : gcd(b,a%b); }
int gcd(int a, int b) { return (b==0) ? a : gcd(b,a%b); }
void run() throws Exception {
is = INPUT.isEmpty() ? System.in : new ByteArrayInputStream(INPUT.getBytes());
out = new PrintWriter(System.out);
long s = System.currentTimeMillis();
int T = TestCases ? ni() : 1;
for(int t=1;t<=T;t++) solve(t);
out.flush();
if(!INPUT.isEmpty())tr(System.currentTimeMillis()-s+"ms");
}
void p(Object o) { out.print(o); }
void pn(Object o) { out.println(o); }
void pni(Object o) { out.println(o);out.flush(); }
double PI = 3.141592653589793238462643383279502884197169399;
int ni() {
int num = 0, b;
boolean minus = false;
while((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-'));
if(b == '-') {
minus = true;
b = readByte();
}
while(true) {
if(b >= '0' && b <= '9'){
num = num * 10 + (b - '0');
} else {
return minus ? -num : num;
}
b = readByte();
}
}
long nl() {
long num = 0;
int b;
boolean minus = false;
while((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-'));
if(b == '-') {
minus = true;
b = readByte();
}
while(true) {
if(b >= '0' && b <= '9') {
num = num * 10 + (b - '0');
} else {
return minus ? -num : num;
}
b = readByte();
}
}
double nd() { return Double.parseDouble(ns()); }
char nc() { return (char)skip(); }
int BUF_SIZE = 1024 * 8;
byte[] inbuf = new byte[BUF_SIZE];
int lenbuf = 0, ptrbuf = 0;
int readByte() {
if(lenbuf == -1)throw new InputMismatchException();
if(ptrbuf >= lenbuf){
ptrbuf = 0;
try { lenbuf = is.read(inbuf); } catch (IOException e) { throw new InputMismatchException(); }
if(lenbuf <= 0)return -1;
} return inbuf[ptrbuf++];
}
boolean isSpaceChar(int c) { return !(c >= 33 && c <= 126); }
int skip() { int b; while((b = readByte()) != -1 && isSpaceChar(b)); return b; }
String ns() {
int b = skip();
StringBuilder sb = new StringBuilder();
while(!(isSpaceChar(b))) {
sb.appendCodePoint(b); b = readByte();
} return sb.toString();
}
char[] ns(int n) {
char[] buf = new char[n];
int b = skip(), p = 0;
while(p < n && !(isSpaceChar(b))){
buf[p++] = (char)b;
b = readByte();
} return n == p ? buf : Arrays.copyOf(buf, p);
}
void tr(Object... o) { if(INPUT.length() > 0)System.out.println(Arrays.deepToString(o)); }
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given two integer N and K, find the minimum number of times N must be multiplied by X to make it divisible by (2<sup>K</sup>).
Where X = Summation (2<sup>(4*i)</sup>) for 1 <= i <= 25.The first and the only line of input contains two space separated integers N and K.
Constraints
1 <= N <= 10^18
1 <= K <= 10^18Print a single integer which is the minimum number of times N must be multiplied by X to make it divisible by (2<sup>K</sup>).Sample Input 1
40 4
Sample Output 1
1
Sample Input
40 3
Sample Output
0, I have written this Solution Code: [n,k]=[int(j) for j in input().split()]
a=0
while n%2==0:
a+=1
n=n//2
if k>a:
print((k-a-1)//4+1)
else:
print(0), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given two integer N and K, find the minimum number of times N must be multiplied by X to make it divisible by (2<sup>K</sup>).
Where X = Summation (2<sup>(4*i)</sup>) for 1 <= i <= 25.The first and the only line of input contains two space separated integers N and K.
Constraints
1 <= N <= 10^18
1 <= K <= 10^18Print a single integer which is the minimum number of times N must be multiplied by X to make it divisible by (2<sup>K</sup>).Sample Input 1
40 4
Sample Output 1
1
Sample Input
40 3
Sample Output
0, I have written this Solution Code: #pragma GCC optimize ("Ofast")
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define VV vector
#define pb push_back
#define bitc __builtin_popcountll
#define m_p make_pair
#define infi 1e18+1
#define eps 0.000000000001
#define fastio ios_base::sync_with_stdio(false);cin.tie(NULL);
string char_to_str(char c){string tem(1,c);return tem;}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
template<class T>//usage rand<long long>()
T rand() {
return uniform_int_distribution<T>()(rng);
}
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
template<class T>
using oset = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
// string to integer stoi()
// string to long long stoll()
// string.substr(position,length);
// integer to string to_string();
//////////////
auto clk=clock();
#define all(x) x.begin(),x.end()
#define S second
#define F first
#define sz(x) ((long long)x.size())
#define int long long
#define f80 __float128
#define pii pair<int,int>
/////////////
signed main()
{
fastio;
#ifdef ANIKET_GOYAL
freopen("inputf.in","r",stdin);
freopen("outputf.in","w",stdout);
#endif
int n,k;
cin>>n>>k;
while(k&&n%2==0){
n/=2;
--k;
}
cout<<(k+3)/4;
#ifdef ANIKET_GOYAL
// cout<<endl<<endl<<endl<<endl<<"Time elapsed: "<<(double)(clock()-clk)/CLOCKS_PER_SEC<<endl;
#endif
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a cubic dice with 6 faces. All the individual faces have a numbers printed on them. The numbers are in the range of 1 to 6, like any ordinary dice.
Given a number on one face of the dice , you need to print the number on the opposite face .
NOTE : In a normal dice sum of numbers on opposite faces is 7 .The first line of the input contains a single integer T, denoting the number of test cases. Then T test case follows. Each test case contains a single line of the input containing a positive integer N.
Constraints:
1 <= T <= 100
1 <= N <= 6For each testcase, print the number that is on the opposite side of the given face.Input:
2
6
2
Output:
1
5
Explanation:
Testcase 1: For dice facing number 6 opposite face will have the number 1., I have written this Solution Code: def get_opposite_face(n):
return 7-n
t = int(input())
for n in range(t):
print(get_opposite_face(int(input()))), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a cubic dice with 6 faces. All the individual faces have a numbers printed on them. The numbers are in the range of 1 to 6, like any ordinary dice.
Given a number on one face of the dice , you need to print the number on the opposite face .
NOTE : In a normal dice sum of numbers on opposite faces is 7 .The first line of the input contains a single integer T, denoting the number of test cases. Then T test case follows. Each test case contains a single line of the input containing a positive integer N.
Constraints:
1 <= T <= 100
1 <= N <= 6For each testcase, print the number that is on the opposite side of the given face.Input:
2
6
2
Output:
1
5
Explanation:
Testcase 1: For dice facing number 6 opposite face will have the number 1., I have written this Solution Code: #include "bits/stdc++.h"
#pragma GCC optimize "03"
using namespace std;
#define int long long int
#define ld long double
#define pi pair<int, int>
#define pb push_back
#define fi first
#define se second
#define IOS ios::sync_with_stdio(false); cin.tie(0); cout.tie(0)
#ifndef LOCAL
#define endl '\n'
#endif
const int N = 2e5 + 5;
const int mod = 1e9 + 7;
const int inf = 1e9 + 9;
signed main() {
IOS;
int t; cin >> t;
while(t--){
int n; cin >> n;
cout << 7-n << endl;
}
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a cubic dice with 6 faces. All the individual faces have a numbers printed on them. The numbers are in the range of 1 to 6, like any ordinary dice.
Given a number on one face of the dice , you need to print the number on the opposite face .
NOTE : In a normal dice sum of numbers on opposite faces is 7 .The first line of the input contains a single integer T, denoting the number of test cases. Then T test case follows. Each test case contains a single line of the input containing a positive integer N.
Constraints:
1 <= T <= 100
1 <= N <= 6For each testcase, print the number that is on the opposite side of the given face.Input:
2
6
2
Output:
1
5
Explanation:
Testcase 1: For dice facing number 6 opposite face will have the number 1., I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args)throws IOException {
BufferedReader br =new BufferedReader(new InputStreamReader(System.in));
int t=Integer.parseInt(br.readLine());
while(t-->0){
int n=Integer.parseInt(br.readLine());
System.out.println(6-n+1);
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer N, you have to print the given below pattern for N ≥ 3.
<b>Example</b>
Pattern for N = 4
*
*^*
*^^*
*****<b>User Task</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>Pattern()</b> that takes integers N as an argument.
<b>Constraints</b>
3 ≤ N ≤ 100Print the given pattern for size N.<b>Sample Input 1</b>
3
<b>Sample Output 1</b>
*
*^*
****
<b>Sample Input 2</b>
6
<b>Sample Output 2</b>
*
*^*
*^^*
*^^^*
*^^^^*
*******, I have written this Solution Code: def Pattern(N):
print('*')
for i in range (0,N-2):
print('*',end='')
for j in range (0,i+1):
print('^',end='')
print('*')
for i in range (0,N+1):
print('*',end='')
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer N, you have to print the given below pattern for N ≥ 3.
<b>Example</b>
Pattern for N = 4
*
*^*
*^^*
*****<b>User Task</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>Pattern()</b> that takes integers N as an argument.
<b>Constraints</b>
3 ≤ N ≤ 100Print the given pattern for size N.<b>Sample Input 1</b>
3
<b>Sample Output 1</b>
*
*^*
****
<b>Sample Input 2</b>
6
<b>Sample Output 2</b>
*
*^*
*^^*
*^^^*
*^^^^*
*******, I have written this Solution Code: void Pattern(int N){
cout<<'*'<<endl;
for(int i=0;i<N-2;i++){
cout<<'*';
for(int j=0;j<=i;j++){
cout<<'^';
}
cout<<'*'<<endl;
}
for(int i=0;i<=N;i++){
cout<<'*';
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer N, you have to print the given below pattern for N ≥ 3.
<b>Example</b>
Pattern for N = 4
*
*^*
*^^*
*****<b>User Task</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>Pattern()</b> that takes integers N as an argument.
<b>Constraints</b>
3 ≤ N ≤ 100Print the given pattern for size N.<b>Sample Input 1</b>
3
<b>Sample Output 1</b>
*
*^*
****
<b>Sample Input 2</b>
6
<b>Sample Output 2</b>
*
*^*
*^^*
*^^^*
*^^^^*
*******, I have written this Solution Code: static void Pattern(int N){
System.out.println('*');
for(int i=0;i<N-2;i++){
System.out.print('*');
for(int j=0;j<=i;j++){
System.out.print('^');
}System.out.println('*');
}
for(int i=0;i<=N;i++){
System.out.print('*');
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer N, you have to print the given below pattern for N ≥ 3.
<b>Example</b>
Pattern for N = 4
*
*^*
*^^*
*****<b>User Task</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>Pattern()</b> that takes integers N as an argument.
<b>Constraints</b>
3 ≤ N ≤ 100Print the given pattern for size N.<b>Sample Input 1</b>
3
<b>Sample Output 1</b>
*
*^*
****
<b>Sample Input 2</b>
6
<b>Sample Output 2</b>
*
*^*
*^^*
*^^^*
*^^^^*
*******, I have written this Solution Code: void Pattern(int N){
printf("*\n");
for(int i=0;i<N-2;i++){
printf("*");
for(int j=0;j<=i;j++){
printf("^");}printf("*\n");
}
for(int i=0;i<=N;i++){
printf("*");
}
}
, In this Programming Language: C, Now tell me if this Code is compilable or not? | Compilable |
For this Question: The time stone is hidden inside a sacred box which has a lock of N different rings and each ring has 0-9 serially on it. Initially, all N- rings together show an N- digit integer but there is a particular code only that can open the lock. You can rotate each ring any number of times in either direction. You have to find the minimum number of rotations done on rings of the lock to open the lock.First line contains the integer N.
Second line contains a string that depicts the digit on rings
Third line contains a string that depicts the unlock code
Constraint :
0 <= digit on each ring <= 9
1 <= |String| <= 100000Print the minimum rotation of rotations done on the rings to unlock the box.Sample Input:-
4
2345
5432
Sample Output:-
8
Explanation:-
1st ring is rotated thrice as 2- >3- >4- >5
2nd ring is rotated once as 3- >4
3rd ring is rotated once as 4- >3
4th ring is rotated thrice as 5- >4- >3- >2
Sample Input:-
4
1919
0000
Sample Output:-
4, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) {
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
String key =sc.next();
String code =sc.next();
int ans =0;
for(int i=0;i<n;i++){
int num1 = key.charAt(i)-'0';
int num2 = code.charAt(i)-'0';
int forward = Math.abs(num1 - num2);
int backward = 10-forward;
if(forward<backward){
ans+=forward;
}
else{
ans+=backward;
}
}
System.out.println(ans);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: The time stone is hidden inside a sacred box which has a lock of N different rings and each ring has 0-9 serially on it. Initially, all N- rings together show an N- digit integer but there is a particular code only that can open the lock. You can rotate each ring any number of times in either direction. You have to find the minimum number of rotations done on rings of the lock to open the lock.First line contains the integer N.
Second line contains a string that depicts the digit on rings
Third line contains a string that depicts the unlock code
Constraint :
0 <= digit on each ring <= 9
1 <= |String| <= 100000Print the minimum rotation of rotations done on the rings to unlock the box.Sample Input:-
4
2345
5432
Sample Output:-
8
Explanation:-
1st ring is rotated thrice as 2- >3- >4- >5
2nd ring is rotated once as 3- >4
3rd ring is rotated once as 4- >3
4th ring is rotated thrice as 5- >4- >3- >2
Sample Input:-
4
1919
0000
Sample Output:-
4, I have written this Solution Code: n=int(input())
a=input()
b=input()
c=0
for i in range(n):
diff=abs(int(a[i])-int(b[i]))
c+=min(diff,10-diff)
print(c), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: The time stone is hidden inside a sacred box which has a lock of N different rings and each ring has 0-9 serially on it. Initially, all N- rings together show an N- digit integer but there is a particular code only that can open the lock. You can rotate each ring any number of times in either direction. You have to find the minimum number of rotations done on rings of the lock to open the lock.First line contains the integer N.
Second line contains a string that depicts the digit on rings
Third line contains a string that depicts the unlock code
Constraint :
0 <= digit on each ring <= 9
1 <= |String| <= 100000Print the minimum rotation of rotations done on the rings to unlock the box.Sample Input:-
4
2345
5432
Sample Output:-
8
Explanation:-
1st ring is rotated thrice as 2- >3- >4- >5
2nd ring is rotated once as 3- >4
3rd ring is rotated once as 4- >3
4th ring is rotated thrice as 5- >4- >3- >2
Sample Input:-
4
1919
0000
Sample Output:-
4, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main(){
int n;
cin>>n;
string a,b;
cin>>a>>b;
int ans=0;
for(int i=0;i<n;i++){
ans+=min(abs(a[i]-b[i]),10-abs(a[i]-b[i]));
}
cout<<ans;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Nutan and Tusla are both students at Newton School. They are both among the best students in the class. In order to know who is better among them, a game was organised. The game consisted of L rounds, where L is an odd integer. The student winning more rounds than the other was declared the winner.
You would be given a string of odd length L in which each character is 'N' or 'T'. If the i<sup>th</sup> character is 'N', then the i<sup>th</sup> round was won by Nutan, else if the character is 'T' it was won by Tusla. Print "Nutan'' if Nutan has won more rounds than Tusla, else print "Tusla'' if Tusla has won more rounds than Nutan.
Note: You have to print everything without quotes.The first line of the input contains a single integer L — the number of rounds (1 ≤ L ≤ 100 and L is odd).
The second line contains a string S of length L. Each character of S is either 'N' or 'T'.Print "Nutan" or "Tusla" according to the input.Sample Input:
3
NNT
Sample Output:
Nutan
Explanation:
Nutan has won two games while Tusla has only won a single game, so the overall winner is Nutan., I have written this Solution Code: import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;
import static java.lang.System.out;
public class Main {
static class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader() {
br = new BufferedReader(new InputStreamReader(System.in));
}
String next() {
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
long nextLong() {
return Long.parseLong(next());
}
double nextDouble() {
return Double.parseDouble(next());
}
String nextLine() {
String str = "";
try {
str = br.readLine();
} catch (IOException e) {
e.printStackTrace();
}
return str;
}
}
public static void main(String[] args) {
FastReader reader = new FastReader();
int n = reader.nextInt();
String S = reader.next();
int ncount = 0;
int tcount = 0;
for (char c : S.toCharArray()) {
if (c == 'N') ncount++;
else tcount++;
}
if (ncount > tcount) {
out.print("Nutan\n");
} else {
out.print("Tusla\n");
}
out.flush();
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Nutan and Tusla are both students at Newton School. They are both among the best students in the class. In order to know who is better among them, a game was organised. The game consisted of L rounds, where L is an odd integer. The student winning more rounds than the other was declared the winner.
You would be given a string of odd length L in which each character is 'N' or 'T'. If the i<sup>th</sup> character is 'N', then the i<sup>th</sup> round was won by Nutan, else if the character is 'T' it was won by Tusla. Print "Nutan'' if Nutan has won more rounds than Tusla, else print "Tusla'' if Tusla has won more rounds than Nutan.
Note: You have to print everything without quotes.The first line of the input contains a single integer L — the number of rounds (1 ≤ L ≤ 100 and L is odd).
The second line contains a string S of length L. Each character of S is either 'N' or 'T'.Print "Nutan" or "Tusla" according to the input.Sample Input:
3
NNT
Sample Output:
Nutan
Explanation:
Nutan has won two games while Tusla has only won a single game, so the overall winner is Nutan., I have written this Solution Code: n = int(input())
s = input()
a1 = s.count('N')
a2 = s.count('T')
if(a1 > a2):
print("Nutan")
else:
print('Tusla'), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Nutan and Tusla are both students at Newton School. They are both among the best students in the class. In order to know who is better among them, a game was organised. The game consisted of L rounds, where L is an odd integer. The student winning more rounds than the other was declared the winner.
You would be given a string of odd length L in which each character is 'N' or 'T'. If the i<sup>th</sup> character is 'N', then the i<sup>th</sup> round was won by Nutan, else if the character is 'T' it was won by Tusla. Print "Nutan'' if Nutan has won more rounds than Tusla, else print "Tusla'' if Tusla has won more rounds than Nutan.
Note: You have to print everything without quotes.The first line of the input contains a single integer L — the number of rounds (1 ≤ L ≤ 100 and L is odd).
The second line contains a string S of length L. Each character of S is either 'N' or 'T'.Print "Nutan" or "Tusla" according to the input.Sample Input:
3
NNT
Sample Output:
Nutan
Explanation:
Nutan has won two games while Tusla has only won a single game, so the overall winner is Nutan., I have written this Solution Code: //Author: Xzirium
//Time and Date: 02:18:28 24 March 2022
//Optional FAST
//#pragma GCC optimize("Ofast")
//#pragma GCC optimize("unroll-loops")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,fma,abm,mmx,avx,avx2,tune=native")
//Required Libraries
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/detail/standard_policies.hpp>
//Required namespaces
using namespace std;
using namespace __gnu_pbds;
//Required defines
#define endl '\n'
#define READ(X) cin>>X;
#define READV(X) long long X; cin>>X;
#define READAR(A,N) long long A[N]; for(long long i=0;i<N;i++) {cin>>A[i];}
#define rz(A,N) A.resize(N);
#define sz(X) (long long)(X.size())
#define pb push_back
#define pf push_front
#define fi first
#define se second
#define FORI(a,b,c) for(long long a=b;a<c;a++)
#define FORD(a,b,c) for(long long a=b;a>c;a--)
//Required typedefs
template <typename T> using ordered_set = tree<T,null_type,less<T>,rb_tree_tag,tree_order_statistics_node_update>;
template <typename T> using ordered_set1 = tree<T,null_type,greater<T>,rb_tree_tag,tree_order_statistics_node_update>;
typedef long long ll;
typedef long double ld;
typedef pair<int,int> pii;
typedef pair<long long,long long> pll;
//Required Constants
const long long inf=(long long)1e18;
const long long MOD=(long long)(1e9+7);
const long long INIT=(long long)(1e6+1);
const long double PI=3.14159265358979;
// Required random number generators
// mt19937 gen_rand_int(chrono::steady_clock::now().time_since_epoch().count());
// mt19937_64 gen_rand_ll(chrono::steady_clock::now().time_since_epoch().count());
//Required Functions
ll power(ll b, ll e)
{
ll r = 1ll;
for(; e > 0; e /= 2, (b *= b) %= MOD)
if(e % 2) (r *= b) %= MOD;
return r;
}
ll modInverse(ll a)
{
return power(a,MOD-2);
}
//Work
int main()
{
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
//-----------------------------------------------------------------------------------------------------------//
READV(N);
string S;
cin>>S;
ll n=0,t=0;
FORI(i,0,N)
{
if(S[i]=='N')
{
n++;
}
else if(S[i]=='T')
{
t++;
}
}
if(n>t)
{
cout<<"Nutan"<<endl;
}
else
{
cout<<"Tusla"<<endl;
}
//-----------------------------------------------------------------------------------------------------------//
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Today is Newton School's first class of this year. Nutan, a student at Newton School, has received his first assignment. He will be given a string as input. His task is to print "Gravity'' if the input is "Apple''; otherwise, he will have to print "Space''.
Can you help Nutan in solving his first assignment? Note that the quotation marks are just for clarity. They are not part of the input string, and should not be a part of your output string.The input consists of a single line that contains a string S (1 ≤ length of S ≤ 10). The string only consists of lowercase and uppercase letters.Print "Gravity'' or "Space'' according to the input.Sample Input 1:
Apple
Sample Output 1:
Gravity
Sample Input 2:
Mango
Sample Output 2:
Space
Sample Input 3:
AppLE
Sample Output 3:
Space, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
static class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader()
{
br = new BufferedReader(
new InputStreamReader(System.in));
}
String next()
{
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
}
catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() { return Integer.parseInt(next()); }
long nextLong() { return Long.parseLong(next()); }
double nextDouble()
{
return Double.parseDouble(next());
}
String nextLine()
{
String str = "";
try {
str = br.readLine();
}
catch (IOException e) {
e.printStackTrace();
}
return str;
}
}
public static void main (String[] args) {
FastReader sc= new FastReader();
String str= sc.nextLine();
String a="Apple";
if(a.equals(str)){
System.out.println("Gravity");
}
else{
System.out.println("Space");
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Today is Newton School's first class of this year. Nutan, a student at Newton School, has received his first assignment. He will be given a string as input. His task is to print "Gravity'' if the input is "Apple''; otherwise, he will have to print "Space''.
Can you help Nutan in solving his first assignment? Note that the quotation marks are just for clarity. They are not part of the input string, and should not be a part of your output string.The input consists of a single line that contains a string S (1 ≤ length of S ≤ 10). The string only consists of lowercase and uppercase letters.Print "Gravity'' or "Space'' according to the input.Sample Input 1:
Apple
Sample Output 1:
Gravity
Sample Input 2:
Mango
Sample Output 2:
Space
Sample Input 3:
AppLE
Sample Output 3:
Space, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
//Work
int main()
{
#ifndef ONLINE_JUDGE
if (fopen("INPUT.txt", "r"))
{
freopen ("INPUT.txt" , "r" , stdin);
//freopen ("OUTPUT.txt" , "w" , stdout);
}
#endif
//-----------------------------------------------------------------------------------------------------------//
string S;
cin>>S;
if(S=="Apple")
{
cout<<"Gravity"<<endl;
}
else
{
cout<<"Space"<<endl;
}
//-----------------------------------------------------------------------------------------------------------//
return 0;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Today is Newton School's first class of this year. Nutan, a student at Newton School, has received his first assignment. He will be given a string as input. His task is to print "Gravity'' if the input is "Apple''; otherwise, he will have to print "Space''.
Can you help Nutan in solving his first assignment? Note that the quotation marks are just for clarity. They are not part of the input string, and should not be a part of your output string.The input consists of a single line that contains a string S (1 ≤ length of S ≤ 10). The string only consists of lowercase and uppercase letters.Print "Gravity'' or "Space'' according to the input.Sample Input 1:
Apple
Sample Output 1:
Gravity
Sample Input 2:
Mango
Sample Output 2:
Space
Sample Input 3:
AppLE
Sample Output 3:
Space, I have written this Solution Code: n=input()
if n=='Apple':print('Gravity')
else:print('Space'), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given marks of a student in 5 subjects. You need to find the grade that a student would get on the basis of the percentage obtained. Grades computed are as follows:
If the percentage is >= 80 then print Grade ‘A’
If the percentage is <80 and >=60 then print Grade ‘B’
If the percentage is <60 and >=40 then print Grade ‘C’
else print Grade ‘D’The input contains 5 integers separated by spaces.
<b>Constraints:</b>
1 ≤ marks ≤ 100You need to print the grade obtained by a student.Sample Input:
75 70 80 90 100
Sample Output:
A
<b>Explanation</b>
((75+70+80+90+100)/5)*100=83%
A grade.
, I have written this Solution Code: import java.io.IOException;
import java.io.InputStreamReader;
import java.util.*;
import java.io.*;
public class Main {
static final int MOD = 1000000007;
public static void main(String args[]) throws IOException {
BufferedReader br
= new BufferedReader(new InputStreamReader(System.in));
String str[] = br.readLine().trim().split(" ");
int a = Integer.parseInt(str[0]);
int b = Integer.parseInt(str[1]);
int c = Integer.parseInt(str[2]);
int d = Integer.parseInt(str[3]);
int e = Integer.parseInt(str[4]);
System.out.println(grades(a, b, c, d, e));
}
static char grades(int a, int b, int c, int d, int e)
{
int sum = a+b+c+d+e;
int per = sum/5;
if(per >= 80)
return 'A';
else if(per >= 60)
return 'B';
else if(per >= 40)
return 'C';
else
return 'D';
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given marks of a student in 5 subjects. You need to find the grade that a student would get on the basis of the percentage obtained. Grades computed are as follows:
If the percentage is >= 80 then print Grade ‘A’
If the percentage is <80 and >=60 then print Grade ‘B’
If the percentage is <60 and >=40 then print Grade ‘C’
else print Grade ‘D’The input contains 5 integers separated by spaces.
<b>Constraints:</b>
1 ≤ marks ≤ 100You need to print the grade obtained by a student.Sample Input:
75 70 80 90 100
Sample Output:
A
<b>Explanation</b>
((75+70+80+90+100)/5)*100=83%
A grade.
, I have written this Solution Code: li = list(map(int,input().strip().split()))
avg=0
for i in li:
avg+=i
avg=avg/5
if(avg>=80):
print("A")
elif(avg>=60 and avg<80):
print("B")
elif(avg>=40 and avg<60):
print("C")
else:
print("D"), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a linked list consisting of <b>L</b> nodes and given a number <b>N</b>. The task is to find the Nth node from the end of the linked list.First line of input contains number of testcase T. For each testcase, first line of input contains number of nodes in the linked list L and the number N. Next line contains N nodes of linked list.
<b>User Task:</b>
The task is to complete the function <b>getNthFromLast()</b> which takes two arguments: reference to head and N and you need to return Nth from end.
<b>Constraints:</b>
1 <= T <= 100
1 <= N <= L <= 10^3
For each testcase, output the data of node which is at Nth distance from end.Input:
2
9 2
1 2 3 4 5 6 7 8 9
4 5
10 5 100 5
Output:
8
-1
Explanation:
Testcase 1: In the first example, there are 9 nodes in linked list and we need to find 2nd node from end. 2nd node from end os 8.
Testcase 2: In the second example, there are 4 nodes in linked list and we need to find 5th from end. Since 'n' is more than number of nodes in linked list, output is -1., I have written this Solution Code: static int getNthFromLast(Node head, int n)
{
int len = 0;
Node temp = head;
while(temp != null) // Traverse temp throught the linked list and find the length
{
temp = temp.next;
len++;
}
if(len < n)
return -1;
//System.out.println(count);
//int r = count - n;
temp = head;
for(int i=1; i<len-n+1; i++) // Traverse the node till the position from begining: length - n +1.
temp = temp.next;
return temp.data;
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a cubic dice with 6 faces. All the individual faces have a numbers printed on them. The numbers are in the range of 1 to 6, like any ordinary dice.
Given a number on one face of the dice , you need to print the number on the opposite face .
NOTE : In a normal dice sum of numbers on opposite faces is 7 .The first line of the input contains a single integer T, denoting the number of test cases. Then T test case follows. Each test case contains a single line of the input containing a positive integer N.
Constraints:
1 <= T <= 100
1 <= N <= 6For each testcase, print the number that is on the opposite side of the given face.Input:
2
6
2
Output:
1
5
Explanation:
Testcase 1: For dice facing number 6 opposite face will have the number 1., I have written this Solution Code: def get_opposite_face(n):
return 7-n
t = int(input())
for n in range(t):
print(get_opposite_face(int(input()))), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a cubic dice with 6 faces. All the individual faces have a numbers printed on them. The numbers are in the range of 1 to 6, like any ordinary dice.
Given a number on one face of the dice , you need to print the number on the opposite face .
NOTE : In a normal dice sum of numbers on opposite faces is 7 .The first line of the input contains a single integer T, denoting the number of test cases. Then T test case follows. Each test case contains a single line of the input containing a positive integer N.
Constraints:
1 <= T <= 100
1 <= N <= 6For each testcase, print the number that is on the opposite side of the given face.Input:
2
6
2
Output:
1
5
Explanation:
Testcase 1: For dice facing number 6 opposite face will have the number 1., I have written this Solution Code: #include "bits/stdc++.h"
#pragma GCC optimize "03"
using namespace std;
#define int long long int
#define ld long double
#define pi pair<int, int>
#define pb push_back
#define fi first
#define se second
#define IOS ios::sync_with_stdio(false); cin.tie(0); cout.tie(0)
#ifndef LOCAL
#define endl '\n'
#endif
const int N = 2e5 + 5;
const int mod = 1e9 + 7;
const int inf = 1e9 + 9;
signed main() {
IOS;
int t; cin >> t;
while(t--){
int n; cin >> n;
cout << 7-n << endl;
}
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a cubic dice with 6 faces. All the individual faces have a numbers printed on them. The numbers are in the range of 1 to 6, like any ordinary dice.
Given a number on one face of the dice , you need to print the number on the opposite face .
NOTE : In a normal dice sum of numbers on opposite faces is 7 .The first line of the input contains a single integer T, denoting the number of test cases. Then T test case follows. Each test case contains a single line of the input containing a positive integer N.
Constraints:
1 <= T <= 100
1 <= N <= 6For each testcase, print the number that is on the opposite side of the given face.Input:
2
6
2
Output:
1
5
Explanation:
Testcase 1: For dice facing number 6 opposite face will have the number 1., I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args)throws IOException {
BufferedReader br =new BufferedReader(new InputStreamReader(System.in));
int t=Integer.parseInt(br.readLine());
while(t-->0){
int n=Integer.parseInt(br.readLine());
System.out.println(6-n+1);
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer N, you have to print the given below pattern for N ≥ 3.
<b>Example</b>
Pattern for N = 4
*
*^*
*^^*
*****<b>User Task</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>Pattern()</b> that takes integers N as an argument.
<b>Constraints</b>
3 ≤ N ≤ 100Print the given pattern for size N.<b>Sample Input 1</b>
3
<b>Sample Output 1</b>
*
*^*
****
<b>Sample Input 2</b>
6
<b>Sample Output 2</b>
*
*^*
*^^*
*^^^*
*^^^^*
*******, I have written this Solution Code: def Pattern(N):
print('*')
for i in range (0,N-2):
print('*',end='')
for j in range (0,i+1):
print('^',end='')
print('*')
for i in range (0,N+1):
print('*',end='')
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer N, you have to print the given below pattern for N ≥ 3.
<b>Example</b>
Pattern for N = 4
*
*^*
*^^*
*****<b>User Task</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>Pattern()</b> that takes integers N as an argument.
<b>Constraints</b>
3 ≤ N ≤ 100Print the given pattern for size N.<b>Sample Input 1</b>
3
<b>Sample Output 1</b>
*
*^*
****
<b>Sample Input 2</b>
6
<b>Sample Output 2</b>
*
*^*
*^^*
*^^^*
*^^^^*
*******, I have written this Solution Code: void Pattern(int N){
cout<<'*'<<endl;
for(int i=0;i<N-2;i++){
cout<<'*';
for(int j=0;j<=i;j++){
cout<<'^';
}
cout<<'*'<<endl;
}
for(int i=0;i<=N;i++){
cout<<'*';
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer N, you have to print the given below pattern for N ≥ 3.
<b>Example</b>
Pattern for N = 4
*
*^*
*^^*
*****<b>User Task</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>Pattern()</b> that takes integers N as an argument.
<b>Constraints</b>
3 ≤ N ≤ 100Print the given pattern for size N.<b>Sample Input 1</b>
3
<b>Sample Output 1</b>
*
*^*
****
<b>Sample Input 2</b>
6
<b>Sample Output 2</b>
*
*^*
*^^*
*^^^*
*^^^^*
*******, I have written this Solution Code: static void Pattern(int N){
System.out.println('*');
for(int i=0;i<N-2;i++){
System.out.print('*');
for(int j=0;j<=i;j++){
System.out.print('^');
}System.out.println('*');
}
for(int i=0;i<=N;i++){
System.out.print('*');
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer N, you have to print the given below pattern for N ≥ 3.
<b>Example</b>
Pattern for N = 4
*
*^*
*^^*
*****<b>User Task</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>Pattern()</b> that takes integers N as an argument.
<b>Constraints</b>
3 ≤ N ≤ 100Print the given pattern for size N.<b>Sample Input 1</b>
3
<b>Sample Output 1</b>
*
*^*
****
<b>Sample Input 2</b>
6
<b>Sample Output 2</b>
*
*^*
*^^*
*^^^*
*^^^^*
*******, I have written this Solution Code: void Pattern(int N){
printf("*\n");
for(int i=0;i<N-2;i++){
printf("*");
for(int j=0;j<=i;j++){
printf("^");}printf("*\n");
}
for(int i=0;i<=N;i++){
printf("*");
}
}
, In this Programming Language: C, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of n integers. Print the second maximum difference between any two numbers in the array.First line contains n.
Next line contains n space separated integers.
<b>Constraints </b>
3 <= n <= 10<sup>5</sup>
1 <= arr[i] <= 10<sup>9</sup>A single integer denoting the answer.Input:
5
3 1 4 6 9
Output:
6
Explanation :
Difference between 1 and 9 is 8.
Difference between 3 and 9 is 6., 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());
int a[] = new int[n];
for(int i=0;i<n;i++){
a[i] = Integer.parseInt(in.next());
}
Arrays.sort(a);
int x=a[n-1] - a[1];
int y=a[n-2] - a[0];
out.print(Math.max(x,y));
out.close();
}
static class InputReader {
BufferedReader reader;
StringTokenizer tokenizer;
public InputReader(InputStream stream) {
reader = new BufferedReader(new InputStreamReader(stream), 32768);
tokenizer = null;
}
public String next() {
while (tokenizer == null || !tokenizer.hasMoreTokens()) {
try {
tokenizer = new StringTokenizer(reader.readLine());
} catch (IOException e) {
throw new RuntimeException(e);
}
}
return tokenizer.nextToken();
}
public int nextInt() {
return Integer.parseInt(next());
}
}
static class OutputWriter {
private final PrintWriter writer;
public OutputWriter(OutputStream outputStream) {
writer = new PrintWriter(new BufferedWriter(new OutputStreamWriter(outputStream)));
}
public OutputWriter(Writer writer) {
this.writer = new PrintWriter(writer);
}
public void print(Object... objects) {
for (int i = 0; i < objects.length; i++) {
if (i != 0) {
writer.print(' ');
}
writer.print(objects[i]);
}
}
public void println(Object... objects) {
print(objects);
writer.println();
}
public void close() {
writer.close();
}
public void println(int i) {
writer.println(i);
}
}
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Nobita wants to become rich so he came up with an idea, So, he buys some gadgets from the future at a price of C and sells them at a price of S to his friends. Now Nobita wants to know how much he gains by selling all gadget. As we all know Nobita is weak in maths help him to find the profit he getsYou don't have to worry about the input, you just have to complete the function <b>Profit()</b>
<b>Constraints:-</b>
1 <= C <= S <= 1000Print the profit Nobita gets from selling one gadget.Sample Input:-
3 5
Sample Output:-
2
Sample Input:-
9 16
Sample Output:-
7, I have written this Solution Code: def profit(C, S):
print(S - C), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Nobita wants to become rich so he came up with an idea, So, he buys some gadgets from the future at a price of C and sells them at a price of S to his friends. Now Nobita wants to know how much he gains by selling all gadget. As we all know Nobita is weak in maths help him to find the profit he getsYou don't have to worry about the input, you just have to complete the function <b>Profit()</b>
<b>Constraints:-</b>
1 <= C <= S <= 1000Print the profit Nobita gets from selling one gadget.Sample Input:-
3 5
Sample Output:-
2
Sample Input:-
9 16
Sample Output:-
7, I have written this Solution Code: static void Profit(int C, int S){
System.out.println(S-C);
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of N elements containing only 0 and 1. Find the length of the largest subarray with an equal number of 0's and 1's.
This problem was asked in Amazon.Each test case contains two lines. The first line of each test case is a number N denoting the size of the array and in the next line are N space-separated values of A [].
Constraints:-
1 < = N < = 100000Print the max length of the subarray.Sample input
4
0 1 0 1
Sample Output
4
Sample Input
5
0 0 1 0 0
Sample Output
2, I have written this Solution Code: def maxLen(arr, n):
hash_map = {}
curr_sum = 0
max_len = 0
ending_index = -1
for i in range (0, n):
if(arr[i] == 0):
arr[i] = -1
else:
arr[i] = 1
for i in range (0, n):
curr_sum = curr_sum + arr[i]
if (curr_sum == 0):
max_len = i + 1
ending_index = i
if curr_sum in hash_map:
if max_len < i - hash_map[curr_sum]:
max_len = i - hash_map[curr_sum]
ending_index = i
else:
hash_map[curr_sum] = i
for i in range (0, n):
if(arr[i] == -1):
arr[i] = 0
else:
arr[i] = 1
return max_len
a=input()
a=input().split()
for i in range(len(a)):
a[i]=int(a[i])
print(maxLen(a, len(a))), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of N elements containing only 0 and 1. Find the length of the largest subarray with an equal number of 0's and 1's.
This problem was asked in Amazon.Each test case contains two lines. The first line of each test case is a number N denoting the size of the array and in the next line are N space-separated values of A [].
Constraints:-
1 < = N < = 100000Print the max length of the subarray.Sample input
4
0 1 0 1
Sample Output
4
Sample Input
5
0 0 1 0 0
Sample Output
2, 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());
Map<Integer, Integer> map = new HashMap<>();
int arr[]=new int[n];
int sum=0;
int Largestsubarray=0;
String []b=br.readLine().split(" ");
boolean flag = false;
for(int i=0;i<n;i++){
int temp=Integer.parseInt(b[i]);
if(temp==0)
arr[i]=1;
else
arr[i]=-1;
sum=sum+arr[i];
if(sum == 0) {
Largestsubarray = i+1;
continue;
}
if(map.containsKey(sum)){
flag=true;
int CurrentLargestSubarray = i-map.get(sum);
if(CurrentLargestSubarray > Largestsubarray) {
Largestsubarray = CurrentLargestSubarray;
}
}
else
map.put(sum,i);
}
if(!flag)
System.out.print(-1);
else
System.out.print(Largestsubarray);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of N elements containing only 0 and 1. Find the length of the largest subarray with an equal number of 0's and 1's.
This problem was asked in Amazon.Each test case contains two lines. The first line of each test case is a number N denoting the size of the array and in the next line are N space-separated values of A [].
Constraints:-
1 < = N < = 100000Print the max length of the subarray.Sample input
4
0 1 0 1
Sample Output
4
Sample Input
5
0 0 1 0 0
Sample Output
2, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main(){
int n,k;
cin>>n;
k=0;
int a[n];
for(int i=0;i<n;i++)
{
cin>>a[i];
if(a[i]==0){
a[i]=-1;}
}
int sum=0;
int ans=0;
unordered_map<int,int> m;
for(int i=0;i<n;i++){
sum+=a[i];
if(sum==k){ans=i+1;}
if(m.find(sum-k)!=m.end()){
ans=max(ans,i-m[sum-k]);
}
if(m.find(sum)==m.end()){m[sum]=i;}
}
cout<<ans;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A of size N, you need to find its maximum, 2<sup>nd</sup> maximum and 3<sup>rd</sup> maximum element.
Try solving it in O(N) per test caseThe first line of the input contains the number of test cases T.
For each test case, the first line of the input contains an integer N denoting the number of elements in the array A. The following line contains N (space-separated) elements of A.
<b>Constraints:</b>
1 <= T <= 100
3 <= N <= 10<sup>6</sup>
1 <= A[i] <= 10<sup>9</sup>
<b>Note</b>:-It is guaranteed that the sum of N over all text cases does not exceed 10<sup>6</sup>For each test case, output the first, second and third maximum elements in the array.Sample Input:
3
5
1 4 2 4 5
6
1 3 5 7 9 8
7
11 22 33 44 55 66 77
Sample Output:
5 4 4
9 8 7
77 66 55
<b>Explanation:</b>
Testcase 1:
[1 4 2 4 5]
First max = 5
Second max = 4
Third max = 4, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) {
Scanner sc =new Scanner(System.in);
int T= sc.nextInt();
for(int i=0;i<T;i++){
int arrsize=sc.nextInt();
int max=0,secmax=0,thirdmax=0,j;
for(int k=0;k<arrsize;k++){
j=sc.nextInt();
if(j>max){
thirdmax=secmax;
secmax=max;
max=j;
}
else if(j>secmax){
thirdmax=secmax;
secmax=j;
}
else if(j>thirdmax){
thirdmax=j;
}
if(k%10000==0){
System.gc();
}
}
System.out.println(max+" "+secmax+" "+thirdmax+" ");
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A of size N, you need to find its maximum, 2<sup>nd</sup> maximum and 3<sup>rd</sup> maximum element.
Try solving it in O(N) per test caseThe first line of the input contains the number of test cases T.
For each test case, the first line of the input contains an integer N denoting the number of elements in the array A. The following line contains N (space-separated) elements of A.
<b>Constraints:</b>
1 <= T <= 100
3 <= N <= 10<sup>6</sup>
1 <= A[i] <= 10<sup>9</sup>
<b>Note</b>:-It is guaranteed that the sum of N over all text cases does not exceed 10<sup>6</sup>For each test case, output the first, second and third maximum elements in the array.Sample Input:
3
5
1 4 2 4 5
6
1 3 5 7 9 8
7
11 22 33 44 55 66 77
Sample Output:
5 4 4
9 8 7
77 66 55
<b>Explanation:</b>
Testcase 1:
[1 4 2 4 5]
First max = 5
Second max = 4
Third max = 4, I have written this Solution Code: t=int(input())
while t>0:
t-=1
n=int(input())
l=list(map(int,input().strip().split()))
li=[0,0,0]
for i in l:
x=i
for j in range(0,3):
y=min(x,li[j])
li[j]=max(x,li[j])
x=y
print(li[0],end=" ")
print(li[1],end=" ")
print(li[2]), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A of size N, you need to find its maximum, 2<sup>nd</sup> maximum and 3<sup>rd</sup> maximum element.
Try solving it in O(N) per test caseThe first line of the input contains the number of test cases T.
For each test case, the first line of the input contains an integer N denoting the number of elements in the array A. The following line contains N (space-separated) elements of A.
<b>Constraints:</b>
1 <= T <= 100
3 <= N <= 10<sup>6</sup>
1 <= A[i] <= 10<sup>9</sup>
<b>Note</b>:-It is guaranteed that the sum of N over all text cases does not exceed 10<sup>6</sup>For each test case, output the first, second and third maximum elements in the array.Sample Input:
3
5
1 4 2 4 5
6
1 3 5 7 9 8
7
11 22 33 44 55 66 77
Sample Output:
5 4 4
9 8 7
77 66 55
<b>Explanation:</b>
Testcase 1:
[1 4 2 4 5]
First max = 5
Second max = 4
Third max = 4, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main(){
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
int t;
cin>>t;
while(t--){
long long n;
cin>>n;
vector<long> a(n);
long ans[3]={0};
long x,y;
for(int i=0;i<n;i++){
cin>>a[i];
x=a[i];
for(int j=0;j<3;j++){
y=min(x,ans[j]);
ans[j]=max(x,ans[j]);
// cout<<ans[j]<<" ";
x=y;
}
}
if(ans[1]<ans[0]){
swap(ans[1],ans[0]);
}
if(ans[2]<ans[1]){
swap(ans[1],ans[2]);
}
if(ans[1]<ans[0]){
swap(ans[1],ans[0]);
}
cout<<ans[2]<<" "<<ans[1]<<" "<<ans[0]<<endl;
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A of size N, you need to find its maximum, 2<sup>nd</sup> maximum and 3<sup>rd</sup> maximum element.
Try solving it in O(N) per test caseThe first line of the input contains the number of test cases T.
For each test case, the first line of the input contains an integer N denoting the number of elements in the array A. The following line contains N (space-separated) elements of A.
<b>Constraints:</b>
1 <= T <= 100
3 <= N <= 10<sup>6</sup>
1 <= A[i] <= 10<sup>9</sup>
<b>Note</b>:-It is guaranteed that the sum of N over all text cases does not exceed 10<sup>6</sup>For each test case, output the first, second and third maximum elements in the array.Sample Input:
3
5
1 4 2 4 5
6
1 3 5 7 9 8
7
11 22 33 44 55 66 77
Sample Output:
5 4 4
9 8 7
77 66 55
<b>Explanation:</b>
Testcase 1:
[1 4 2 4 5]
First max = 5
Second max = 4
Third max = 4, I have written this Solution Code: function maxNumbers(arr,n) {
// write code here
// do not console.log the answer
// return the answer as an array of 3 numbers
return arr.sort((a,b)=>b-a).slice(0,3)
};
, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Palindrome is a word, phrase, or sequence that reads the same backwards as forwards. Use recursion to check if a given string is palindrome or not.User Task:
Since this is a functional problem, you don't have to worry about the input, you just have to complete the function check_Palindrome() where you will get input string, starting index of string (which is 0) and the end index of string( which is str.length-1) as argument.
Constraints:
1 ≤ T ≤ 100
1 ≤ N ≤ 10000Return true if given string is palindrome else return falseSample Input
2
ab
aba
Sample Output
false
true, I have written this Solution Code:
static boolean check_Palindrome(String str,int s, int e)
{
// If there is only one character
if (s == e)
return true;
// If first and last
// characters do not match
if ((str.charAt(s)) != (str.charAt(e)))
return false;
// If there are more than
// two characters, check if
// middle substring is also
// palindrome or not.
if (s < e + 1)
return check_Palindrome(str, s + 1, e - 1);
return true;
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array, sort the array in reverse order by simply swapping its adjacent elements.First line of the input contains an integer, N, which denotes the length of the array. Next N inputs are elements of the array that is to be sorted in descending order.
Constraints
1<=N<=1000
-10000<=Arr[i]<=100000Output sorted array in descending order where each element is space separated.Sample Input:
6
3 1 2 7 9 87
Sample Output:
87 9 7 3 2 1, I have written this Solution Code: function bubbleSort(arr, n) {
// write code here
// do not console.log the answer
// return sorted array
return arr.sort((a, b) => b - a)
}
, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array, sort the array in reverse order by simply swapping its adjacent elements.First line of the input contains an integer, N, which denotes the length of the array. Next N inputs are elements of the array that is to be sorted in descending order.
Constraints
1<=N<=1000
-10000<=Arr[i]<=100000Output sorted array in descending order where each element is space separated.Sample Input:
6
3 1 2 7 9 87
Sample Output:
87 9 7 3 2 1, I have written this Solution Code: def bubbleSort(arr):
arr.sort(reverse = True)
return arr
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array, sort the array in reverse order by simply swapping its adjacent elements.First line of the input contains an integer, N, which denotes the length of the array. Next N inputs are elements of the array that is to be sorted in descending order.
Constraints
1<=N<=1000
-10000<=Arr[i]<=100000Output sorted array in descending order where each element is space separated.Sample Input:
6
3 1 2 7 9 87
Sample Output:
87 9 7 3 2 1, I have written this Solution Code:
import java.util.*;
import java.lang.*;
import java.io.*;
class Main
{
public static void main (String[] args) throws java.lang.Exception
{
Scanner sc = new Scanner(System.in);
int n=sc.nextInt();
int a[] = new int[n];
for(int i=0;i<n;i++){
a[i]=sc.nextInt();
}
int t;
for(int i=1;i<n;i++){
if(a[i]>a[i-1]){
for(int j=i;j>0;j--){
if(a[j]>a[j-1]){
t=a[j];
a[j]=a[j-1];
a[j-1]=t;
}
}
}
}
for(int i=0;i<n;i++){
System.out.print(a[i]+" ");
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array, sort the array in reverse order by simply swapping its adjacent elements.First line of the input contains an integer, N, which denotes the length of the array. Next N inputs are elements of the array that is to be sorted in descending order.
Constraints
1<=N<=1000
-10000<=Arr[i]<=100000Output sorted array in descending order where each element is space separated.Sample Input:
6
3 1 2 7 9 87
Sample Output:
87 9 7 3 2 1, I have written this Solution Code:
// author-Shivam gupta
#include <bits/stdc++.h>
using namespace std;
#define MEM(a, b) memset(a, (b), sizeof(a))
#define FOREACH(it, l) for (auto it = l.begin(); it != l.end(); it++)
#define IN(A, B, C) assert( B <= A && A <= C)
#define MP make_pair
#define FOR(i,a) for(int i=0;i<a;i++)
#define FOR1(i,j,a) for(int i=j;i<a;i++)
#define EB emplace_back
#define INF (int)1e9
#define EPS 1e-9
#define PI 3.1415926535897932384626433832795
#define MOD 1000000007
const int N = 3e5+5;
#define read(type) readInt<type>()
#define max1 100001
#define out(x) cout<<x<<'\n'
#define out1(x) cout<<x<<" "
#define END cout<<'\n'
const double pi=acos(-1.0);
typedef pair<int, int> PII;
typedef vector<int> VI;
typedef vector<string> VS;
typedef vector<PII> VII;
typedef vector<VI> VVI;
typedef map<int,int> MPII;
typedef set<int> SETI;
typedef multiset<int> MSETI;
typedef long int li;
typedef unsigned long int uli;
typedef long long int ll;
typedef unsigned long long int ull;
const ll inf = 0x3f3f3f3f3f3f3f3f;
const ll mod = 998244353;
using vl = vector<ll>;
bool isPowerOfTwo (int x)
{
/* First x in the below expression is
for the case when x is 0 */
return x && (!(x&(x-1)));
}
void fast(){
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
}
ll power(ll x, ll y, ll p)
{
ll res = 1; // Initialize result
x = x % p; // Update x if it is more than or
// equal to p
while (y > 0)
{
// If y is odd, multiply x with result
if (y & 1)
res = (res*x) % p;
// y must be even now
y = y>>1; // y = y/2
x = (x*x) % p;
}
return res;
}
long long phi[max1], result[max1],F[max1];
// Precomputation of phi[] numbers. Refer below link
// for details : https://goo.gl/LUqdtY
void computeTotient()
{
// Refer https://goo.gl/LUqdtY
phi[1] = 1;
for (int i=2; i<max1; i++)
{
if (!phi[i])
{
phi[i] = i-1;
for (int j = (i<<1); j<max1; j+=i)
{
if (!phi[j])
phi[j] = j;
phi[j] = (phi[j]/i)*(i-1);
}
}
}
for(int i=1;i<=100000;i++)
{
for(int j=i;j<=100000;j+=i)
{ int p=j/i;
F[j]+=(i*phi[p])%mod;
F[j]%=mod;
}
}
}
int gcd(int a, int b, int& x, int& y) {
if (b == 0) {
x = 1;
y = 0;
return a;
}
int x1, y1;
int d = gcd(b, a % b, x1, y1);
x = y1;
y = x1 - y1 * (a / b);
return d;
}
bool find_any_solution(int a, int b, int c, int &x0, int &y0, int &g) {
g = gcd(abs(a), abs(b), x0, y0);
if (c % g) {
return false;
}
x0 *= c / g;
y0 *= c / g;
if (a < 0) x0 = -x0;
if (b < 0) y0 = -y0;
return true;
}
int main() {
int n;
cin>>n;
int a[n];
for(int i=0;i<n;i++){
cin>>a[i];
}
sort(a,a+n,greater<int>());
FOR(i,n){
out1(a[i]);}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara has A number of Pokeballs with her and there are B pokemons in front of Sara. Considering each pokemon takes one Pokeball, your task is to tell Sara if she can catch all the pokemons or not.
Sara can catch a pokemon if she is having at least one pokeball for that pokemon.<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>PokemonMaster()</b> that takes integers A and B as arguments.
Constraints:-
1 <= A, B <= 8Return 1 if Sara can catch all the pokemon else return 0.Sample Input:-
4 3
Sample Output:-
1
Sample Input:-
4 6
Sample Output:-
0, I have written this Solution Code: function PokemonMaster(a,b) {
// write code here
// do no console.log the answer
// return the output using return keyword
const ans = (a - b >= 0) ? 1 : 0
return ans
}, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara has A number of Pokeballs with her and there are B pokemons in front of Sara. Considering each pokemon takes one Pokeball, your task is to tell Sara if she can catch all the pokemons or not.
Sara can catch a pokemon if she is having at least one pokeball for that pokemon.<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>PokemonMaster()</b> that takes integers A and B as arguments.
Constraints:-
1 <= A, B <= 8Return 1 if Sara can catch all the pokemon else return 0.Sample Input:-
4 3
Sample Output:-
1
Sample Input:-
4 6
Sample Output:-
0, I have written this Solution Code: def PokemonMaster(A,B):
if(A>=B):
return 1
return 0
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara has A number of Pokeballs with her and there are B pokemons in front of Sara. Considering each pokemon takes one Pokeball, your task is to tell Sara if she can catch all the pokemons or not.
Sara can catch a pokemon if she is having at least one pokeball for that pokemon.<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>PokemonMaster()</b> that takes integers A and B as arguments.
Constraints:-
1 <= A, B <= 8Return 1 if Sara can catch all the pokemon else return 0.Sample Input:-
4 3
Sample Output:-
1
Sample Input:-
4 6
Sample Output:-
0, I have written this Solution Code: int PokemonMaster(int A, int B){
if(A>=B){return 1;}
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara has A number of Pokeballs with her and there are B pokemons in front of Sara. Considering each pokemon takes one Pokeball, your task is to tell Sara if she can catch all the pokemons or not.
Sara can catch a pokemon if she is having at least one pokeball for that pokemon.<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>PokemonMaster()</b> that takes integers A and B as arguments.
Constraints:-
1 <= A, B <= 8Return 1 if Sara can catch all the pokemon else return 0.Sample Input:-
4 3
Sample Output:-
1
Sample Input:-
4 6
Sample Output:-
0, I have written this Solution Code: int PokemonMaster(int A, int B){
if(A>=B){return 1;}
return 0;
}, In this Programming Language: C, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara has A number of Pokeballs with her and there are B pokemons in front of Sara. Considering each pokemon takes one Pokeball, your task is to tell Sara if she can catch all the pokemons or not.
Sara can catch a pokemon if she is having at least one pokeball for that pokemon.<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>PokemonMaster()</b> that takes integers A and B as arguments.
Constraints:-
1 <= A, B <= 8Return 1 if Sara can catch all the pokemon else return 0.Sample Input:-
4 3
Sample Output:-
1
Sample Input:-
4 6
Sample Output:-
0, I have written this Solution Code: static int PokemonMaster(int A, int B){
if(A>=B){return 1;}
return 0;
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given the number N. You need to convert it to 1 in the minimum number of operations.
The operations allowed are as follows:
1. If N is even then divide the number by 2.
2. If N is odd then you can either add 1 to it or subtract 1 from it.
Using the above operations, find the minimum number of operations required to convert N to 1.The first line of input contains an integer T denoting the number of test cases. T test cases follow. Each test case contains 1 line of input containing integer N.
<b>Constraints:</b>
1 ≤ T ≤ 100
1 ≤ N ≤ 10<sup>7</sup>
For each testcase, in a new line, print the minimum number of steps required.Sample Input:
4
1
2
3
4
Sample Output:
0
1
2
2
<b>Explanation:</b>
1 can be converted into 1 in 0 steps.
2 can be converted into 1 in 1 step: 2/2=1
3 can be converted into 1 in 3 steps: 3-1= 2 then 2/2=1
4 can be converted into 1 in 2 steps: 4/2=2 then 2/2=1, I have written this Solution Code: def stepCount(n):
count = 0
while n > 1:
if n % 2 == 0:
n = n // 2
elif n == 3 or n % 4 == 1:
n = n - 1
else:
n = n + 1
count += 1
return count
t=int(input())
for _ in range(t):
n=int(input())
print(stepCount(n)), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given the number N. You need to convert it to 1 in the minimum number of operations.
The operations allowed are as follows:
1. If N is even then divide the number by 2.
2. If N is odd then you can either add 1 to it or subtract 1 from it.
Using the above operations, find the minimum number of operations required to convert N to 1.The first line of input contains an integer T denoting the number of test cases. T test cases follow. Each test case contains 1 line of input containing integer N.
<b>Constraints:</b>
1 ≤ T ≤ 100
1 ≤ N ≤ 10<sup>7</sup>
For each testcase, in a new line, print the minimum number of steps required.Sample Input:
4
1
2
3
4
Sample Output:
0
1
2
2
<b>Explanation:</b>
1 can be converted into 1 in 0 steps.
2 can be converted into 1 in 1 step: 2/2=1
3 can be converted into 1 in 3 steps: 3-1= 2 then 2/2=1
4 can be converted into 1 in 2 steps: 4/2=2 then 2/2=1, I have written this Solution Code: #include<bits/stdc++.h>
using namespace std;
#define int long long
#define f(i,n) for(int i=0;i<(n);++i)
#define fA(i,a,n) for(int i=a;i<=(n);++i)
#define fD(i,a,n) for(int i=a;i>=(n);--i)
#define tc int t;cin>>t;f(testcase,t)
#define pii pair<int,int>
void c_p_c() {
ios_base::sync_with_stdio(false);
cin.tie(0); cout.tie(0);
#ifndef ONLINE_JUDGE
freopen("inputf.in", "r", stdin);
freopen("outputf.in", "w", stdout);
#endif
}
int dp[10000001];
int minsteps(int n , int dp[]) {
if (n == 1)
return 0;
if (dp[n] != 0)
return dp[n];
int op1, op2, op3;
op1 = op2 = op3 = INT_MAX;
if (n % 2 == 0) {
op1 = 1 + minsteps(n / 2, dp);
}
else {
op2 = 1 + minsteps(n - 1, dp);
op3 = 1 + minsteps(n + 1, dp);
}
int ans = min(op1, min(op2, op3));
dp[n] = ans;
return dp[n];
}
int32_t main() {
memset(dp, 0, sizeof(dp));
tc {
int n;
cin >> n;
cout << minsteps(n, dp) << "\n";
}
return 0;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given the number N. You need to convert it to 1 in the minimum number of operations.
The operations allowed are as follows:
1. If N is even then divide the number by 2.
2. If N is odd then you can either add 1 to it or subtract 1 from it.
Using the above operations, find the minimum number of operations required to convert N to 1.The first line of input contains an integer T denoting the number of test cases. T test cases follow. Each test case contains 1 line of input containing integer N.
<b>Constraints:</b>
1 ≤ T ≤ 100
1 ≤ N ≤ 10<sup>7</sup>
For each testcase, in a new line, print the minimum number of steps required.Sample Input:
4
1
2
3
4
Sample Output:
0
1
2
2
<b>Explanation:</b>
1 can be converted into 1 in 0 steps.
2 can be converted into 1 in 1 step: 2/2=1
3 can be converted into 1 in 3 steps: 3-1= 2 then 2/2=1
4 can be converted into 1 in 2 steps: 4/2=2 then 2/2=1, I have written this Solution Code: import java.lang.*;
import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) {
Scanner sc=new Scanner(System.in);
int testcases=sc.nextInt();
while(testcases-->0)
{
long n=sc.nextLong();
System.out.println(NS.minOperations(n));
}
}
}
class NS
{
public static long minOperations(long n)
{
if(n==1)
return 0; //since 1 is already 1
if(n==2)
return 1; //convert 2 to 1. 1 step
if(n==3)
return 2; //convert 3 to 2. Then 2 to 1. 2 steps
long total=0; //save total steps
if(n%2!=0) //if odd
{
total=1+Math.min(minOperations(n-1),minOperations(n+1));
//convert n to n-1 or n+1 then minimum of those conversions
}
else
total=1+minOperations(n/2); //convert n to n/2 then count operations required for n/2 to 1
return total; //returning total at the end
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given three positive integers A, B and C. If all the three integers are equal, print "Yes". Otherwise, print "No".
Note that the quotation marks are not part of the string, thus you should print your answer without the quotation marks.The input consists of three space separated integers A, B and C.
Constraints:
1 ≤ A, B, C ≤ 100Print a single word "Yes" if all the integers are equal, otherwise print "No" (without the quotes). Note that the output is case- sensitive.Sample Input 1:
1 3 1
Sample Output 1:
No
Sample Input 2:
5 5 5
Sample Output 2:
Yes, 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));
String[] input = in.readLine().split(" ");
int a = Integer.parseInt(input[0]);
int b = Integer.parseInt(input[1]);
int c = Integer.parseInt(input[2]);
if(a==b && b==c && c == a) {
System.out.println("Yes");
} else {
System.out.println("No");
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given three positive integers A, B and C. If all the three integers are equal, print "Yes". Otherwise, print "No".
Note that the quotation marks are not part of the string, thus you should print your answer without the quotation marks.The input consists of three space separated integers A, B and C.
Constraints:
1 ≤ A, B, C ≤ 100Print a single word "Yes" if all the integers are equal, otherwise print "No" (without the quotes). Note that the output is case- sensitive.Sample Input 1:
1 3 1
Sample Output 1:
No
Sample Input 2:
5 5 5
Sample Output 2:
Yes, I have written this Solution Code: a,b,c=map(int,input().split())
if a==b and b==c:
print("Yes")
else:
print("No"), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given three positive integers A, B and C. If all the three integers are equal, print "Yes". Otherwise, print "No".
Note that the quotation marks are not part of the string, thus you should print your answer without the quotation marks.The input consists of three space separated integers A, B and C.
Constraints:
1 ≤ A, B, C ≤ 100Print a single word "Yes" if all the integers are equal, otherwise print "No" (without the quotes). Note that the output is case- sensitive.Sample Input 1:
1 3 1
Sample Output 1:
No
Sample Input 2:
5 5 5
Sample Output 2:
Yes, I have written this Solution Code: #include <iostream>
using namespace std;
int main() {
int a, b, c;
cin >> a >> b >> c;
if (a == b && b == c) cout << "Yes";
else cout << "No";
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Nutan and Tusla are both students at Newton School. They are both among the best students in the class. In order to know who is better among them, a game was organised. The game consisted of L rounds, where L is an odd integer. The student winning more rounds than the other was declared the winner.
You would be given a string of odd length L in which each character is 'N' or 'T'. If the i<sup>th</sup> character is 'N', then the i<sup>th</sup> round was won by Nutan, else if the character is 'T' it was won by Tusla. Print "Nutan'' if Nutan has won more rounds than Tusla, else print "Tusla'' if Tusla has won more rounds than Nutan.
Note: You have to print everything without quotes.The first line of the input contains a single integer L — the number of rounds (1 ≤ L ≤ 100 and L is odd).
The second line contains a string S of length L. Each character of S is either 'N' or 'T'.Print "Nutan" or "Tusla" according to the input.Sample Input:
3
NNT
Sample Output:
Nutan
Explanation:
Nutan has won two games while Tusla has only won a single game, so the overall winner is Nutan., I have written this Solution Code: import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;
import static java.lang.System.out;
public class Main {
static class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader() {
br = new BufferedReader(new InputStreamReader(System.in));
}
String next() {
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
long nextLong() {
return Long.parseLong(next());
}
double nextDouble() {
return Double.parseDouble(next());
}
String nextLine() {
String str = "";
try {
str = br.readLine();
} catch (IOException e) {
e.printStackTrace();
}
return str;
}
}
public static void main(String[] args) {
FastReader reader = new FastReader();
int n = reader.nextInt();
String S = reader.next();
int ncount = 0;
int tcount = 0;
for (char c : S.toCharArray()) {
if (c == 'N') ncount++;
else tcount++;
}
if (ncount > tcount) {
out.print("Nutan\n");
} else {
out.print("Tusla\n");
}
out.flush();
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Nutan and Tusla are both students at Newton School. They are both among the best students in the class. In order to know who is better among them, a game was organised. The game consisted of L rounds, where L is an odd integer. The student winning more rounds than the other was declared the winner.
You would be given a string of odd length L in which each character is 'N' or 'T'. If the i<sup>th</sup> character is 'N', then the i<sup>th</sup> round was won by Nutan, else if the character is 'T' it was won by Tusla. Print "Nutan'' if Nutan has won more rounds than Tusla, else print "Tusla'' if Tusla has won more rounds than Nutan.
Note: You have to print everything without quotes.The first line of the input contains a single integer L — the number of rounds (1 ≤ L ≤ 100 and L is odd).
The second line contains a string S of length L. Each character of S is either 'N' or 'T'.Print "Nutan" or "Tusla" according to the input.Sample Input:
3
NNT
Sample Output:
Nutan
Explanation:
Nutan has won two games while Tusla has only won a single game, so the overall winner is Nutan., I have written this Solution Code: n = int(input())
s = input()
a1 = s.count('N')
a2 = s.count('T')
if(a1 > a2):
print("Nutan")
else:
print('Tusla'), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Nutan and Tusla are both students at Newton School. They are both among the best students in the class. In order to know who is better among them, a game was organised. The game consisted of L rounds, where L is an odd integer. The student winning more rounds than the other was declared the winner.
You would be given a string of odd length L in which each character is 'N' or 'T'. If the i<sup>th</sup> character is 'N', then the i<sup>th</sup> round was won by Nutan, else if the character is 'T' it was won by Tusla. Print "Nutan'' if Nutan has won more rounds than Tusla, else print "Tusla'' if Tusla has won more rounds than Nutan.
Note: You have to print everything without quotes.The first line of the input contains a single integer L — the number of rounds (1 ≤ L ≤ 100 and L is odd).
The second line contains a string S of length L. Each character of S is either 'N' or 'T'.Print "Nutan" or "Tusla" according to the input.Sample Input:
3
NNT
Sample Output:
Nutan
Explanation:
Nutan has won two games while Tusla has only won a single game, so the overall winner is Nutan., I have written this Solution Code: //Author: Xzirium
//Time and Date: 02:18:28 24 March 2022
//Optional FAST
//#pragma GCC optimize("Ofast")
//#pragma GCC optimize("unroll-loops")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,fma,abm,mmx,avx,avx2,tune=native")
//Required Libraries
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/detail/standard_policies.hpp>
//Required namespaces
using namespace std;
using namespace __gnu_pbds;
//Required defines
#define endl '\n'
#define READ(X) cin>>X;
#define READV(X) long long X; cin>>X;
#define READAR(A,N) long long A[N]; for(long long i=0;i<N;i++) {cin>>A[i];}
#define rz(A,N) A.resize(N);
#define sz(X) (long long)(X.size())
#define pb push_back
#define pf push_front
#define fi first
#define se second
#define FORI(a,b,c) for(long long a=b;a<c;a++)
#define FORD(a,b,c) for(long long a=b;a>c;a--)
//Required typedefs
template <typename T> using ordered_set = tree<T,null_type,less<T>,rb_tree_tag,tree_order_statistics_node_update>;
template <typename T> using ordered_set1 = tree<T,null_type,greater<T>,rb_tree_tag,tree_order_statistics_node_update>;
typedef long long ll;
typedef long double ld;
typedef pair<int,int> pii;
typedef pair<long long,long long> pll;
//Required Constants
const long long inf=(long long)1e18;
const long long MOD=(long long)(1e9+7);
const long long INIT=(long long)(1e6+1);
const long double PI=3.14159265358979;
// Required random number generators
// mt19937 gen_rand_int(chrono::steady_clock::now().time_since_epoch().count());
// mt19937_64 gen_rand_ll(chrono::steady_clock::now().time_since_epoch().count());
//Required Functions
ll power(ll b, ll e)
{
ll r = 1ll;
for(; e > 0; e /= 2, (b *= b) %= MOD)
if(e % 2) (r *= b) %= MOD;
return r;
}
ll modInverse(ll a)
{
return power(a,MOD-2);
}
//Work
int main()
{
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
//-----------------------------------------------------------------------------------------------------------//
READV(N);
string S;
cin>>S;
ll n=0,t=0;
FORI(i,0,N)
{
if(S[i]=='N')
{
n++;
}
else if(S[i]=='T')
{
t++;
}
}
if(n>t)
{
cout<<"Nutan"<<endl;
}
else
{
cout<<"Tusla"<<endl;
}
//-----------------------------------------------------------------------------------------------------------//
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer N and M check if N candies can be divided in M people such that each person get equal number of candies.Input contains two integers N and M.
Constraints:
1 <= N <= 10^18
1 <= M <= 10^18Print "Yes" if it is possible otherwise "No".Sample Input
10 5
Sample Output
Yes
Explanation: Give 2 candies to all.
Sample Input:
4 3
Sample Output:
No, I have written this Solution Code: m,n = map(int , input().split())
if (m%n==0):
print("Yes")
else:
print("No");, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer N and M check if N candies can be divided in M people such that each person get equal number of candies.Input contains two integers N and M.
Constraints:
1 <= N <= 10^18
1 <= M <= 10^18Print "Yes" if it is possible otherwise "No".Sample Input
10 5
Sample Output
Yes
Explanation: Give 2 candies to all.
Sample Input:
4 3
Sample Output:
No, I have written this Solution Code: #pragma GCC optimize ("Ofast")
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define VV vector
#define pb push_back
#define bitc __builtin_popcountll
#define m_p make_pair
#define infi 1e18+1
#define eps 0.000000000001
#define fastio ios_base::sync_with_stdio(false);cin.tie(NULL);
string char_to_str(char c){string tem(1,c);return tem;}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
template<class T>//usage rand<long long>()
T rand() {
return uniform_int_distribution<T>()(rng);
}
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
template<class T>
using oset = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
// string to integer stoi()
// string to long long stoll()
// string.substr(position,length);
// integer to string to_string();
//////////////
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,m;
cin>>n>>m;
if(n%m==0)
cout<<"Yes";
else
cout<<"No";
#ifdef ANIKET_GOYAL
// cout<<endl<<endl<<endl<<endl<<"Time elapsed: "<<(double)(clock()-clk)/CLOCKS_PER_SEC<<endl;
#endif
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer N and M check if N candies can be divided in M people such that each person get equal number of candies.Input contains two integers N and M.
Constraints:
1 <= N <= 10^18
1 <= M <= 10^18Print "Yes" if it is possible otherwise "No".Sample Input
10 5
Sample Output
Yes
Explanation: Give 2 candies to all.
Sample Input:
4 3
Sample Output:
No, I have written this Solution Code:
import java.util.*;
import java.lang.*;
import java.io.*;
class Main
{
public static void main (String[] args) throws java.lang.Exception
{
Scanner sc = new Scanner(System.in);
long n = sc.nextLong();
Long m = sc.nextLong();
if(n%m==0){
System.out.print("Yes");
}
else{
System.out.print("No");
}
}
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a string S. The task is to print all permutations of the characters in the given string.The only line of input contains a string S with all distinct uppercase letters of the English alphabet.
Constraints:-
1<=|S|<=8Print all permutations of a given string S with single space and all permutations should be in lexicographically increasing order.Sample Input:
ABC
Sample Output:
ABC ACB BAC BCA CAB CBA, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
static ArrayList<String> solution = new ArrayList<String>();
public static char[] swap(char[] charArray, int i, int j){
char temp;
temp = charArray[i];
charArray[i] = charArray[j];
charArray[j] = temp;
return charArray;
}
public static void permute(char[] charArray, int left, int right){
if(left == right){
String str = new String(charArray);
solution.add(str);
}
else{
for(int i=left; i<=right; i++)
{
charArray = swap(charArray, left, i);
permute(charArray, left+1, right);
charArray = swap(charArray, left, i);
}
}
}
public static void main(String[] args)
{
Scanner s = new Scanner(System.in);
String str = s.nextLine();
int n = str.length();
char[] charArray = str.toCharArray();
permute(charArray, 0, n-1);
Collections.sort(solution);
for(int i = 0; i < solution.size(); i++){
System.out.print(solution.get(i)+" ");
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a string S. The task is to print all permutations of the characters in the given string.The only line of input contains a string S with all distinct uppercase letters of the English alphabet.
Constraints:-
1<=|S|<=8Print all permutations of a given string S with single space and all permutations should be in lexicographically increasing order.Sample Input:
ABC
Sample Output:
ABC ACB BAC BCA CAB CBA, I have written this Solution Code: #include "bits/stdc++.h"
#pragma GCC optimize "03"
using namespace std;
#define int long long int
#define ld long double
#define pi pair<int, int>
#define pb push_back
#define fi first
#define se second
#define IOS ios::sync_with_stdio(false); cin.tie(0); cout.tie(0)
#ifndef LOCAL
#define endl '\n'
#endif
const int N = 2e5 + 5;
const int mod = 1e9 + 7;
const int inf = 1e9 + 9;
vector<string> v;
void permute(string a, int l, int r)
{
// Base case
if (l == r)
v.push_back(a);
else
{
// Permutations made
for (int i = l; i <= r; i++)
{
// Swapping done
swap(a[l], a[i]);
// Recursion called
permute(a, l+1, r);
//backtrack
swap(a[l], a[i]);
}
}
}
signed main() {
IOS;
string s;
cin >> s;
sort(s.begin(), s.end());
permute(s, 0, s.length()-1);
sort(v.begin(),v.end());
for(int i=0;i<v.size();i++)
cout<<v[i]<<" ";
return 0;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a string S. The task is to print all permutations of the characters in the given string.The only line of input contains a string S with all distinct uppercase letters of the English alphabet.
Constraints:-
1<=|S|<=8Print all permutations of a given string S with single space and all permutations should be in lexicographically increasing order.Sample Input:
ABC
Sample Output:
ABC ACB BAC BCA CAB CBA, I have written this Solution Code: from itertools import permutations
s = input()
l1 = list(s)
s1 = "".join(sorted(l1))
l = permutations(s1,len(s1))
for i in l:
print("".join(i),end=" "), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Today is Newton School's first class of this year. Nutan, a student at Newton School, has received his first assignment. He will be given a string as input. His task is to print "Gravity'' if the input is "Apple''; otherwise, he will have to print "Space''.
Can you help Nutan in solving his first assignment? Note that the quotation marks are just for clarity. They are not part of the input string, and should not be a part of your output string.The input consists of a single line that contains a string S (1 ≤ length of S ≤ 10). The string only consists of lowercase and uppercase letters.Print "Gravity'' or "Space'' according to the input.Sample Input 1:
Apple
Sample Output 1:
Gravity
Sample Input 2:
Mango
Sample Output 2:
Space
Sample Input 3:
AppLE
Sample Output 3:
Space, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
static class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader()
{
br = new BufferedReader(
new InputStreamReader(System.in));
}
String next()
{
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
}
catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() { return Integer.parseInt(next()); }
long nextLong() { return Long.parseLong(next()); }
double nextDouble()
{
return Double.parseDouble(next());
}
String nextLine()
{
String str = "";
try {
str = br.readLine();
}
catch (IOException e) {
e.printStackTrace();
}
return str;
}
}
public static void main (String[] args) {
FastReader sc= new FastReader();
String str= sc.nextLine();
String a="Apple";
if(a.equals(str)){
System.out.println("Gravity");
}
else{
System.out.println("Space");
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Today is Newton School's first class of this year. Nutan, a student at Newton School, has received his first assignment. He will be given a string as input. His task is to print "Gravity'' if the input is "Apple''; otherwise, he will have to print "Space''.
Can you help Nutan in solving his first assignment? Note that the quotation marks are just for clarity. They are not part of the input string, and should not be a part of your output string.The input consists of a single line that contains a string S (1 ≤ length of S ≤ 10). The string only consists of lowercase and uppercase letters.Print "Gravity'' or "Space'' according to the input.Sample Input 1:
Apple
Sample Output 1:
Gravity
Sample Input 2:
Mango
Sample Output 2:
Space
Sample Input 3:
AppLE
Sample Output 3:
Space, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
//Work
int main()
{
#ifndef ONLINE_JUDGE
if (fopen("INPUT.txt", "r"))
{
freopen ("INPUT.txt" , "r" , stdin);
//freopen ("OUTPUT.txt" , "w" , stdout);
}
#endif
//-----------------------------------------------------------------------------------------------------------//
string S;
cin>>S;
if(S=="Apple")
{
cout<<"Gravity"<<endl;
}
else
{
cout<<"Space"<<endl;
}
//-----------------------------------------------------------------------------------------------------------//
return 0;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Today is Newton School's first class of this year. Nutan, a student at Newton School, has received his first assignment. He will be given a string as input. His task is to print "Gravity'' if the input is "Apple''; otherwise, he will have to print "Space''.
Can you help Nutan in solving his first assignment? Note that the quotation marks are just for clarity. They are not part of the input string, and should not be a part of your output string.The input consists of a single line that contains a string S (1 ≤ length of S ≤ 10). The string only consists of lowercase and uppercase letters.Print "Gravity'' or "Space'' according to the input.Sample Input 1:
Apple
Sample Output 1:
Gravity
Sample Input 2:
Mango
Sample Output 2:
Space
Sample Input 3:
AppLE
Sample Output 3:
Space, I have written this Solution Code: n=input()
if n=='Apple':print('Gravity')
else:print('Space'), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: A multi-set S of N integers, along with an integer K, is provided to Geetha. Given that Geetha can only add a maximum of K integers to the multi-set, she needs to determine the MEX, or minimal excluded non-negative integer, of the multi-set. Help her to determine the highest value of MEX she can get. Here are a few examples of finding the MEX of a multi-set. MEX of multi-set {0} is 1, {1} is 0, {0, 1, 3} is 2, {0, 1, 2, 3, 5, 6} is 4.The first line of the input contains an integer T denoting the number of test cases. The first line of each test case contains two space-separated integers N and K denoting the size of the multi-set and the maximum number of extra integers that you can add to the multi-set respectively. The second line contains N space-separated integers denoting the multi-set S: S<sub>1</sub>, S<sub>2</sub>,..... S<sub>N</sub>.
<b>Constraints</b>
1 ≤ T ≤ 10
1 ≤ N ≤ 10<sup>5</sup>
0 ≤ K ≤ 10<sup>5</sup>
0 ≤ S<sub>i</sub> ≤ 2x10<sup>5</sup>For each test case, output the answer in a single line.Sample Input :
4
3 0
1 0 2
3 1
1 0 2
4 3
2 5 4 9
2 0
3 4
Sample Output :
3
4
6
0
Explanation :
<ul><li>As K = 0, so we can't add any element to the multi-set. Elements of the set are {1, 0, 2}. The MEX value of this set is 3. </li><li>As K = 1, you are allowed to add at most 1 element to the multi-set. The multi-set is {1, 0, 2}. You can add element 3 to the multi-set, and it becomes {1, 0, 2, 3}. The MEX value of this multi-set is 4. There is no other way to have a higher value of MEX of the set by adding at most one element to the multi-set. </li></ul>, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main() {
int t;
cin>>t;
while(t--){
int n,k;
cin>>n>>k;
int arr[n];
for(int i=0;i<n;i++){
cin>>arr[i];
}
sort(arr,arr+n);
int cnt=-1;
for(int i=0;i<n;i++){
if(cnt<arr[i]-1){
if(k>0){
while(cnt<arr[i]-1 && k>0){
cnt++;
k--;
}
}
else
break;
}
if(cnt ==arr[i]-1){
cnt++;
}
}
cout<<(cnt+k+1)<<endl;
}
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: A multi-set S of N integers, along with an integer K, is provided to Geetha. Given that Geetha can only add a maximum of K integers to the multi-set, she needs to determine the MEX, or minimal excluded non-negative integer, of the multi-set. Help her to determine the highest value of MEX she can get. Here are a few examples of finding the MEX of a multi-set. MEX of multi-set {0} is 1, {1} is 0, {0, 1, 3} is 2, {0, 1, 2, 3, 5, 6} is 4.The first line of the input contains an integer T denoting the number of test cases. The first line of each test case contains two space-separated integers N and K denoting the size of the multi-set and the maximum number of extra integers that you can add to the multi-set respectively. The second line contains N space-separated integers denoting the multi-set S: S<sub>1</sub>, S<sub>2</sub>,..... S<sub>N</sub>.
<b>Constraints</b>
1 ≤ T ≤ 10
1 ≤ N ≤ 10<sup>5</sup>
0 ≤ K ≤ 10<sup>5</sup>
0 ≤ S<sub>i</sub> ≤ 2x10<sup>5</sup>For each test case, output the answer in a single line.Sample Input :
4
3 0
1 0 2
3 1
1 0 2
4 3
2 5 4 9
2 0
3 4
Sample Output :
3
4
6
0
Explanation :
<ul><li>As K = 0, so we can't add any element to the multi-set. Elements of the set are {1, 0, 2}. The MEX value of this set is 3. </li><li>As K = 1, you are allowed to add at most 1 element to the multi-set. The multi-set is {1, 0, 2}. You can add element 3 to the multi-set, and it becomes {1, 0, 2, 3}. The MEX value of this multi-set is 4. There is no other way to have a higher value of MEX of the set by adding at most one element to the multi-set. </li></ul>, I have written this Solution Code: import java.util.*;
import java.lang.*;
import java.io.*;
/* Name of the class has to be "Main" only if the class is public. */
class Main
{
public static void main (String[] args) throws java.lang.Exception
{
// your code goes here
Scanner sc=new Scanner(System.in);
int t=sc.nextInt();
while(t-->0){
int n=sc.nextInt();
int k=sc.nextInt();
int kk=k;
HashSet<Integer> hs=new HashSet<>();
int max=0;
for(int i=0;i<n;i++){
int entry=sc.nextInt();
hs.add(entry);
}
int i=0;
System.gc();
for(;i<=n+kk;i++){
if(!hs.contains(i)){
//System.out.println(i+" "+k);
if(k==0)
break;
else
k--;
}
}
System.out.println(i);
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A[], size N containing positive integers. You need to arrange the elements of array in increasing order using selection sort.First line of the input denotes number of test cases 'T'. First line of the test case is the size of array and second line consists of array elements.
Constraints:
1 <= T <= 100
1 <= N <= 10^3
1 <= A[i] <= 10^3For each testcase print the sorted array in a new line.Input:
2
5
4 1 3 9 7
10
10 9 8 7 6 5 4 3 2 1
Output:
1 3 4 7 9
1 2 3 4 5 6 7 8 9 10, 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 n;
while(t>0) {
n = Integer.parseInt(br.readLine());
int a[] = new int[n];
String s = br.readLine();
String arr[] = s.split(" ");
for(int i=0;i<n;i++) {
a[i] = Integer.parseInt(arr[i]);
}
int b[] = sort(a);
for(int i=0;i<n;i++) {
System.out.print(b[i] + " ");
}
System.out.println();
t--;
}
}
static int[] sort(int[] a) {
int temp, index;
for(int i=0;i<a.length-1;i++) {
index = i;
for(int j=i+1;j<a.length;j++) {
if(a[j]<a[index]) {
index = j;
}
}
temp = a[i];
a[i] = a[index];
a[index] = temp;
}
return a;
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A[], size N containing positive integers. You need to arrange the elements of array in increasing order using selection sort.First line of the input denotes number of test cases 'T'. First line of the test case is the size of array and second line consists of array elements.
Constraints:
1 <= T <= 100
1 <= N <= 10^3
1 <= A[i] <= 10^3For each testcase print the sorted array in a new line.Input:
2
5
4 1 3 9 7
10
10 9 8 7 6 5 4 3 2 1
Output:
1 3 4 7 9
1 2 3 4 5 6 7 8 9 10, I have written this Solution Code: for _ in range(int(input())):
n = input()
res = map(str, sorted(list( map(int,input().split()))))
print(' '.join(res)), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A[], size N containing positive integers. You need to arrange the elements of array in increasing order using selection sort.First line of the input denotes number of test cases 'T'. First line of the test case is the size of array and second line consists of array elements.
Constraints:
1 <= T <= 100
1 <= N <= 10^3
1 <= A[i] <= 10^3For each testcase print the sorted array in a new line.Input:
2
5
4 1 3 9 7
10
10 9 8 7 6 5 4 3 2 1
Output:
1 3 4 7 9
1 2 3 4 5 6 7 8 9 10, I have written this Solution Code: #include "bits/stdc++.h"
#pragma GCC optimize "03"
using namespace std;
#define int long long int
#define ld long double
#define pi pair<int, int>
#define pb push_back
#define fi first
#define se second
#define IOS ios::sync_with_stdio(false); cin.tie(0); cout.tie(0)
#ifndef LOCAL
#define endl '\n'
#endif
const int N = 2e5 + 5;
const int mod = 1e9 + 7;
const int inf = 1e9 + 9;
int a[N];
signed main() {
IOS;
int t; cin >> t;
while(t--){
int n; cin >> n;
for(int i = 1; i <= n; i++)
cin >> a[i];
sort(a + 1, a + n + 1);
for(int i = 1; i <= n; i++)
cout << a[i] << " ";
cout << endl;
}
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer n , your task is to print the lowercase English word corresponding to the number if it is <=5 else print "Greater than 5".
Numbers <=5 and their corresponding words :
1 = one
2 = two
3 = three
4 = four
5 = fiveThe input contains a single integer N.
Constraint:
1 <= n <= 100Print a string consisting of the lowercase English word corresponding to the number if it is <=5 else print the string "Greater than 5"Sample Input:
4
Sample Output
four
Sample Input:
6
Sample Output:
Greater than 5, I have written this Solution Code: N = int(input())
if N > 5:
print("Greater than 5")
elif(N == 1):
print("one")
elif(N == 2):
print("two")
elif(N == 3):
print("three")
elif(N == 4):
print("four")
elif(N == 5):
print("five"), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer n , your task is to print the lowercase English word corresponding to the number if it is <=5 else print "Greater than 5".
Numbers <=5 and their corresponding words :
1 = one
2 = two
3 = three
4 = four
5 = fiveThe input contains a single integer N.
Constraint:
1 <= n <= 100Print a string consisting of the lowercase English word corresponding to the number if it is <=5 else print the string "Greater than 5"Sample Input:
4
Sample Output
four
Sample Input:
6
Sample Output:
Greater than 5, I have written this Solution Code: import java.util.Scanner;
class Main {
public static void main (String[] args)
{
//Capture the user's input
Scanner scanner = new Scanner(System.in);
//Storing the captured value in a variable
int side = scanner.nextInt();
String area = conditional(side);
System.out.println(area);
}static String conditional(int n){
if(n==1){return "one";}
else if(n==2){return "two";}
else if(n==3){return "three";}
else if(n==4){return "four";}
else if(n==5){return "five";}
else{
return "Greater than 5";}
}}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Complete the function <code>promiseMe</code>
Such that it takes a number as the first argument (time) and a string as the second argument(data).
It returns a promise which resolves after time milliseconds and data is returned.
Note:- You only have to implement the function, in the example it
shows your implemented question will be run.Function should take number as first argument and data to be returned as second.Resolves to the data given as inputpromiseMe(200, 'hi').then(data=>{
console.log(data) // prints hi
}), I have written this Solution Code: function promiseMe(number, dat) {
return new Promise((res,rej)=>{
setTimeout(()=>{
res(dat)
},number)
})
// return the output using return keyword
// do not console.log it
}, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a string S consisting of uppercase and lowercase characters. The task is to sort uppercase and lowercase characters separately such that if the ith place in the original string had an uppercase character then it should not have a lowercase character after being sorted and vice versa.The first line contains a single integer N denoting the length of the string. The second line contains a string S of length N, consisting of uppercase and lowercase characters.
1 <= N <= 100000Print a single integer containing the sorted string.Sample Input:
Valak
Sample Output:
Vaakl
Sample Input:
mwYifAbSQV
Sample Output:
bfAimQwSVY, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static String caseSort(String str) {
StringBuilder sb = new StringBuilder(str);
int lower[] = new int[26];
int upper[] = new int[26];
for(int i=0; i<str.length(); i++) {
if(Character.isLowerCase(str.charAt(i))) {
lower[str.charAt(i) - 'a']++;
}
else {
upper[str.charAt(i) - 'A']++;
}
}
int j = 0;
int k = 0;
while(j<26 && lower[j] == 0) {
j++;
}
while(k<26 && upper[k] == 0) {
k++;
}
for(int i=0; i<str.length(); i++) {
if(Character.isLowerCase(str.charAt(i))) {
while(lower[j] == 0) {
j++;
}
sb.setCharAt(i,(char)(j + 'a'));
lower[j]--;
}
else {
while(upper[k] == 0) {
k++;
}
sb.setCharAt(i,(char)(k + 'A'));
upper[k]--;
}
}
return sb.toString();
}
public static void main (String[] args) throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
String str = br.readLine();
System.out.println(caseSort(str));
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a string S consisting of uppercase and lowercase characters. The task is to sort uppercase and lowercase characters separately such that if the ith place in the original string had an uppercase character then it should not have a lowercase character after being sorted and vice versa.The first line contains a single integer N denoting the length of the string. The second line contains a string S of length N, consisting of uppercase and lowercase characters.
1 <= N <= 100000Print a single integer containing the sorted string.Sample Input:
Valak
Sample Output:
Vaakl
Sample Input:
mwYifAbSQV
Sample Output:
bfAimQwSVY, I have written this Solution Code: a=input()
upparcase=""
lowercase=""
for c in a:
if(ord(c)<97):
upparcase+=c
else:
lowercase+=c
upparcase = ''.join(sorted(upparcase))
lowercase = ''.join(sorted(lowercase))
i,j=0,0
for c in a:
if(ord(c)<97):
print(upparcase[i],end="")
i+=1
else:
print(lowercase[j],end="")
j+=1, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a string S consisting of uppercase and lowercase characters. The task is to sort uppercase and lowercase characters separately such that if the ith place in the original string had an uppercase character then it should not have a lowercase character after being sorted and vice versa.The first line contains a single integer N denoting the length of the string. The second line contains a string S of length N, consisting of uppercase and lowercase characters.
1 <= N <= 100000Print a single integer containing the sorted string.Sample Input:
Valak
Sample Output:
Vaakl
Sample Input:
mwYifAbSQV
Sample Output:
bfAimQwSVY, I have written this Solution Code: #include "bits/stdc++.h"
using namespace std;
#define int long long int
#define ld long double
#define pi pair<int, int>
#define pb push_back
#define fi first
#define se second
#define IOS ios::sync_with_stdio(false); cin.tie(0); cout.tie(0)
#ifndef LOCAL
#define endl '\n'
#endif
const int N = 1e5 + 5;
const int mod = 1e9 + 7;
const int inf = 1e9 + 9;
int a[N];
void solve(){
string s; cin >> s;
vector<char> v, u;
for(int i = 0; i < (int)s.length(); i++){
if(s[i] >= 'A' && s[i] <= 'Z')
a[i] = 1, v.push_back(s[i]);
else
u.push_back(s[i]);
}
sort(v.begin(), v.end());
sort(u.begin(), u.end());
int p = 0, q = 0;
for(int i = 0; i < (int)s.length(); i++){
if(a[i])
cout << v[p++];
else
cout << u[q++];
}
}
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: Ram is studying in Class V and has four subjects, each subject carry 100 marks. He passed with flying colors in his exam, but when his neighbour asked how much percentage did he got in exam, he got stuck in calculation. Ram is a good student but he forgot how to calculate percentage. Help Ram to get him out of this problem.
Given four numbers a , b , c and d denoting the marks in four subjects . Calculate the overall percentage (floor value ) Ram got in exam .First line contains four variables a, b, c and d.
<b>Constraints</b>
1<= a, b, c, d <= 100
Print single line containing the percentage.Sample Input 1:
25 25 25 25
Sample Output 1:
25
Sample Input 2:
75 25 75 25
Sample Output 2:
50, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args)throws Exception {
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
String str[]=br.readLine().split(" ");
int a[]=new int[str.length];
int sum=0;
for(int i=0;i<str.length;i++)
{
a[i]=Integer.parseInt(str[i]);
sum=sum+a[i];
}
System.out.println(sum/4);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Ram is studying in Class V and has four subjects, each subject carry 100 marks. He passed with flying colors in his exam, but when his neighbour asked how much percentage did he got in exam, he got stuck in calculation. Ram is a good student but he forgot how to calculate percentage. Help Ram to get him out of this problem.
Given four numbers a , b , c and d denoting the marks in four subjects . Calculate the overall percentage (floor value ) Ram got in exam .First line contains four variables a, b, c and d.
<b>Constraints</b>
1<= a, b, c, d <= 100
Print single line containing the percentage.Sample Input 1:
25 25 25 25
Sample Output 1:
25
Sample Input 2:
75 25 75 25
Sample Output 2:
50, I have written this Solution Code: a,b,c,d = map(int,input().split())
print((a+b+c+d)*100//400), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: It is a moment of triumph in Stepstones, as they have captured all the ships from Westeros. To celebrate this occasion, they have prepared a problem for you to solve:
You are given an integer R, and an array A consisting of positive integers. You can do this operation at most once (possibly zero times):
• Choose any element of the array. Replace it with any integer X, such that 1 ≤ X ≤ R.
Find the maximum possible GCD of the entire array.The first line consists of two space-separated integers N and R.
The second line consists of N space-separated integers – A<sub>1</sub>, A<sub>2</sub> ... A<sub>N</sub>.
<b>Constraints:</b>
2 ≤ N ≤ 10<sup>5</sup>
1 ≤ R ≤ 10<sup>5</sup>
1 ≤ A<sub>i</sub> ≤ 10<sup>5</sup>Print a single integer – the maximum possible GCD of the array.Sample Input 1:
3 10
2 3 4
Sample Output 1:
2
Sample Explanation 1:
We can change the second element from 3 to 2. Then the GCD of the array becomes gcd(2, 2, 4) = 2, which is the maximum possible.
Sample Input 2:
2 3
10 15
Sample Output 2:
5
Sample Explanation 2:
We do need to perform any operation. The GCD of the array is gcd(10, 15) = 5, which is the maximum possible.
Sample Input 3:
3 5
11 6 12
Sample Output 3:
3
Sample Explanation 3:
We can change the first element from 11 to 3. Then the GCD of the array becomes gcd(3, 6, 12) = 3, which is the maximum possible., I have written this Solution Code: def find_gcd(x, y):
while(y):
x, y = y, x % y
return x
nr = list(map(int,input().split()))
n = nr[0]
r = nr[1]
_ = list(map(int,input().split()))
num1 = _[0]
num2 = _[1]
gcd = find_gcd(num1, num2)
for i in range(2,n):
gcd = find_gcd(gcd, _[i])
if gcd>=r:
print(gcd)
else:
while r:
num = 0
idx = None
for i in range(n):
if _[i]%r != 0:
num += 1
idx = i
if num > 1:
break
if idx!= None:
temp = _[idx]
if num == 1:
_[idx] = r
num1 = _[0]
num2 = _[1]
tgcd = find_gcd(num1, num2)
for i in range(2,n):
tgcd = find_gcd(tgcd, _[i])
gcd = max(tgcd,gcd)
_[idx] = temp
break
r -= 1
print(gcd), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: It is a moment of triumph in Stepstones, as they have captured all the ships from Westeros. To celebrate this occasion, they have prepared a problem for you to solve:
You are given an integer R, and an array A consisting of positive integers. You can do this operation at most once (possibly zero times):
• Choose any element of the array. Replace it with any integer X, such that 1 ≤ X ≤ R.
Find the maximum possible GCD of the entire array.The first line consists of two space-separated integers N and R.
The second line consists of N space-separated integers – A<sub>1</sub>, A<sub>2</sub> ... A<sub>N</sub>.
<b>Constraints:</b>
2 ≤ N ≤ 10<sup>5</sup>
1 ≤ R ≤ 10<sup>5</sup>
1 ≤ A<sub>i</sub> ≤ 10<sup>5</sup>Print a single integer – the maximum possible GCD of the array.Sample Input 1:
3 10
2 3 4
Sample Output 1:
2
Sample Explanation 1:
We can change the second element from 3 to 2. Then the GCD of the array becomes gcd(2, 2, 4) = 2, which is the maximum possible.
Sample Input 2:
2 3
10 15
Sample Output 2:
5
Sample Explanation 2:
We do need to perform any operation. The GCD of the array is gcd(10, 15) = 5, which is the maximum possible.
Sample Input 3:
3 5
11 6 12
Sample Output 3:
3
Sample Explanation 3:
We can change the first element from 11 to 3. Then the GCD of the array becomes gcd(3, 6, 12) = 3, which is the maximum possible., I have written this Solution Code: import java.util.*;
import java.lang.*;
import java.io.*;
class Main {
public static long mod = 1000000007;
public static long mod2 = 998244353;
public static void main(String[] args) throws java.lang.Exception {
Reader sc = new Reader();
FastNum in = new FastNum(System.in);
Writer out = new Writer(System.out);
int n = in.nextInt();
int r = in.nextInt();
int[] a = in.nextIntArray(n);
int ans = a[0];
for (int i = 1; i < n; i++) {
ans = gcd(ans, a[i]);
}
int max = MaxGCD(a, n);
for (int i = 1; i <= r; i++) {
ans = Math.max(ans, gcd(max, i));
}
out.println(ans);
out.flush();
}
static int MaxGCD(int a[], int n) {
int[] Prefix = new int[n + 2];
int[] Suffix = new int[n + 2];
Prefix[1] = a[0];
for (int i = 2; i <= n; i += 1) {
Prefix[i] = gcd(Prefix[i - 1],
a[i - 1]);
}
Suffix[n] = a[n - 1];
for (int i = n - 1; i >= 1; i -= 1) {
Suffix[i] = gcd(Suffix[i + 1],
a[i - 1]);
}
int ans = Math.max(Suffix[2], Prefix[n - 1]);
for (int i = 2; i < n; i += 1) {
ans = Math.max(ans, gcd(Prefix[i - 1],
Suffix[i + 1]));
}
return ans;
}
static long modSum(long p, long q) {
if (q > 0) {
return (p + q) % mod;
} else {
return (p + q + mod) % mod;
}
}
static long invMod(long p, long q, long m) {
long expo = m - 2;
while (expo != 0) {
if ((expo & 1) == 1) {
p = (p * q) % m;
}
q = (q * q) % m;
expo >>= 1;
}
return p;
}
public static int gcd(int a, int b) {
if (a == 0)
return b;
return gcd(b % a, a);
}
public static long power(long a, long b) {
if (b == 0)
return 1;
long answer = power(a, b / 2) % mod;
answer = (answer * answer) % mod;
if (b % 2 != 0)
answer = (answer * a) % mod;
return answer;
}
public static void swap(int x, int y) {
int t = x;
x = y;
y = t;
}
public static long min(long a, long b) {
if (a < b) return a;
return b;
}
public static long divide(long a, long b) {
return (a % mod * (power(b, mod - 2) % mod)) % mod;
}
public static long nCr(long n, long r) {
long answer = 1;
long k = min(r, n - r);
for (long i = 0; i < k; i++) {
answer = (answer % mod * (n - i) % mod) % mod;
answer = divide(answer, i + 1);
}
return answer % mod;
}
public static boolean plaindrome(String str) {
StringBuilder sb = new StringBuilder();
sb.append(str);
return (str.equals((sb.reverse()).toString()));
}
public static class Pair {
int a;
int b;
Pair(int s, int e) {
a = s;
b = e;
}
@Override
public boolean equals(Object o) {
if (this == o)
return true;
if (o == null || getClass() != o.getClass())
return false;
Pair pair = (Pair) o;
return a == pair.a && b == pair.b;
}
@Override
public int hashCode() {
return Objects.hash(a, b);
}
}
static class Assert {
static void check(boolean e) {
if (!e) {
throw new AssertionError();
}
}
}
static class FastNum implements AutoCloseable {
InputStream is;
byte buffer[] = new byte[1 << 16];
int size = 0;
int pos = 0;
FastNum(InputStream is) {
this.is = is;
}
int nextChar() {
if (pos >= size) {
try {
size = is.read(buffer);
} catch (IOException e) {
throw new IOError(e);
}
pos = 0;
if (size == -1) {
return -1;
}
}
Assert.check(pos < size);
int c = buffer[pos] & 0xFF;
pos++;
return c;
}
int nextInt() {
int c = nextChar();
while (c == ' ' || c == '\r' || c == '\n' || c == '\t') {
c = nextChar();
}
if (c == '-') {
c = nextChar();
Assert.check('0' <= c && c <= '9');
int n = -(c - '0');
c = nextChar();
while ('0' <= c && c <= '9') {
int d = c - '0';
c = nextChar();
Assert.check(n > Integer.MIN_VALUE / 10
|| n == Integer.MIN_VALUE / 10 && d <= -(Integer.MIN_VALUE % 10));
n = n * 10 - d;
}
return n;
} else {
Assert.check('0' <= c && c <= '9');
int n = c - '0';
c = nextChar();
while ('0' <= c && c <= '9') {
int d = c - '0';
c = nextChar();
Assert.check(
n < Integer.MAX_VALUE / 10 || n == Integer.MAX_VALUE / 10 && d <= Integer.MAX_VALUE % 10);
n = n * 10 + d;
}
return n;
}
}
char nextCh() {
int c = nextChar();
while (c == ' ' || c == '\r' || c == '\n' || c == '\t') {
c = nextChar();
}
return (char) c;
}
long nextLong() {
int c = nextChar();
while (c == ' ' || c == '\r' || c == '\n' || c == '\t') {
c = nextChar();
}
if (c == '-') {
c = nextChar();
Assert.check('0' <= c && c <= '9');
long n = -(c - '0');
c = nextChar();
while ('0' <= c && c <= '9') {
int d = c - '0';
c = nextChar();
Assert.check(n > Long.MIN_VALUE / 10
|| n == Long.MIN_VALUE / 10 && d <= -(Long.MIN_VALUE % 10));
n = n * 10 - d;
}
return n;
} else {
Assert.check('0' <= c && c <= '9');
long n = c - '0';
c = nextChar();
while ('0' <= c && c <= '9') {
int d = c - '0';
c = nextChar();
Assert.check(
n < Long.MAX_VALUE / 10 || n == Long.MAX_VALUE / 10 && d <= Long.MAX_VALUE % 10);
n = n * 10 + d;
}
return n;
}
}
int[] nextIntArray(int n) {
int[] arr = new int[n];
for (int i = 0; i < n; i++) {
arr[i] = nextInt();
}
return arr;
}
long[] nextLongArray(int n) {
long[] arr = new long[n];
for (int i = 0; i < n; i++) {
arr[i] = nextInt();
}
return arr;
}
char[] nextCharArray(int n) {
char[] arr = new char[n];
for (int i = 0; i < n; i++) {
arr[i] = nextCh();
}
return arr;
}
@Override
public void close() {
}
}
static class Writer extends PrintWriter {
public Writer(java.io.Writer out) {
super(out);
}
public Writer(java.io.Writer out, boolean autoFlush) {
super(out, autoFlush);
}
public Writer(OutputStream out) {
super(out);
}
public Writer(OutputStream out, boolean autoFlush) {
super(out, autoFlush);
}
public void printArray(int[] arr) {
for (int j : arr) {
print(j);
print(' ');
}
println();
}
public void printArray(long[] arr) {
for (long j : arr) {
print(j);
print(' ');
}
println();
}
}
static class Reader {
final private int BUFFER_SIZE = 1 << 16;
private final DataInputStream din;
private final byte[] buffer;
private int bufferPointer, bytesRead;
public Reader() {
din = new DataInputStream(System.in);
buffer = new byte[BUFFER_SIZE];
bufferPointer = bytesRead = 0;
}
public Reader(String file_name) throws IOException {
din = new DataInputStream(new FileInputStream(file_name));
buffer = new byte[BUFFER_SIZE];
bufferPointer = bytesRead = 0;
}
public String readLine() throws IOException {
byte[] buf = new byte[64];
int cnt = 0, c;
while ((c = read()) != -1) {
if (c == '\n')
break;
buf[cnt++] = (byte) c;
}
return new String(buf, 0, cnt);
}
public int nextInt() throws IOException {
int ret = 0;
byte c = read();
while (c <= ' ')
c = read();
boolean neg = (c == '-');
if (neg)
c = read();
do {
ret = ret * 10 + c - '0';
} while ((c = read()) >= '0' && c <= '9');
if (neg)
return -ret;
return ret;
}
public long nextLong() throws IOException {
long ret = 0;
byte c = read();
while (c <= ' ')
c = read();
boolean neg = (c == '-');
if (neg)
c = read();
do {
ret = ret * 10 + c - '0';
}
while ((c = read()) >= '0' && c <= '9');
if (neg)
return -ret;
return ret;
}
public double nextDouble() throws IOException {
double ret = 0, div = 1;
byte c = read();
while (c <= ' ')
c = read();
boolean neg = (c == '-');
if (neg)
c = read();
do {
ret = ret * 10 + c - '0';
}
while ((c = read()) >= '0' && c <= '9');
if (c == '.') {
while ((c = read()) >= '0' && c <= '9') {
ret += (c - '0') / (div *= 10);
}
}
if (neg)
return -ret;
return ret;
}
private void fillBuffer() throws IOException {
bytesRead = din.read(buffer, bufferPointer = 0, BUFFER_SIZE);
if (bytesRead == -1)
buffer[0] = -1;
}
private byte read() throws IOException {
if (bufferPointer == bytesRead)
fillBuffer();
return buffer[bufferPointer++];
}
public void close() throws IOException {
if (din == null)
return;
din.close();
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: It is a moment of triumph in Stepstones, as they have captured all the ships from Westeros. To celebrate this occasion, they have prepared a problem for you to solve:
You are given an integer R, and an array A consisting of positive integers. You can do this operation at most once (possibly zero times):
• Choose any element of the array. Replace it with any integer X, such that 1 ≤ X ≤ R.
Find the maximum possible GCD of the entire array.The first line consists of two space-separated integers N and R.
The second line consists of N space-separated integers – A<sub>1</sub>, A<sub>2</sub> ... A<sub>N</sub>.
<b>Constraints:</b>
2 ≤ N ≤ 10<sup>5</sup>
1 ≤ R ≤ 10<sup>5</sup>
1 ≤ A<sub>i</sub> ≤ 10<sup>5</sup>Print a single integer – the maximum possible GCD of the array.Sample Input 1:
3 10
2 3 4
Sample Output 1:
2
Sample Explanation 1:
We can change the second element from 3 to 2. Then the GCD of the array becomes gcd(2, 2, 4) = 2, which is the maximum possible.
Sample Input 2:
2 3
10 15
Sample Output 2:
5
Sample Explanation 2:
We do need to perform any operation. The GCD of the array is gcd(10, 15) = 5, which is the maximum possible.
Sample Input 3:
3 5
11 6 12
Sample Output 3:
3
Sample Explanation 3:
We can change the first element from 11 to 3. Then the GCD of the array becomes gcd(3, 6, 12) = 3, which is the maximum possible., I have written this Solution Code: #include <bits/stdc++.h>
#define int long long
#define endl '\n'
using namespace std;
typedef long long ll;
typedef long double ld;
#define db(x) cerr << #x << ": " << x << '\n';
#define read(a) int a; cin >> a;
#define reads(s) string s; cin >> s;
#define readb(a, b) int a, b; cin >> a >> b;
#define readc(a, b, c) int a, b, c; cin >> a >> b >> c;
#define readarr(a, n) int a[(n) + 1] = {}; FOR(i, 1, (n)) {cin >> a[i];}
#define readmat(a, n, m) int a[n + 1][m + 1] = {}; FOR(i, 1, n) {FOR(j, 1, m) cin >> a[i][j];}
#define print(a) cout << a << endl;
#define printarr(a, n) FOR (i, 1, n) cout << a[i] << " "; cout << endl;
#define printv(v) for (int i: v) cout << i << " "; cout << endl;
#define printmat(a, n, m) FOR (i, 1, n) {FOR (j, 1, m) cout << a[i][j] << " "; cout << endl;}
#define all(v) v.begin(), v.end()
#define sz(v) (int)(v.size())
#define rz(v, n) v.resize((n) + 1);
#define pb push_back
#define fi first
#define se second
#define vi vector <int>
#define pi pair <int, int>
#define vpi vector <pi>
#define vvi vector <vi>
#define setprec cout << fixed << showpoint << setprecision(20);
#define FOR(i, a, b) for (int i = (a); i <= (b); i++)
#define FORD(i, a, b) for (int i = (a); i >= (b); i--)
const ll inf = 1e18;
const ll mod = 1e9 + 7;
//const ll mod = 998244353;
const ll N = 1e5 + 1;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
int power (int a, int b = mod - 2)
{
int res = 1;
while (b > 0) {
if (b & 1)
res = res * a % mod;
a = a * a % mod;
b >>= 1;
}
return res;
}
signed main()
{
ios_base::sync_with_stdio(false);
cin.tie(0);
vi divisor(N + 1);
readb(n, r);
FORD (i, r, 1)
{
for (int j = i; j <= N; j += i)
if (!divisor[j]) divisor[j] = i;
}
readarr(a, n);
int pre[n + 1] = {}, suf[n + 2] = {};
FOR (i, 1, n) pre[i] = __gcd(pre[i - 1], a[i]);
FORD (i, n, 1) suf[i] = __gcd(suf[i + 1], a[i]);
int ans = pre[n];
FOR (i, 1, n)
{
int x = __gcd(pre[i - 1], suf[i + 1]);
ans = max(ans, divisor[x]);
}
print(ans);
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given three integers your task is to calculate the maximum integer among the given integers.The input contains three integers a, b, and c
<b>Constraint:</b>
1<=integers<=10000Print the maximum integer among the given integers.Sample Input:-
2 6 3
Sample Output:-
6
Sample Input:-
48 100 100
Sample Output:
100, I have written this Solution Code: a,b,c=[int(a) for a in input().split()]
print(max(a,b,c)), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given three integers your task is to calculate the maximum integer among the given integers.The input contains three integers a, b, and c
<b>Constraint:</b>
1<=integers<=10000Print the maximum integer among the given integers.Sample Input:-
2 6 3
Sample Output:-
6
Sample Input:-
48 100 100
Sample Output:
100, I have written this Solution Code: import java.util.Scanner;
class Main {
public static void main (String[] args)
{
//Capture the user's input
Scanner scanner = new Scanner(System.in);
//Storing the captured value in a variable
int p = scanner.nextInt();
int tm = scanner.nextInt();
int r = scanner.nextInt();
int intrst = MaxInteger(p,tm,r);
System.out.println(intrst);
}
static int MaxInteger(int a ,int b, int c){
if(a>=b && a>=c){return a;}
if(b>=a && b>=c){return b;}
return c;}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Complete the function <code>countBy</code>
Such that it takes a initial number which is the defualt value of out counter. And returns
a function which also takes a number and returns the initialCount + number supplied to second function.
Ex:-
<code>
const count = countBy(4) // initial value of counter 4, returns a function <br>
console. log(count(2)) // prints 6 because 4 + 2 <br>
console. log(count(-4)) // prints 2 because 6 - 4 <br>
console. log(count(8)) // prints 10 because 2 + 8 <br>
</code>
You have to return implement countBy function such that it can be run like that.<code>countBy</code> will take one number as input which will be the initial count.<code>countBy</code> will return a function which can be run many times and takes a number as input and returns the sum of it with previously maintained counter values<code>
const count = countBy(4) // initial value of counter 4, returns a function <br>
console. log(count(2)) // prints 6 because 4 + 2 <br>
console. log(count(-4)) // prints 2 because 6 - 4 <br>
console. log(count(8)) // prints 10 because 2 + 8 <br>
</code>, I have written this Solution Code: function countBy(initial){
let copy = initial
return (y)=>{
copy += y
return copy
}
// return the output using return keyword
// do not console.log it
}, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer N, the task is to find the number of divisors of N which are divisible by 2.The input line contains T, denoting the number of testcases. First line of each testcase contains integer N
Constraints:
1 <= T <= 50
1 <= N <= 10^9For each testcase in new line, you need to print the number of divisors of N which are exactly divisble by 2Input:
2
9
8
Output
0
3, I have written this Solution Code: import math
n = int(input())
for i in range(n):
x = int(input())
count = 0
for i in range(1, int(math.sqrt(x))+1):
if x % i == 0:
if (i%2 == 0):
count+=1
if ((x/i) %2 == 0):
count+=1
print(count), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer N, the task is to find the number of divisors of N which are divisible by 2.The input line contains T, denoting the number of testcases. First line of each testcase contains integer N
Constraints:
1 <= T <= 50
1 <= N <= 10^9For each testcase in new line, you need to print the number of divisors of N which are exactly divisble by 2Input:
2
9
8
Output
0
3, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws IOException{
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int t = Integer.parseInt(br.readLine());
while(t-->0){
int n = Integer.parseInt(br.readLine());
int count=0;
for(int i=1;i<=Math.sqrt(n);i++){
if(n%i == 0)
{
if(i%2==0) {
count++;
}
if(i*i != n && (n/i)%2==0) {
count++;
}
}
}
System.out.println(count);
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer N, the task is to find the number of divisors of N which are divisible by 2.The input line contains T, denoting the number of testcases. First line of each testcase contains integer N
Constraints:
1 <= T <= 50
1 <= N <= 10^9For each testcase in new line, you need to print the number of divisors of N which are exactly divisble by 2Input:
2
9
8
Output
0
3, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main(){
int t;
cin>>t;
while(t--){
long long n;
cin>>n;
if(n&1){cout<<0<<endl;continue;}
long x=sqrt(n);
int cnt=0;
for(long long i=1;i<=x;i++){
if(!(n%i)){
if(!(i%2)){cnt++;}
if(i*i!=n){
if(!((n/i)%2)){cnt++;}
}
}
}
cout<<cnt<<endl;}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Two strings are completely different if the characters of one string cannot be rearranged to form the other string. Given a string find the number of completely different substrings of the string.The first line of the input contains a string S.
<b>Constraints:</b>
1 <= |S| <= 100
The string contains only lowercase English letters.The output should contain number of completely different substrings.Sample Input
abc
Sample Output
6
Sample Input
aaa
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 br=new BufferedReader(new InputStreamReader(System.in));
String str=br.readLine();
int len=str.length();
HashSet<String> set=new HashSet<>();
char arr[]=null;
for(int i=0;i<len;++i)
{
for(int j=i+1;j<=len;++j)
{
arr=str.substring(i,j).toCharArray();
Arrays.sort(arr);
set.add(new String(arr));
}
}
System.out.print(set.size());
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Two strings are completely different if the characters of one string cannot be rearranged to form the other string. Given a string find the number of completely different substrings of the string.The first line of the input contains a string S.
<b>Constraints:</b>
1 <= |S| <= 100
The string contains only lowercase English letters.The output should contain number of completely different substrings.Sample Input
abc
Sample Output
6
Sample Input
aaa
Sample Output
3, I have written this Solution Code: s = input()
Set = set()
l = []
for i in range(len(s)):
l.append(s[i])
for i in range(len(s)+1):
for j in range(i+1,len(s)+1):
a = ''.join(sorted(l[i:j]))
#print(a)
Set.add(a)
print(len(Set)), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Two strings are completely different if the characters of one string cannot be rearranged to form the other string. Given a string find the number of completely different substrings of the string.The first line of the input contains a string S.
<b>Constraints:</b>
1 <= |S| <= 100
The string contains only lowercase English letters.The output should contain number of completely different substrings.Sample Input
abc
Sample Output
6
Sample Input
aaa
Sample Output
3, I have written this Solution Code: #include "bits/stdc++.h"
#pragma GCC optimize "03"
using namespace std;
#define int long long int
#define ld long double
#define pi pair<int, int>
#define pb push_back
#define fi first
#define se second
#define IOS ios::sync_with_stdio(false); cin.tie(0); cout.tie(0)
#ifndef LOCAL
#define endl '\n'
#endif
const int N = 2e5 + 5;
const int mod = 1e9 + 7;
const int inf = 1e9 + 9;
signed main() {
IOS;
string s; cin >> s;
int n = s.length();
set<string> p;
for(int i = 0; i < n; i++){
string t = "";
for(int j = i; j < n; j++){
t += s[j];
sort(t.begin(), t.end());
p.insert(t);
}
}
cout << p.size();
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number N, you need to check whether the given number is Armstrong number or not. Now what is Armstrong number let us see below:
<b>A number is said to be Armstrong if it is equal to sum of cube of its digits. </b>User task:
Since this is a functional problem you don't have to worry about the input. You just have to complete the function <b>isArmstrong()</b> which contains N as a parameter.
Constraints:
1 <= N <= 10^4You need to return "true" if the given number is an Armstrong number otherwise "false"Sample Input 1:
1
Sample Output 1:
true
Sample Input 2:
147
Sample Output 2:
false
Sample Input 3:
371
Sample Output 3:
true
, I have written this Solution Code: static boolean isArmstrong(int N)
{
int num = N;
int sum = 0;
while(N > 0)
{
int digit = N%10;
sum += digit*digit*digit;
N = N/10;
}
if(num == sum)
return true;
else return false;
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number N, you need to check whether the given number is Armstrong number or not. Now what is Armstrong number let us see below:
<b>A number is said to be Armstrong if it is equal to sum of cube of its digits. </b>User task:
Since this is a functional problem you don't have to worry about the input. You just have to complete the function <b>isArmstrong()</b> which contains N as a parameter.
Constraints:
1 <= N <= 10^4You need to return "true" if the given number is an Armstrong number otherwise "false"Sample Input 1:
1
Sample Output 1:
true
Sample Input 2:
147
Sample Output 2:
false
Sample Input 3:
371
Sample Output 3:
true
, I have written this Solution Code: function isArmstrong(n) {
// write code here
// do no console.log the answer
// return the output using return keyword
let sum = 0
let k = n;
while(k !== 0){
sum += Math.pow( k%10,3)
k = Math.floor(k/10)
}
return sum === n
}, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of heights of N buildings in a row. You can start from any building and jump to the <b>adjacent</b> right building till the height of the building to the right is less than or equal to the height of your current building. Find the maximum number of jumps you can make.The first line of input contains a single integer N.
The second line of input contains N integers, denoting the array height.
<b>Constraints:</b>
1 <= N <= 10<sup>5</sup>
1 <= height[i] <= 10<sup>9</sup>Print the maximum number of jumps you can make.Sample Input:-
5
5 4 1 2 1
Sample Output:-
2
<b>Explanation:</b>
We start from building with height 5 then jump right to building with height 4 then again to building with height 1 making a total of 2 jumps., 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>
#define rep(i,n) for (int i=0; i<(n); i++)
/////////////
signed main()
{
fastio;
#ifdef ANIKET_GOYAL
freopen("inputf.in","r",stdin);
freopen("outputf.in","w",stdout);
#endif
int n;
cin>>n;
int a[n];
for(int i=0;i<n;++i){
cin>>a[i];
}
int ans=0;
int v=0;
for(int i=1;i<n;++i){
if(a[i]>a[i-1])
v=0;
else
++v;
ans=max(ans,v);
}
cout<<ans;
#ifdef ANIKET_GOYAL
// cout<<endl<<endl<<endl<<endl<<"Time elapsed: "<<(double)(clock()-clk)/CLOCKS_PER_SEC<<endl;
#endif
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of heights of N buildings in a row. You can start from any building and jump to the <b>adjacent</b> right building till the height of the building to the right is less than or equal to the height of your current building. Find the maximum number of jumps you can make.The first line of input contains a single integer N.
The second line of input contains N integers, denoting the array height.
<b>Constraints:</b>
1 <= N <= 10<sup>5</sup>
1 <= height[i] <= 10<sup>9</sup>Print the maximum number of jumps you can make.Sample Input:-
5
5 4 1 2 1
Sample Output:-
2
<b>Explanation:</b>
We start from building with height 5 then jump right to building with height 4 then again to building with height 1 making a total of 2 jumps., 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 num = Integer.parseInt(br.readLine());
String s= br.readLine();
int[] arr= new int[num];
String[] s1 = s.split(" ");
int ccount = 0, pcount = 0;
for(int i=0;i<(num);i++)
{
arr[i]=Integer.parseInt(s1[i]);
if(i+1 < num){
arr[i+1] = Integer.parseInt(s1[i+1]);}
if(((i+1)< num) && (arr[i]>=arr[i+1]) ){
ccount++;
}else{
if(ccount > pcount){
pcount = ccount;
}
ccount = 0;
}
}
System.out.print(pcount);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of heights of N buildings in a row. You can start from any building and jump to the <b>adjacent</b> right building till the height of the building to the right is less than or equal to the height of your current building. Find the maximum number of jumps you can make.The first line of input contains a single integer N.
The second line of input contains N integers, denoting the array height.
<b>Constraints:</b>
1 <= N <= 10<sup>5</sup>
1 <= height[i] <= 10<sup>9</sup>Print the maximum number of jumps you can make.Sample Input:-
5
5 4 1 2 1
Sample Output:-
2
<b>Explanation:</b>
We start from building with height 5 then jump right to building with height 4 then again to building with height 1 making a total of 2 jumps., I have written this Solution Code: N = int(input())
arr = iter(map(int,input().strip().split()))
jumps = []
jump = 0
temp = next(arr)
for i in arr:
if i<=temp:
jump += 1
temp = i
continue
temp = i
jumps.append(jump)
jump = 0
jumps.append(jump)
print(max(jumps)), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
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