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For this Question: Take input from standard input while you do not get 0 as an input. Print all the inputs separated by space. It is guaranteed that the number of integers are less than 100000.The input will contain a series of integers in one line each. Input should be taken while you have not get a 0 as an input.
0 <= input <= 10Print the input integers seperated by space.Sample Input
6
5
5
0
Sample Output
6 5 5 0
Sample Input
9
3
5
7
6
9
8
3
2
7
7
3
5
0
Sample Output
9 3 5 7 6 9 8 3 2 7 7 3 5 0, I have written this Solution Code:
import java.util.*;
import java.lang.*;
import java.io.*;
class Main
{
public static void main (String[] args) throws java.lang.Exception
{
Scanner sc = new Scanner(System.in);
int n=100001;
int a;
for(int i=0;i<n;i++){
a=sc.nextInt();
System.out.print(a+" ");
if(a==0){break;}
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Take input from standard input while you do not get 0 as an input. Print all the inputs separated by space. It is guaranteed that the number of integers are less than 100000.The input will contain a series of integers in one line each. Input should be taken while you have not get a 0 as an input.
0 <= input <= 10Print the input integers seperated by space.Sample Input
6
5
5
0
Sample Output
6 5 5 0
Sample Input
9
3
5
7
6
9
8
3
2
7
7
3
5
0
Sample Output
9 3 5 7 6 9 8 3 2 7 7 3 5 0, I have written this Solution Code: #include "bits/stdc++.h"
#pragma GCC optimize "03"
using namespace std;
#define int long long int
#define ld long double
#define pi pair<int, int>
#define pb push_back
#define fi first
#define se second
#define IOS ios::sync_with_stdio(false); cin.tie(0); cout.tie(0)
#ifndef LOCAL
#define endl '\n'
#endif
const int N = 2e5 + 5;
const int mod = 1e9 + 7;
const int inf = 1e9 + 9;
signed main() {
IOS;
int n;
while(cin >> n){
cout << n << " ";
if(n == 0) break;
}
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a string S (containing only lowercase characters). You need to print the repeated character whose first appearance is leftmost.The first line of input contains T denoting the number of testcases. T testcases follow.
Each testcase contains one line of input containing the string S.
Constraints:
1 <= T <= 100
1 <= |S| <= 1000For each testcase, in a new line, print the character. If not found then print "-1".Input:
2
geeksforgeeks
abcd
Output:
g
-1
Explanation:
Testcase1: We see that both e and g repeat as we move from left to right. But the leftmost is g so we print g.
Testcase2: No character repeats so we print -1.
, I have written this Solution Code:
function leftMostOcurringChar(str) {
// write code here
// do not console.log answer
// return the answer using return keyword
let firstIndex = new Array(256).fill(-1);
let res = Number.MAX_VALUE;
for (let i = 0; i < str.length; i++)
{
if (firstIndex[str[i].charCodeAt()] == -1)
{
firstIndex[str[i].charCodeAt()] = i;
}
else
{
res = Math.min(res, firstIndex[str[i].charCodeAt()]);
}
}
const ans = (res == Number.MAX_VALUE) ? -1 : res;
if (ans === -1) return -1;
return str[ans]
}, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a string S (containing only lowercase characters). You need to print the repeated character whose first appearance is leftmost.The first line of input contains T denoting the number of testcases. T testcases follow.
Each testcase contains one line of input containing the string S.
Constraints:
1 <= T <= 100
1 <= |S| <= 1000For each testcase, in a new line, print the character. If not found then print "-1".Input:
2
geeksforgeeks
abcd
Output:
g
-1
Explanation:
Testcase1: We see that both e and g repeat as we move from left to right. But the leftmost is g so we print g.
Testcase2: No character repeats so we print -1.
, I have written this Solution Code: n=int(input())
while(n):
n-=1
string=input()
temp=True
for i in range(len(string)):
if(string[i] in string[i+1:] and i!=len(string)-1):
print(string[i])
temp=False
break
if(temp):
print(-1), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a string S (containing only lowercase characters). You need to print the repeated character whose first appearance is leftmost.The first line of input contains T denoting the number of testcases. T testcases follow.
Each testcase contains one line of input containing the string S.
Constraints:
1 <= T <= 100
1 <= |S| <= 1000For each testcase, in a new line, print the character. If not found then print "-1".Input:
2
geeksforgeeks
abcd
Output:
g
-1
Explanation:
Testcase1: We see that both e and g repeat as we move from left to right. But the leftmost is g so we print g.
Testcase2: No character repeats so we print -1.
, I have written this Solution Code: #include<bits/stdc++.h>
using namespace std;
#define pu push_back
#define fi first
#define se second
#define mp make_pair
#define int long long
#define pii pair<int,int>
#define mm (s+e)/2
#define all(x) x.begin(), x.end()
#define For(i, st, en) for(int i=st; i<en; i++)
#define tr(x) for(auto it=x.begin(); it!=x.end(); it++)
#define fast std::ios::sync_with_stdio(false);cin.tie(NULL);
#define sz 200000
signed main()
{
int t;
cin>>t;
while(t>0)
{
t--;
string s;
cin>>s;
map<char,int> ss;
int ch=-1;
for(int i=0;i<s.size();i++)
{
ss[s[i]]++;
}
for(int i=0;i<s.size();i++)
{
if(ss[s[i]]>=2)
{ ch=0;
cout<<s[i]<<endl;
break;
}
}
if(ch==-1) cout<<-1<<endl;
}
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a string S (containing only lowercase characters). You need to print the repeated character whose first appearance is leftmost.The first line of input contains T denoting the number of testcases. T testcases follow.
Each testcase contains one line of input containing the string S.
Constraints:
1 <= T <= 100
1 <= |S| <= 1000For each testcase, in a new line, print the character. If not found then print "-1".Input:
2
geeksforgeeks
abcd
Output:
g
-1
Explanation:
Testcase1: We see that both e and g repeat as we move from left to right. But the leftmost is g so we print g.
Testcase2: No character repeats so we print -1.
, I have written this Solution Code: import java.io.*;
import java.util.*;
import java.lang.*;
class Main
{
public static void main(String args[])
{
Scanner sc = new Scanner(System.in);
int t = sc.nextInt();
while(t-- > 0)
{
String str = sc.next();
int index = repeatedCharacter(str);
if(index == -1)
System.out.println("-1");
else
System.out.println(str.charAt(index));
}
}
static int repeatedCharacter(String S)
{
// Initialize leftmost index of every
// character as -1.
int firstIndex[]= new int[256];
for (int i = 0; i <256; i++)
firstIndex[i] = -1;
// Traverse from left and update result
// if we see a repeating character whose
// first index is smaller than current
// result.
int res = Integer.MAX_VALUE;
for (int i = 0; i < S.length(); i++) {
if (firstIndex[S.charAt(i)] == -1)
firstIndex[S.charAt(i)] = i;
else
res = Math.min(res, firstIndex[S.charAt(i)]);
}
return (res == Integer.MAX_VALUE) ? - 1 : res;
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a number n. You need to find the count of digits in n.The first line of input contains T denoting the number of test cases. Each testcase contains one line of input containing n.
For Python, Language Just Completes the function.
<b>Constraints:</b>
1 ≤ T ≤ 100
1 ≤ n ≤ 10<sup>8</sup>For each testcase, in a newline, print the count of digits in n.Input:
2
1
99999
Output:
1
5
<b>Explanation:</b>
Testcase 1: The number of digits in 1 is 1.
Testcase 2: The number of digits in 99999 is 5, I have written this Solution Code: #include<bits/stdc++.h>
using namespace std;
#define pu push_back
#define fi first
#define se second
#define mp make_pair
#define int long long
#define pii pair<int,int>
#define mm (s+e)/2
#define all(x) x.begin(), x.end()
#define For(i, st, en) for(int i=st; i<en; i++)
#define tr(x) for(auto it=x.begin(); it!=x.end(); it++)
#define fast std::ios::sync_with_stdio(false);cin.tie(NULL);
#define sz 200000
signed main()
{
int t;
cin>>t;
while(t>0)
{
t--;
int n;
cin>>n;
int cnt=0;
while(n>0)
{
cnt++;
n/=10LL;
}
cout<<cnt<<endl;
}
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a number n. You need to find the count of digits in n.The first line of input contains T denoting the number of test cases. Each testcase contains one line of input containing n.
For Python, Language Just Completes the function.
<b>Constraints:</b>
1 ≤ T ≤ 100
1 ≤ n ≤ 10<sup>8</sup>For each testcase, in a newline, print the count of digits in n.Input:
2
1
99999
Output:
1
5
<b>Explanation:</b>
Testcase 1: The number of digits in 1 is 1.
Testcase 2: The number of digits in 99999 is 5, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static int countDigit(int a)
{
if(a==0)
{
return 0;
}
else
{
return 1 + countDigit(a/10);
}
}
public static void main (String[] args)
{
Scanner sc = new Scanner(System.in);
int T = sc.nextInt();
for(int i=0;i<T;i++)
{
int n = sc.nextInt();
System.out.println(countDigit(n));
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a number n. You need to find the count of digits in n.The first line of input contains T denoting the number of test cases. Each testcase contains one line of input containing n.
For Python, Language Just Completes the function.
<b>Constraints:</b>
1 ≤ T ≤ 100
1 ≤ n ≤ 10<sup>8</sup>For each testcase, in a newline, print the count of digits in n.Input:
2
1
99999
Output:
1
5
<b>Explanation:</b>
Testcase 1: The number of digits in 1 is 1.
Testcase 2: The number of digits in 99999 is 5, I have written this Solution Code: def totalDigits(n):
cnt=0
while(n>0):
cnt +=1
n//=10
return cnt, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an infix expression, your task is to convert it into postfix.
<b>Infix expression</b>: The expression of the form a operator b. When an operator is in- between every pair of operands.
<b>Postfix expression</b>: The expression of the form a b operator. When an operator is followed for every pair of operands.
Valid operators are +, - , *, /, ^. Each operand is an uppercase english alphabet. Infix expression may contains parenthesis as '(' and ')'.
See example for better understanding.The input contains a single string of infix expression.
<b>Constraints:-</b>
1 <= |String| <= 40Output the Postfix expression.Sample Input:-
((A-(B/C))*((A/K)-L))
Sample Output:-
ABC/-AK/L-*
Sample Input:-
A+B
Sample Output:-
AB+, 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 s = in.readLine().trim();
char[] express = s.toCharArray();
StringBuilder ans = new StringBuilder("");
Stack<Character> stack = new Stack<>();
for(int i=0; i<express.length; i++){
if(express[i]>=65 && express[i]<=90){
ans.append(express[i]);
}else if(express[i]==')'){
while(stack.peek()!='('){
ans.append(stack.peek());
stack.pop();
}
stack.pop();
}else if(stack.empty() || express[i]=='(' || precedence(express[i])>=precedence(stack.peek())){
stack.push(express[i]);
}else{
while(!stack.empty() && stack.peek()!='(' && precedence(stack.peek())<=express[i]){
ans.append(stack.peek());
stack.pop();
}
stack.push(express[i]);
}
}
while(!stack.empty()){
ans.append(stack.peek());
stack.pop();
}
System.out.println(ans);
}
static int precedence(char operator){
int precedenceValue;
if(operator=='+' || operator=='-')
precedenceValue = 1;
else if(operator=='*' || operator=='/')
precedenceValue = 2;
else if(operator == '^')
precedenceValue = 3;
else
precedenceValue = -1;
return precedenceValue;
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an infix expression, your task is to convert it into postfix.
<b>Infix expression</b>: The expression of the form a operator b. When an operator is in- between every pair of operands.
<b>Postfix expression</b>: The expression of the form a b operator. When an operator is followed for every pair of operands.
Valid operators are +, - , *, /, ^. Each operand is an uppercase english alphabet. Infix expression may contains parenthesis as '(' and ')'.
See example for better understanding.The input contains a single string of infix expression.
<b>Constraints:-</b>
1 <= |String| <= 40Output the Postfix expression.Sample Input:-
((A-(B/C))*((A/K)-L))
Sample Output:-
ABC/-AK/L-*
Sample Input:-
A+B
Sample Output:-
AB+, I have written this Solution Code: string=input()
stack=list()
preced={'+':1,'-':1,'*':2,'/':2,'^':3}
for i in string:
if i!=')':
if i=='(':
stack.append(i)
elif i in ['*','+','/','-','^']:
while(stack and stack[-1]!='(' and preced[i]<=preced[stack[-1]]):
print(stack.pop(),end="")
stack.append(i)
else:
print(i,end="")
else:
while(stack and stack[-1]!='('):
print(stack.pop(),end="")
stack.pop()
while(stack):
print(stack.pop(),end=""), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an infix expression, your task is to convert it into postfix.
<b>Infix expression</b>: The expression of the form a operator b. When an operator is in- between every pair of operands.
<b>Postfix expression</b>: The expression of the form a b operator. When an operator is followed for every pair of operands.
Valid operators are +, - , *, /, ^. Each operand is an uppercase english alphabet. Infix expression may contains parenthesis as '(' and ')'.
See example for better understanding.The input contains a single string of infix expression.
<b>Constraints:-</b>
1 <= |String| <= 40Output the Postfix expression.Sample Input:-
((A-(B/C))*((A/K)-L))
Sample Output:-
ABC/-AK/L-*
Sample Input:-
A+B
Sample Output:-
AB+, I have written this Solution Code: #include<bits/stdc++.h>
using namespace std;
//Function to return precedence of operators
int prec(char c)
{
if(c == '^')
return 3;
else if(c == '*' || c == '/')
return 2;
else if(c == '+' || c == '-')
return 1;
else
return -1;
}
// The main function to convert infix expression
//to postfix expression
void infixToPostfix(string s)
{
std::stack<char> st;
st.push('N');
int l = s.length();
string ns;
for(int i = 0; i < l; i++)
{
// If the scanned character is an operand, add it to output string.
if((s[i] >= 'a' && s[i] <= 'z')||(s[i] >= 'A' && s[i] <= 'Z'))
ns+=s[i];
// If the scanned character is an ‘(‘, push it to the stack.
else if(s[i] == '(')
st.push('(');
// If the scanned character is an ‘)’, pop and to output string from the stack
// until an ‘(‘ is encountered.
else if(s[i] == ')')
{
while(st.top() != 'N' && st.top() != '(')
{
char c = st.top();
st.pop();
ns += c;
}
if(st.top() == '(')
{
char c = st.top();
st.pop();
}
}
//If an operator is scanned
else{
while(st.top() != 'N' && prec(s[i]) <= prec(st.top()))
{
char c = st.top();
st.pop();
ns += c;
}
st.push(s[i]);
}
}
//Pop all the remaining elements from the stack
while(st.top() != 'N')
{
char c = st.top();
st.pop();
ns += c;
}
cout << ns << endl;
}
//Driver program to test above functions
int main()
{
string exp;
cin>>exp;
infixToPostfix(exp);
return 0;
} , In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: In number theory, the totient Q(N) of a positive integer N is defined as the number of positive integers less than or equal to N that are co-prime to N.
Given an integer N. Compute the value of the totient Q(N) .First line contain integer t denoting number of test cases.
Next t lines contains an integer x.
Constraints
1<=t<=5
1<=x<=10^12For each test case print Q(n) in separate lineSample Input
4
1
2
3
6
Sample output
1
1
2
2, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
static long CoPrime(long n)
{
double result = n;
for (long p = 2; p * p <= n; ++p) {
if (n % p == 0) {
while (n % p == 0)
n /= p;
result *= (1.0 - (1.0 / (double)p));
}
}
if (n > 1)
result *= (1.0 - (1.0 / (double)n));
return (long)result;
}
public static void main (String[] args) throws IOException{
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
long t=Long.parseLong(br.readLine());
while(t-->0){
long ans=1;
long k=Long.parseLong(br.readLine());
System.out.println(CoPrime(k));
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: In number theory, the totient Q(N) of a positive integer N is defined as the number of positive integers less than or equal to N that are co-prime to N.
Given an integer N. Compute the value of the totient Q(N) .First line contain integer t denoting number of test cases.
Next t lines contains an integer x.
Constraints
1<=t<=5
1<=x<=10^12For each test case print Q(n) in separate lineSample Input
4
1
2
3
6
Sample output
1
1
2
2, I have written this Solution Code: def totient(n):
result = n;
p = 2;
while(p * p <= n):
if (n % p == 0):
while (n % p == 0):
n = int(n / p);
result -= int(result / p);
p += 1;
if (n > 1):
result -= int(result / n);
return result;
for n in range(int(input())):
print(totient(int(input()))), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: In number theory, the totient Q(N) of a positive integer N is defined as the number of positive integers less than or equal to N that are co-prime to N.
Given an integer N. Compute the value of the totient Q(N) .First line contain integer t denoting number of test cases.
Next t lines contains an integer x.
Constraints
1<=t<=5
1<=x<=10^12For each test case print Q(n) in separate lineSample Input
4
1
2
3
6
Sample output
1
1
2
2, I have written this Solution Code: #include<bits/stdc++.h>
using namespace std;
#define pu push_back
#define fi first
#define se second
#define mp make_pair
#define int long long
#define pii pair<int,int>
#define mm (s+e)/2
#define all(x) x.begin(), x.end()
#define For(i, st, en) for(int i=st; i<en; i++)
#define tr(x) for(auto it=x.begin(); it!=x.end(); it++)
#define fast std::ios::sync_with_stdio(false);cin.tie(NULL);
#define sz 200000
int mod=1000000007;
int fact[sz];
int ifact[sz];
int lpf[2000000];
int power(int a,int b)
{
if(a==0) return 0;
if(a==1) return 1;
if(b==0) return 1;
if(b==1) return a;
int xx=power(a,b/2);
int yy=(xx*xx)%mod;
if(b%2==0) return xx;
else return (xx*a)%mod;
}
int inverse(int a)
{
return power(a,mod-2LL);
}
int nCr(int a,int b)
{
if(b>a) return 0;
int x=fact[a];
x=(x*ifact[b])%mod;
x=(x*ifact[a-b])%mod;
return x;
}
void seiv()
{
lpf[1]=1;
for(int i=2;i<=1000000;i++)
{
if(lpf[i]==0)
{
for(int j=i;j<=1000000;j+=i)
{
if(lpf[j]==0) lpf[j]=i;
}
}
}
}
int totient(int n)
{ if(n==1) return 1;
int vh=1;
for(int i=2;i*i<=n;i++)
{ if(n%i==0)
{
int x=i;
int cnt=0;
while(n%x==0) {
n/=x;
cnt++;
}
vh*=(pow(x,cnt)-pow(x,cnt-1));
}
}
if(n>1) vh*=(n-1LL);
return vh;
}
signed main()
{
int t;
cin>>t;
while(t>0)
{t--;
int n;
cin>>n;
cout<<totient(n)<<endl;
}
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a 2X2 square matrix. You need to find the determinant of the matrix.The input contains two-line, each line contain two integers separated by spaces.
Each element of the matrix is from 1 to 100.Output a single integer, the determinant of the matrix.Sample Input
4 5
2 3
Sample Output
2
Sample Input
2 10
10 40
Sample Output
-20, 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[][]arr=new int[2][2];
for(int i=0;i<2;i++){
for(int j=0;j<2;j++){
arr[i][j]=sc.nextInt();
}
}
int determinent=(arr[0][0]*arr[1][1])-(arr[1][0]*arr[0][1]);
System.out.print(determinent);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a 2X2 square matrix. You need to find the determinant of the matrix.The input contains two-line, each line contain two integers separated by spaces.
Each element of the matrix is from 1 to 100.Output a single integer, the determinant of the matrix.Sample Input
4 5
2 3
Sample Output
2
Sample Input
2 10
10 40
Sample Output
-20, I have written this Solution Code: arr1=list(map(int,input().split()))
arr2=list(map(int,input().split()))
lis=[]
a11=arr1[0]
a12=arr1[1]
a21=arr2[0]
a22=arr2[1]
det=(a11*a22)-(a12*a21)
print(det), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a 2X2 square matrix. You need to find the determinant of the matrix.The input contains two-line, each line contain two integers separated by spaces.
Each element of the matrix is from 1 to 100.Output a single integer, the determinant of the matrix.Sample Input
4 5
2 3
Sample Output
2
Sample Input
2 10
10 40
Sample Output
-20, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main(){
int a,b,c,d;
cin>>a>>b>>c>>d;
cout<<a*d-b*c;
}, 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, give the number with maximum frequency. If multiple numbers have maximum frequency print the maximum number among them.The first line of the input contains an integer N, and the Second line contains N space-separated integers of the array.
<b>Constraints:</b>
3 <= N <= 1000
1 <= Arr[i] <= 100The output should contain single integer, the number with maximum frequency.If multiple numbers have maximum frequency print the maximum number among them.Sample Input
5
1 4 2 4 5
Sample Output
4
<b>Explanation:-</b>
4 has max frequency=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());
String [] str=br.readLine().trim().split(" ");
int a[]=new int[n];
for(int i=0;i<n;i++){
a[i]=Integer.parseInt(str[i]);
}
Arrays.sort(a);
int size=a[n-1]+1;
int c[]=new int[size];
for(int i=0;i<size;i++) c[i]=0;
for(int i=0;i<n;i++) c[a[i]]++;
int max=0,freq=c[1];
for(int i=2;i<size;i++){
if(freq<=c[i]){
freq=c[i];
max=i;
}
}
System.out.println(max);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of N integers, give the number with maximum frequency. If multiple numbers have maximum frequency print the maximum number among them.The first line of the input contains an integer N, and the Second line contains N space-separated integers of the array.
<b>Constraints:</b>
3 <= N <= 1000
1 <= Arr[i] <= 100The output should contain single integer, the number with maximum frequency.If multiple numbers have maximum frequency print the maximum number among them.Sample Input
5
1 4 2 4 5
Sample Output
4
<b>Explanation:-</b>
4 has max frequency=2, I have written this Solution Code: n = int(input())
a = [int(x) for x in input().split()]
freq = {}
for x in a:
if x not in freq:
freq[x] = 1
else:
freq[x] += 1
mx = max(freq.values())
rf = sorted(freq)
for i in range(len(rf) - 1, -1, -1):
if freq[rf[i]] == mx:
print(rf[i])
break, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of N integers, give the number with maximum frequency. If multiple numbers have maximum frequency print the maximum number among them.The first line of the input contains an integer N, and the Second line contains N space-separated integers of the array.
<b>Constraints:</b>
3 <= N <= 1000
1 <= Arr[i] <= 100The output should contain single integer, the number with maximum frequency.If multiple numbers have maximum frequency print the maximum number among them.Sample Input
5
1 4 2 4 5
Sample Output
4
<b>Explanation:-</b>
4 has max frequency=2, I have written this Solution Code: #include "bits/stdc++.h"
#pragma GCC optimize "03"
using namespace std;
#define int long long int
#define ld long double
#define pi pair<int, int>
#define pb push_back
#define fi first
#define se second
#define IOS ios::sync_with_stdio(false); cin.tie(0); cout.tie(0)
#ifndef LOCAL
#define endl '\n'
#endif
const int N = 2e5 + 5;
const int mod = 1e9 + 7;
const int inf = 1e9 + 9;
int a[N];
signed main() {
IOS;
int n; cin >> n;
for(int i = 1; i <= n; i++){
int p; cin >> p;
a[p]++;
}
int mx = 0, id = -1;
for(int i = 1; i <= 100; i++){
if(a[i] >= mx)
mx = a[i], id = i;
}
cout << id;
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Modify the <code>takeMultipleNumbersAndAdd </code> such that it can take any number of arguments
and return its sum.
This is JS only question.Function should be able to take any number of argsSum of the numberstakeMultipleNumbersAndAdd(1, 2, 2) should return 5 because 1 + 2 + 2
takeMultipleNumbersAndAdd(-1, 2, -1, 5) should return 5, I have written this Solution Code:
function takeMultipleNumbersAndAdd (...nums){
// write your code here
return nums.reduce((prev,cur)=>prev+cur,0)
// 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 sorted array arr[] of N integers and a number K is given. The task is to check if the element K is present in the array or not.
<b>Note:</b> Use binary search to solve the problemThe first line of input contains a number of test cases T. For each test case, the first line of input contains a number of elements in the array, and the number K is separated by space. The next line contains N elements.
<b>Constraints:</b>
1 ≤ T ≤ 10
1 ≤ N ≤ 10<sup>5</sup>
1 ≤ K ≤ 10<sup>9</sup>
1 ≤ arr[i] ≤ 10<sup>9</sup>
<b>Sum of N over all test cases doesn't exceed 10<sup>6</sup></b>If the element is present in the array print "1" else print "-1".Sample Input:
2
5 6
1 2 3 4 6
5 2
1 3 4 5 6
Sample Output:
1
-1, I have written this Solution Code: static int isPresent(long arr[], int n, long k)
{
int left = 0;
int right = n-1;
int res = -1;
while(left<=right){
int mid = (left+right)/2;
if(arr[mid] == k){
res = 1;
break;
}else if(arr[mid] < k){
left = mid + 1;
}else{
right = mid - 1;
}
}
return res;
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a sorted array arr[] of N integers and a number K is given. The task is to check if the element K is present in the array or not.
<b>Note:</b> Use binary search to solve the problemThe first line of input contains a number of test cases T. For each test case, the first line of input contains a number of elements in the array, and the number K is separated by space. The next line contains N elements.
<b>Constraints:</b>
1 ≤ T ≤ 10
1 ≤ N ≤ 10<sup>5</sup>
1 ≤ K ≤ 10<sup>9</sup>
1 ≤ arr[i] ≤ 10<sup>9</sup>
<b>Sum of N over all test cases doesn't exceed 10<sup>6</sup></b>If the element is present in the array print "1" else print "-1".Sample Input:
2
5 6
1 2 3 4 6
5 2
1 3 4 5 6
Sample Output:
1
-1, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main(){
int t;
cin>>t;
while(t--){
int n;
cin>>n;
unordered_map<long long,int> m;
long k;
cin>>k;
long long a;
for(int i=0;i<n;i++){
cin>>a;
m[a]++;
}
if(m.find(k)!=m.end()){
cout<<1<<endl;
}
else{
cout<<-1<<endl;
}
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a sorted array arr[] of N integers and a number K is given. The task is to check if the element K is present in the array or not.
<b>Note:</b> Use binary search to solve the problemThe first line of input contains a number of test cases T. For each test case, the first line of input contains a number of elements in the array, and the number K is separated by space. The next line contains N elements.
<b>Constraints:</b>
1 ≤ T ≤ 10
1 ≤ N ≤ 10<sup>5</sup>
1 ≤ K ≤ 10<sup>9</sup>
1 ≤ arr[i] ≤ 10<sup>9</sup>
<b>Sum of N over all test cases doesn't exceed 10<sup>6</sup></b>If the element is present in the array print "1" else print "-1".Sample Input:
2
5 6
1 2 3 4 6
5 2
1 3 4 5 6
Sample Output:
1
-1, I have written this Solution Code: def binary_search(arr, low, high, x):
if high >= low:
mid = (high + low) // 2
if arr[mid] == x:
return 1
elif arr[mid] > x:
return binary_search(arr, low, mid - 1, x)
else:
return binary_search(arr, mid + 1, high, x)
else:
return -1
def position(n,arr,x):
return binary_search(arr,0,n-1,x)
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a sorted array arr[] of N integers and a number K is given. The task is to check if the element K is present in the array or not.
<b>Note:</b> Use binary search to solve the problemThe first line of input contains a number of test cases T. For each test case, the first line of input contains a number of elements in the array, and the number K is separated by space. The next line contains N elements.
<b>Constraints:</b>
1 ≤ T ≤ 10
1 ≤ N ≤ 10<sup>5</sup>
1 ≤ K ≤ 10<sup>9</sup>
1 ≤ arr[i] ≤ 10<sup>9</sup>
<b>Sum of N over all test cases doesn't exceed 10<sup>6</sup></b>If the element is present in the array print "1" else print "-1".Sample Input:
2
5 6
1 2 3 4 6
5 2
1 3 4 5 6
Sample Output:
1
-1, I have written this Solution Code: // arr is they array to search from
// x is target
function binSearch(arr, x) {
// write code here
// do not console.log
// return the 1 or -1
let l = 0;
let r = arr.length - 1;
let mid;
while (r >= l) {
mid = l + Math.floor((r - l) / 2);
// If the element is present at the middle
// itself
if (arr[mid] == x)
return 1;
// If element is smaller than mid, then
// it can only be present in left subarray
if (arr[mid] > x)
r = mid - 1;
// Else the element can only be present
// in right subarray
else
l = mid + 1;
}
// We reach here when element is not
// present in array
return -1;
}, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer <b>N</b>, you need to typecast this integer to String. If the typecasting is done successfully then we will print "<b>Nice Job</b>" otherwise "<b>Wrong answer</b>".User task:
Since this is a functional problem you don't have to worry about the input. You just have to complete the function <b>checkConvertion()</b>, which contains N as a parameter.You need to return the typecasted string value. The driver code will print "<b>Nice Job</b>" otherwise "<b>Wrong answer</b>".Sample Input:
5
Sample Output:
Nice Job
Sample Input:
6
Sample Output:
Nice Job, I have written this Solution Code: def checkConevrtion(a):
return str(a)
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer <b>N</b>, you need to typecast this integer to String. If the typecasting is done successfully then we will print "<b>Nice Job</b>" otherwise "<b>Wrong answer</b>".User task:
Since this is a functional problem you don't have to worry about the input. You just have to complete the function <b>checkConvertion()</b>, which contains N as a parameter.You need to return the typecasted string value. The driver code will print "<b>Nice Job</b>" otherwise "<b>Wrong answer</b>".Sample Input:
5
Sample Output:
Nice Job
Sample Input:
6
Sample Output:
Nice Job, I have written this Solution Code: static String checkConevrtion(int a)
{
return String.valueOf(a);
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of size N, your task is to calculate the total sum of maximum and minimum elements in each subarray of size K.
See example for better understanding.First line of input contains an two space separated integers depicting values of N and K, next line contains N space separated integers depicting values of Arr[i].
Constraints:-
1 < = k < = N < = 100000
-100000 < = Arr[i] < = 100000Print the required sumSample Input:-
5 3
1 2 3 4 5
Sample Output:-
18
Explanation:-
For subarray 1 2 3 :- 1 + 3 = 4
For subarray 2 3 4 :- 2 + 4 = 6
For subarray 3 4 5 :- 3 + 5 = 8
total sum = 4+6+8 = 18
Sample Input:-
7 4
2 5 -1 7 -3 -1 -2
Sample Output:-
18, 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 a[]=br.readLine().split(" ");
int n=Integer.parseInt(a[0]);
int k=Integer.parseInt(a[1]);
String s[]=br.readLine().split(" ");
int arr[]=new int[n];
for(int i=0;i<n;i++)
arr[i]=Integer.parseInt(s[i]);
int min=100000,max=-100000;
long sum=0;
for(int i=0;i<n-k+1;i++){
min=100000;
max=-100000;
for(int j=i;j<k+i;j++){
min=Math.min(arr[j],min);
max=Math.max(arr[j],max);
}
sum=sum+max+min;
}
System.out.print(sum);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of size N, your task is to calculate the total sum of maximum and minimum elements in each subarray of size K.
See example for better understanding.First line of input contains an two space separated integers depicting values of N and K, next line contains N space separated integers depicting values of Arr[i].
Constraints:-
1 < = k < = N < = 100000
-100000 < = Arr[i] < = 100000Print the required sumSample Input:-
5 3
1 2 3 4 5
Sample Output:-
18
Explanation:-
For subarray 1 2 3 :- 1 + 3 = 4
For subarray 2 3 4 :- 2 + 4 = 6
For subarray 3 4 5 :- 3 + 5 = 8
total sum = 4+6+8 = 18
Sample Input:-
7 4
2 5 -1 7 -3 -1 -2
Sample Output:-
18, I have written this Solution Code: from collections import deque
def solve(arr,n,k):
Sum = 0
S = deque()
G = deque()
for i in range(k):
while ( len(S) > 0 and arr[S[-1]] >= arr[i]):
S.pop()
while ( len(G) > 0 and arr[G[-1]] <= arr[i]):
G.pop()
G.append(i)
S.append(i)
for i in range(k, n):
Sum += arr[S[0]] + arr[G[0]]
while ( len(S) > 0 and S[0] <= i - k):
S.popleft()
while ( len(G) > 0 and G[0] <= i - k):
G.popleft()
while ( len(S) > 0 and arr[S[-1]] >= arr[i]):
S.pop()
while ( len(G) > 0 and arr[G[-1]] <= arr[i]):
G.pop()
G.append(i)
S.append(i)
Sum += arr[S[0]] + arr[G[0]]
return Sum
n,k=map(int,input().split())
l=list(map(int,input().split()))
print(solve(l,n,k)), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of size N, your task is to calculate the total sum of maximum and minimum elements in each subarray of size K.
See example for better understanding.First line of input contains an two space separated integers depicting values of N and K, next line contains N space separated integers depicting values of Arr[i].
Constraints:-
1 < = k < = N < = 100000
-100000 < = Arr[i] < = 100000Print the required sumSample Input:-
5 3
1 2 3 4 5
Sample Output:-
18
Explanation:-
For subarray 1 2 3 :- 1 + 3 = 4
For subarray 2 3 4 :- 2 + 4 = 6
For subarray 3 4 5 :- 3 + 5 = 8
total sum = 4+6+8 = 18
Sample Input:-
7 4
2 5 -1 7 -3 -1 -2
Sample Output:-
18, I have written this Solution Code: // C++ program to find sum of all minimum and maximum
// elements Of Sub-array Size k.
#include<bits/stdc++.h>
using namespace std;
#define int long long int
int SumOfKsubArray(int arr[] , int n , int k)
{
int sum = 0; // Initialize result
// The queue will store indexes of useful elements
// in every window
// In deque 'G' we maintain decreasing order of
// values from front to rear
// In deque 'S' we maintain increasing order of
// values from front to rear
deque< int > S(k), G(k);
// Process first window of size K
int i = 0;
for (i = 0; i < k; i++)
{
// Remove all previous greater elements
// that are useless.
while ( (!S.empty()) && arr[S.back()] >= arr[i])
S.pop_back(); // Remove from rear
// Remove all previous smaller that are elements
// are useless.
while ( (!G.empty()) && arr[G.back()] <= arr[i])
G.pop_back(); // Remove from rear
// Add current element at rear of both deque
G.push_back(i);
S.push_back(i);
}
// Process rest of the Array elements
for ( ; i < n; i++ )
{
// Element at the front of the deque 'G' & 'S'
// is the largest and smallest
// element of previous window respectively
sum += arr[S.front()] + arr[G.front()];
// Remove all elements which are out of this
// window
while ( !S.empty() && S.front() <= i - k)
S.pop_front();
while ( !G.empty() && G.front() <= i - k)
G.pop_front();
// remove all previous greater element that are
// useless
while ( (!S.empty()) && arr[S.back()] >= arr[i])
S.pop_back(); // Remove from rear
// remove all previous smaller that are elements
// are useless
while ( (!G.empty()) && arr[G.back()] <= arr[i])
G.pop_back(); // Remove from rear
// Add current element at rear of both deque
G.push_back(i);
S.push_back(i);
}
// Sum of minimum and maximum element of last window
sum += arr[S.front()] + arr[G.front()];
return sum;
}
// Driver program to test above functions
signed main()
{
int n,k;
cin>>n>>k;
int a[n];
for(int i=0;i<n;i++){
cin>>a[i];
}
cout << SumOfKsubArray(a, n, k) ;
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, find the number of occurrences of its smallest character.Input contains a single string S.
Constraints:
1 <= |S| <= 100000
S contains english lowercase letters.Print the number of occurrences of smallest character of S.Sample Input
abcda
Sample Output
2
Explanation: Smallest character is a and it occurs 2 times., I have written this Solution Code: import java.io.*;
import java.util.*;
class Main
{
public static void main (String[] args) throws Exception
{
InputStreamReader ir = new InputStreamReader(System.in);
BufferedReader br = new BufferedReader(ir);
String s = br.readLine();
int[] arr = new int[26];
for (int i=0; i<s.length(); i++) {
arr[(s.charAt(i)-97)]++;
}
for(int i=0; i<26; i++) {
if(arr[i] != 0) {
System.out.println(arr[i]);
break;
}
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a string S, find the number of occurrences of its smallest character.Input contains a single string S.
Constraints:
1 <= |S| <= 100000
S contains english lowercase letters.Print the number of occurrences of smallest character of S.Sample Input
abcda
Sample Output
2
Explanation: Smallest character is a and it occurs 2 times., I have written this Solution Code: #pragma GCC optimize ("Ofast")
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define VV vector
#define pb push_back
#define bitc __builtin_popcountll
#define m_p make_pair
#define infi 1e18+1
#define eps 0.000000000001
#define fastio ios_base::sync_with_stdio(false);cin.tie(NULL);
string char_to_str(char c){string tem(1,c);return tem;}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
template<class T>//usage rand<long long>()
T rand() {
return uniform_int_distribution<T>()(rng);
}
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
template<class T>
using oset = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
// string to integer stoi()
// string to long long stoll()
// string.substr(position,length);
// integer to string to_string();
//////////////
auto clk=clock();
#define all(x) x.begin(),x.end()
#define S second
#define F first
#define sz(x) ((long long)x.size())
#define int long long
#define f80 __float128
#define pii pair<int,int>
/////////////
signed main()
{
fastio;
#ifdef ANIKET_GOYAL
freopen("inputf.in","r",stdin);
freopen("outputf.in","w",stdout);
#endif
string s;
cin>>s;
int n=s.length();
int f[26]={};
for(auto r:s){
f[r-'a']++;
}
for(int i=0;i<26;++i)
if(f[i]!=0){
cout<<f[i];
return 0;
}
#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 a string S, find the number of occurrences of its smallest character.Input contains a single string S.
Constraints:
1 <= |S| <= 100000
S contains english lowercase letters.Print the number of occurrences of smallest character of S.Sample Input
abcda
Sample Output
2
Explanation: Smallest character is a and it occurs 2 times., I have written this Solution Code: import numpy as np
from collections import defaultdict
d=defaultdict(int)
s=input()
for i in s:
d[i]+=1
d=sorted(d.items())
for i in d:
print(i[1])
break, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given an array a that contains N integers. All the integers in the array may not be distinct. The number of repetitions of each integer in the array is represented by ri. Your task is to print the integers in the decreasing order of their occurrence in the array.
Note
1. If ri > rj, then ai must be printed before aj.
2. If ri == rj, then among ai and aj whichever has a larger value, is printed first.
Here ri and rj is the number of repetitions of integers ai and aj in the array.The first line of the input contains an integer N.
The second line contains N space- separated integers representing the elements of array a.
<b>Constraints</b>
1<= N <= 1000
1<= ai <= 1000Print the space separated integers in the decreasing order of their occurrence in the array. The output must be printed in a single line.Sample input:
6
1 2 3 3 2 1
Sample Output
3 2 1, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws IOException{
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int n = Integer.parseInt(br.readLine());
Map<Integer,Integer> map = new HashMap<>();
String []str = br.readLine().split(" ");
for(int i = 0; i < n; i++){
int e = Integer.parseInt(str[i]);
map.put(e, map.getOrDefault(e,0)+1);
}
List<Map.Entry<Integer,Integer>> list = new ArrayList<>(map.entrySet());
Collections.sort(list, new Comparator<Map.Entry<Integer,Integer>>(){
public int compare(Map.Entry<Integer,Integer> o1, Map.Entry<Integer,Integer> o2){
if(o1.getValue() == o2.getValue()){
return o2.getKey() - o1.getKey();
}
return o2.getValue() - o1.getValue();
}
});
for(Map.Entry<Integer,Integer> entry: list){
System.out.print(entry.getKey() + " ");
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given an array a that contains N integers. All the integers in the array may not be distinct. The number of repetitions of each integer in the array is represented by ri. Your task is to print the integers in the decreasing order of their occurrence in the array.
Note
1. If ri > rj, then ai must be printed before aj.
2. If ri == rj, then among ai and aj whichever has a larger value, is printed first.
Here ri and rj is the number of repetitions of integers ai and aj in the array.The first line of the input contains an integer N.
The second line contains N space- separated integers representing the elements of array a.
<b>Constraints</b>
1<= N <= 1000
1<= ai <= 1000Print the space separated integers in the decreasing order of their occurrence in the array. The output must be printed in a single line.Sample input:
6
1 2 3 3 2 1
Sample Output
3 2 1, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
bool static comp(pair<int, int> p, pair<int, int> q) {
if (p.second == q.second) {
return p.first > q.first;
}
return p.second > q.second;
}
vector<int> group_Sol(int N, vector<int> Arr) {
unordered_map<int, int> m;
for (auto &it : Arr) {
m[it] += 1;
}
vector<pair<int, int>> v;
for (auto &it : m) {
v.push_back(it);
}
sort(v.begin(), v.end(), comp);
vector<int> res;
for (auto &it : v) {
res.push_back(it.first);
}
return res;
}
int main() {
ios::sync_with_stdio(0);
cin.tie(0);
int N;
cin >> N;
vector<int> a(N);
for (int i_a = 0; i_a < N; i_a++) {
cin >> a[i_a];
}
vector<int> out_;
out_ = group_Sol(N, a);
cout << out_[0];
for (int i_out_ = 1; i_out_ < out_.size(); i_out_++) {
cout << " " << out_[i_out_];
}
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given an array a that contains N integers. All the integers in the array may not be distinct. The number of repetitions of each integer in the array is represented by ri. Your task is to print the integers in the decreasing order of their occurrence in the array.
Note
1. If ri > rj, then ai must be printed before aj.
2. If ri == rj, then among ai and aj whichever has a larger value, is printed first.
Here ri and rj is the number of repetitions of integers ai and aj in the array.The first line of the input contains an integer N.
The second line contains N space- separated integers representing the elements of array a.
<b>Constraints</b>
1<= N <= 1000
1<= ai <= 1000Print the space separated integers in the decreasing order of their occurrence in the array. The output must be printed in a single line.Sample input:
6
1 2 3 3 2 1
Sample Output
3 2 1, I have written this Solution Code: from collections import defaultdict
import numpy as np
n=int(input())
val=np.array([input().strip().split()],int).flatten()
d=defaultdict(int)
for i in val:
d[i]+=1
a=[]
for i in d:
a.append([d[i],i])
a.sort()
for i in range(len(a)-1,-1,-1):
print(a[i][1],end=" "), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a string S of length N. Find the character which occurs the most number of times in this string. If there are mutliple such characters present, print -1.The first line of the input contains a single integer N.
The second line of the input contains a string S.
<b>Constraints:</b>
2 <b>≤</b> N <b>≤</b> 10<sup>5</sup>
All characters in string S are lowercase english letters.Print the character which occurs the most number of times in this string. If there are mutliple such characters present, print -1.Sample Input:
7
abadcar
Sample Output:
a, 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 s = sc.next();
int arr[]= new int[26];
for(int i=0;i<n;i++){
char ch = s.charAt(i);
int j = ch - 'a';
arr[j]++;
}
int max = 0;
int sec = 0;
int fr = 0;
int j =0;
for(int i=0;i<26; i++){
if(arr[i]> max){
sec = max;
max = arr[i];
j=i;
}else if(arr[i]>sec){
sec = arr[i];
}
}
if(sec == max){
System.out.print(-1);
}else{
for(int i=0;i<n; i++){
char c = s.charAt(i);
if(c-'a' == j){
System.out.print(c);
break;
}
}
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a string S of length N. Find the character which occurs the most number of times in this string. If there are mutliple such characters present, print -1.The first line of the input contains a single integer N.
The second line of the input contains a string S.
<b>Constraints:</b>
2 <b>≤</b> N <b>≤</b> 10<sup>5</sup>
All characters in string S are lowercase english letters.Print the character which occurs the most number of times in this string. If there are mutliple such characters present, print -1.Sample Input:
7
abadcar
Sample Output:
a, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define int long long
signed main(){
int n;
cin >> n;
string s;
cin >> s;
map<char, int> mp;
for(auto i : s) mp[i]++;
int mx = 0, f = 0;
char ch;
for(auto i : mp){
if(i.second > mx){
mx = i.second;
f = 1;
ch = i.first;
}
else if(i.second == mx){
f++;
}
}
if(f == 1){
cout << ch;
}
else cout << -1;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given N and K, find the lexicographically smallest string of length N using only the first K lowercase letters of the alphabet such that each letter is used at least once and no two adjacent characters are equal.
If such a string doesn't exist, print -1.The first line of input contains a single integer, T (1 <= T <= 100).
Then T lines follow, each containing two space-separated integers, N (1 <= N <= 10<sup>5</sup>) and K (1 <= K <= 26).
It is guaranteed that sum of N over all test cases does not exceed 10<sup>6</sup>For each test case, output its answer in a new line.Sample Input:
2
2 3
3 2
Sample Output:
-1
aba, 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));
BufferedWriter bo=new BufferedWriter(new OutputStreamWriter(System.out));
int t;
try{
t=Integer.parseInt(br.readLine());
}
catch(Exception e)
{
return;
}
while(t-->0)
{
String[] g=br.readLine().split(" ");
int n=Integer.parseInt(g[0]);
int k=Integer.parseInt(g[1]);
if(k>n || (k==1) || (k>26))
{
if(n==1 && k==1)
bo.write("a\n");
else
bo.write(-1+"\n");
}
else
{
int extra=k-2;
boolean check=true;
while(n>extra)
{
if(check==true)
bo.write("a");
else
bo.write("b");
if(check==true)
check=false;
else
check=true;
n--;
}
for(int i=0;i<extra;i++)
bo.write((char)(i+99));
bo.write("\n");
}
}
bo.close();
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given N and K, find the lexicographically smallest string of length N using only the first K lowercase letters of the alphabet such that each letter is used at least once and no two adjacent characters are equal.
If such a string doesn't exist, print -1.The first line of input contains a single integer, T (1 <= T <= 100).
Then T lines follow, each containing two space-separated integers, N (1 <= N <= 10<sup>5</sup>) and K (1 <= K <= 26).
It is guaranteed that sum of N over all test cases does not exceed 10<sup>6</sup>For each test case, output its answer in a new line.Sample Input:
2
2 3
3 2
Sample Output:
-1
aba, I have written this Solution Code: t=int(input())
for tt in range(t):
n,k=map(int,input().split())
if (k==1 and n>1) or (k>n):
print(-1)
continue
s="abcdefghijklmnopqrstuvwxyz"
ss="ab"
if (n-k)%2==0:
a=ss*((n-k)//2)+s[:k]
else:
a=ss*((n-k)//2)+s[:2]+"a"+s[2:k]
print(a), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given N and K, find the lexicographically smallest string of length N using only the first K lowercase letters of the alphabet such that each letter is used at least once and no two adjacent characters are equal.
If such a string doesn't exist, print -1.The first line of input contains a single integer, T (1 <= T <= 100).
Then T lines follow, each containing two space-separated integers, N (1 <= N <= 10<sup>5</sup>) and K (1 <= K <= 26).
It is guaranteed that sum of N over all test cases does not exceed 10<sup>6</sup>For each test case, output its answer in a new line.Sample Input:
2
2 3
3 2
Sample Output:
-1
aba, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define fast ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
typedef long long int ll;
typedef unsigned long long int ull;
const long double PI = acos(-1);
const ll mod=1e9+7;
const ll mod1=998244353;
const int inf = 1e9;
const ll INF=1e18;
void precompute(){
}
void TEST_CASE(){
int n,k;
cin >> n >> k;
if(k==1){
if(n>1){
cout << -1 << endl;
}else{
cout << 'a' << endl;
}
}else if(n<k){
cout << -1 << endl;
}else if(n==k){
string s="";
for(int i=0 ; i<k ; i++){
s+=('a'+i);
}
cout << s << endl;
}else{
string s="";
for(int i=0 ; i<(n-k+2) ; i++){
if(i%2){
s+="b";
}else{
s+="a";
}
}
for(int i=2 ; i<k ; i++){
s+=('a'+i);
}
cout << s << endl;
}
}
signed main(){
fast;
//freopen ("INPUT.txt","r",stdin);
//freopen ("OUTPUT.txt","w",stdout);
int test=1,TEST=1;
precompute();
cin >> test;
while(test--){
TEST_CASE();
}
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A[] of size N containing non-negative integers. You have to move all 0's to the end of array while maintaining the relative order of the non-zero elements.
Note:
You must do this in-place without making a copy of the array.
Minimize the total number of operations.The first line of input line contains T, denoting the number of testcases. Each test cases contains 2 lines. First-line contains N, size of array. Second-line contains elements of array separated by space.
Constraints:
1 <= T <= 100
1 <= N <= 10^5
1 <= A[i] <= 10^5
<b>Sum of N over all testcases does not exceed 10^6</b>
For each testcase you need to print the updated array.Sample Input:
2
5
0 1 0 3 12
8
0 0 0 0 1 2 3 4
Sample Output:
1 3 12 0 0
1 2 3 4 0 0 0 0, I have written this Solution Code:
import java.util.*;
import java.lang.*;
import java.io.*;
class Main
{
public static void main (String[] args) throws java.lang.Exception
{
Scanner sc = new Scanner(System.in);
int t;
t=sc.nextInt();
while(t-->0){
int n=sc.nextInt();
int a[] = new int[n];
for(int i=0;i<n;i++){
a[i]=sc.nextInt();
}
int j=-1;
for(int i=0;i<n;i++){
if(a[i]==0 && j==-1){
j=i;
}
if(j!=-1 && a[i]!=0){
a[j]=a[i];
a[i]=0;
j++;
}
}
for(int i=0;i<n;i++){
System.out.print(a[i]+" ");
}
System.out.println();
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A[] of size N containing non-negative integers. You have to move all 0's to the end of array while maintaining the relative order of the non-zero elements.
Note:
You must do this in-place without making a copy of the array.
Minimize the total number of operations.The first line of input line contains T, denoting the number of testcases. Each test cases contains 2 lines. First-line contains N, size of array. Second-line contains elements of array separated by space.
Constraints:
1 <= T <= 100
1 <= N <= 10^5
1 <= A[i] <= 10^5
<b>Sum of N over all testcases does not exceed 10^6</b>
For each testcase you need to print the updated array.Sample Input:
2
5
0 1 0 3 12
8
0 0 0 0 1 2 3 4
Sample Output:
1 3 12 0 0
1 2 3 4 0 0 0 0, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main(){
int t;
cin>>t;
while(t--){
int n,k;
cin>>n;
int a[n];
int cnt=0;
for(int i=0;i<n;i++){
cin>>a[i];\
if(a[i]==0){cnt++;}
}
for(int i=0;i<n;i++){
if(a[i]!=0){
cout<<a[i]<<" ";
}
}
while(cnt--){
cout<<0<<" ";
}
cout<<endl;
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Write a program to check whether a number is even or odd using switch case.First Line of the input contains the number n.
<b>Constraints</b>
1 <= n <= 1e9If the number is even print "Even" otherwise "Odd"Sample Input :
23
Sample Output :
Odd
Sample Input :
24
Sample Output :
Even, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) {
try{
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int t= Integer.parseInt(br.readLine());
if(t%2==0)
System.out.println("Even");
else
System.out.println("Odd");
}
catch (Exception e){
return;
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Write a program to check whether a number is even or odd using switch case.First Line of the input contains the number n.
<b>Constraints</b>
1 <= n <= 1e9If the number is even print "Even" otherwise "Odd"Sample Input :
23
Sample Output :
Odd
Sample Input :
24
Sample Output :
Even, I have written this Solution Code: n = int(input())
if n % 2 == 0:
print("Even")
else:
print("Odd"), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Write a program to check whether a number is even or odd using switch case.First Line of the input contains the number n.
<b>Constraints</b>
1 <= n <= 1e9If the number is even print "Even" otherwise "Odd"Sample Input :
23
Sample Output :
Odd
Sample Input :
24
Sample Output :
Even, I have written this Solution Code: #include <stdio.h>
int main()
{
int num;
scanf("%d", &num);
switch(num % 2)
{
case 0:
printf("Even");
break;
/* Else if n%2 == 1 */
case 1:
printf("Odd");
break;
}
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given an unsorted array of length n and must sort it using merge sort while also printing the amount of merges that occur throughout the sorting process.The first line of input will be n, which represents the array's length, followed by the n array items in the second line.
Constraints:
0<= n <=100000First- line should be the sorted array and the second should be the number of mergers that occurs when the array is sorted using merge sort.Sample Input:
5
5 1 2 7 3
Output:
1 2 3 5 7
4, I have written this Solution Code: def mergeSort(arr,count):
if len(arr)>1:
mid= len(arr)//2
a=arr[:mid]
b=arr[mid:]
count=mergeSort(a,count)
count=mergeSort(b,count)
count+=1
i = j = k = 0
l1 = len(a)
l2 = len(b)
while i< l1 and j <l2:
if a[i]<b[j]:
arr[k]=a[i]
i +=1
else:
arr[k]=b[j]
j+=1
k+=1
while i <l1:
arr[k]=a[i]
i+=1
k+=1
while j<l2:
arr[k]=b[j]
j+=1
k+=1
return count
N=int(input())
arr=list(map(int,input().split()))
count=mergeSort(arr,0)
print(' '.join(map(str,arr)))
print(count), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given an unsorted array of length n and must sort it using merge sort while also printing the amount of merges that occur throughout the sorting process.The first line of input will be n, which represents the array's length, followed by the n array items in the second line.
Constraints:
0<= n <=100000First- line should be the sorted array and the second should be the number of mergers that occurs when the array is sorted using merge sort.Sample Input:
5
5 1 2 7 3
Output:
1 2 3 5 7
4, I have written this Solution Code: import java.util.Scanner;
public class Main
{
int noOfMerge=0;
void merge(int arr[], int l, int m, int r)
{
int n1 = m - l + 1;
int n2 = r - m;
int L[] = new int [n1];
int R[] = new int [n2];
for (int i=0; i<n1; ++i)
L[i] = arr[l + i];
for (int j=0; j<n2; ++j)
R[j] = arr[m + 1+ j];
int i = 0, j = 0,k=l;
while (i < n1 && j < n2) {
if (L[i] <= R[j]) {
arr[k] = L[i];
i++;
} else {
arr[k] = R[j];
j++;
}
k++;
}
while (i < n1)
{
arr[k] = L[i];
i++;
k++;
}
while (j < n2)
{
arr[k] = R[j];
j++;
k++;
}
}
void sort(int arr[], int l, int r)
{
if (l < r)
{
int m = (l+r)/2;
sort(arr, l, m);
sort(arr , m+1, r);
merge(arr, l, m, r);
noOfMerge+=1;
}
}
public static void main(String args[])
{
Scanner scanner = new Scanner(System.in);
int n = scanner.nextInt();
int[] array=new int[n];
for(int i=0;i<n;i++)
array[i]=scanner.nextInt();
Main ob = new Main();
ob.sort(array, 0, n-1);
for (int i=0; i<n; ++i)
System.out.print(array[i] + " ");
System.out.println();
System.out.println(ob.noOfMerge);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a binary tree containing with N nodes and an integer X. Your task is to complete the function countSubtreesWithSumX() that returns the count of the number of subtress having total node’s data sum equal to a value X.
Example: A tree given below<b>User Task:</b>
Since this will be a functional problem. You don't have to take input. You just have to complete the function <b>countSubtreesWithSumX()</b> that takes "root" node and the integer x as parameter.
<b>Constraints:</b>
1 <= T <= 100
1 <= N <= 10^3
1 <= node values <= 10^4
<b>Sum of "N" over all testcases does not exceed 10^5</b>Return the number of subtrees with sum X. The driver code will take care of printing it.Sample Input:
1
3 5
1 2 3
Sum=5
Tree:-
1
/ \
2 3
Sample Output:
0
Explanation:
No subtree has a sum equal to 5.
Sample Input:-
1
5 5
2 1 3 4 5
Sum=5
Tree:-
2
/ \
1 3
/ \
4 5
Sample Output:-
1, I have written this Solution Code: static int c = 0;
static int countSubtreesWithSumXUtil(Node root,int x)
{
// if tree is empty
if (root==null)return 0;
// sum of nodes in the left subtree
int ls = countSubtreesWithSumXUtil(root.left,x);
// sum of nodes in the right subtree
int rs = countSubtreesWithSumXUtil(root.right, x);
int sum = ls + rs + root.data;
// if tree's nodes sum == x
if (sum == x)c++;
return sum;
}
static int countSubtreesWithSumX(Node root, int x)
{
c = 0;
// if tree is empty
if (root==null)return 0;
// sum of nodes in the left subtree
int ls = countSubtreesWithSumXUtil(root.left, x);
// sum of nodes in the right subtree
int rs = countSubtreesWithSumXUtil(root.right, x);
// check if above sum is equal to x
if ((ls + rs + root.data) == x)c++;
return c;
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You're given a string S of lowercase letters of the english alphabet. Find whether you can choose some characters of the string S in any order to create the string "red".The first and the only line of input contains the string S.
Constraints
1 <= |S| <= 100
All the characters in S are lowercase letters of the english alphabet.Output "Yes" (without quotes) if you can create the string "red", else output "No" (without quotes).Sample Input
damngrey
Sample Output
Yes
Explanation: We choose character at position 6, then position 7, then position 1.
Sample Input
newtonschool
Sample Output
No, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
String str = br.readLine();
boolean r=false,e=false,d=false;
for(int i=0;i<str.length();i++){
char ch = str.charAt(i);
if(ch=='r')r=true;
if(ch=='e')e=true;
if(ch=='d')d=true;
}
String ans = (r && e && d)?"Yes":"No";
System.out.print(ans);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You're given a string S of lowercase letters of the english alphabet. Find whether you can choose some characters of the string S in any order to create the string "red".The first and the only line of input contains the string S.
Constraints
1 <= |S| <= 100
All the characters in S are lowercase letters of the english alphabet.Output "Yes" (without quotes) if you can create the string "red", else output "No" (without quotes).Sample Input
damngrey
Sample Output
Yes
Explanation: We choose character at position 6, then position 7, then position 1.
Sample Input
newtonschool
Sample Output
No, I have written this Solution Code: S = input()
if ("r" in S and "e" in S and "d" in S ) :
print("Yes")
else:
print("No"), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You're given a string S of lowercase letters of the english alphabet. Find whether you can choose some characters of the string S in any order to create the string "red".The first and the only line of input contains the string S.
Constraints
1 <= |S| <= 100
All the characters in S are lowercase letters of the english alphabet.Output "Yes" (without quotes) if you can create the string "red", else output "No" (without quotes).Sample Input
damngrey
Sample Output
Yes
Explanation: We choose character at position 6, then position 7, then position 1.
Sample Input
newtonschool
Sample Output
No, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define sd(x) scanf("%d", &x)
#define sz(v) (int) v.size()
#define pr(v) For(i, 0, sz(v)) {cout<<v[i]<<" ";} cout<<endl;
#define slld(x) scanf("%lld", &x)
#define all(x) x.begin(), x.end()
#define For(i, st, en) for(int i=st; i<en; i++)
#define tr(x) for(auto it=x.begin(); it!=x.end(); it++)
#define fast std::ios::sync_with_stdio(false);cin.tie(NULL);
#define pb push_back
#define ll long long
#define ld long double
#define int long long
#define double long double
#define mp make_pair
#define F first
#define S second
typedef pair<int, int> pii;
typedef vector<int> vi;
#define pi 3.141592653589793238
const int MOD = 1e9+7;
const int INF = 1LL<<60;
const int N = 2e5+5;
// it's swapnil07 ;)
#ifdef SWAPNIL07
#define trace(...) __f(#__VA_ARGS__, __VA_ARGS__)
template <typename Arg1>
void __f(const char* name, Arg1&& arg1){
cout << name << " : " << arg1 << endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args){
const char* comma = strchr(names + 1, ',');cout.write(names, comma - names) << " : " << arg1<<" | ";__f(comma+1, args...);
}
int begtime = clock();
#define end_routine() cout << "\n\nTime elapsed: " << (clock() - begtime)*1000/CLOCKS_PER_SEC << " ms\n\n";
#else
#define endl '\n'
#define trace(...)
#define end_routine()
#endif
void solve(){
string s; cin>>s;
map<char, int> m;
For(i, 0, sz(s)){
m[s[i]]++;
}
if(m['r'] && m['e'] && m['d']){
cout<<"Yes";
}
else{
cout<<"No";
}
}
signed main()
{
fast
#ifdef SWAPNIL07
freopen("input.txt","r",stdin);
freopen("output.txt","w",stdout);
#endif
int t=1;
// cin>>t;
while(t--){
solve();
cout<<"\n";
}
return 0;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: A string is called amicable if there is no character in the string which appears exactly once. For example, "abcccba" is amicable, whereas "abcba" is not amicable as the character 'c' appears only once. Your task is to find the total number of amicable substrings of a given string S.
Note:
Two substrings are considered different if either their starting position or their ending position are different, even if their contents may be equal.The first line consists of a single integer T – the number of test cases.
Then T lines follow, the i<sup>th</sup> line containing a string S for the i<sup>th</sup> test case.
<b>Constraints:</b>
1 ≤ T ≤ 10<sup>5</sup>
The sum of lengths of strings across all test cases is less than or equal to 10<sup>5</sup>.For each test case, print a single line containing a single integer – the number of amicable substrings of string S.Sample Input 1:
3
abbac
bvb
sasa
Sample Output 1:
2
0
1
Explanation:
For the first test case, the amicable substrings are "bb" and "abba".
For the second test case, there are no amicable substrings.
For the third test case, there is only one amicable substring "sasa".
Sample Input 2:
3
meeoowww
idqaislac
eommrejv
Sample Output 2:
10
0
1, 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 = 2e5 + 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);
read(t);
while (t--)
{
reads(s);
int n = sz(s), ans = 0, valid = 0;
vi l(26), sl(26), cnt(n + 1);
FOR (i, 1, n)
{
int ch = s[i - 1] - 'a';
valid++;
FOR (j, sl[ch] + 1, l[ch])
{
cnt[j]--;
if (j > 0 && cnt[j] == 0) valid++;
}
FOR (j, l[ch] + 1, i)
{
cnt[j]++;
if (j > 0 && cnt[j] == 1) valid--;
}
sl[ch] = l[ch];
l[ch] = i;
ans += valid;
}
print(ans);
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: A string is called amicable if there is no character in the string which appears exactly once. For example, "abcccba" is amicable, whereas "abcba" is not amicable as the character 'c' appears only once. Your task is to find the total number of amicable substrings of a given string S.
Note:
Two substrings are considered different if either their starting position or their ending position are different, even if their contents may be equal.The first line consists of a single integer T – the number of test cases.
Then T lines follow, the i<sup>th</sup> line containing a string S for the i<sup>th</sup> test case.
<b>Constraints:</b>
1 ≤ T ≤ 10<sup>5</sup>
The sum of lengths of strings across all test cases is less than or equal to 10<sup>5</sup>.For each test case, print a single line containing a single integer – the number of amicable substrings of string S.Sample Input 1:
3
abbac
bvb
sasa
Sample Output 1:
2
0
1
Explanation:
For the first test case, the amicable substrings are "bb" and "abba".
For the second test case, there are no amicable substrings.
For the third test case, there is only one amicable substring "sasa".
Sample Input 2:
3
meeoowww
idqaislac
eommrejv
Sample Output 2:
10
0
1, I have written this Solution Code: import java.util.*;
import java.io.*;
public class Main {
static long mod = 1000000007;
static PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out));
public static void main(String[] args) throws IOException {
FastReader sc = new FastReader();
int t = sc.nextInt();
while( t-- > 0) {
char arr[] = sc.next().toCharArray();
int n = arr.length;
long total = ((1L)*(n)*(n+1))/2;
long ans = 0;
for( int i = 0 ;i < n ;i++) {
int a[] = new int[26];
int two = 0;
int one = 0;
int zero = 26;
for( int j = i ; j < n && two != 26 ;j++) {
int idx = arr[j]-'a';
if( a[idx] == 1) {
a[idx]++;
one--;
two++;
}
else if( a[idx] == 0) {
one++;
zero--;
a[idx]++;
}
if( one != 0) {
ans++;
}
}
for( int j = 0 ;j < 26 ; j++) {
a[j] = 0;
}
}
out.println(total - ans);
}
out.flush();
}
public static boolean ifpowof2(long n ) {
return ((n&(n-1)) == 0);
}
static boolean isprime(long x ) {
if( x== 2) {
return true;
}
if( x%2 == 0) {
return false;
}
for( long i = 3 ;i*i <= x ;i+=2) {
if( x%i == 0) {
return false;
}
}
return true;
}
static boolean[] sieveOfEratosthenes(long n) {
boolean prime[] = new boolean[(int)n + 1];
for (int i = 0; i <= n; i++) {
prime[i] = true;
}
for (long p = 2; p * p <= n; p++) {
if (prime[(int)p] == true) {
for (long i = p * p; i <= n; i += p)
prime[(int)i] = false;
}
}
return prime;
}
public static int[] nextLargerElement(int[] arr, int n) {
Stack<Integer> stack = new Stack<>();
int rtrn[] = new int[n];
rtrn[n-1] = -1;
stack.push( n-1);
for( int i = n-2 ;i >= 0 ; i--){
int temp = arr[i];
int lol = -1;
while( !stack.isEmpty() && arr[stack.peek()] <= temp){
if(arr[stack.peek()] == temp ) {
lol = stack.peek();
}
stack.pop();
}
if( stack.isEmpty()){
if( lol != -1) {
rtrn[i] = lol;
}
else {
rtrn[i] = -1;
}
}
else{
rtrn[i] = stack.peek();
}
stack.push( i);
}
return rtrn;
}
static void mysort(int[] arr) {
for(int i=0;i<arr.length;i++) {
int rand = (int) (Math.random() * arr.length);
int loc = arr[rand];
arr[rand] = arr[i];
arr[i] = loc;
}
Arrays.sort(arr);
}
static void mySort(long[] arr) {
for(int i=0;i<arr.length;i++) {
int rand = (int) (Math.random() * arr.length);
long loc = arr[rand];
arr[rand] = arr[i];
arr[i] = loc;
}
Arrays.sort(arr);
}
static long gcd(long a, long b)
{
if (a == 0)
return b;
return gcd(b % a, a);
}
static long lcm(long a, long b)
{
return (a / gcd(a, b)) * b;
}
static long rightmostsetbit(long n) {
return n&-n;
}
static long leftmostsetbit(long n)
{
long k = (long)(Math.log(n) / Math.log(2));
return k;
}
static HashMap<Long,Long> primefactor( long n){
HashMap<Long ,Long> hm = new HashMap<>();
long temp = 0;
while( n%2 == 0) {
temp++;
n/=2;
}
if( temp!= 0) {
hm.put( 2L, temp);
}
long c = (long)Math.sqrt(n);
for( long i = 3 ; i <= c ; i+=2) {
temp = 0;
while( n% i == 0) {
temp++;
n/=i;
}
if( temp!= 0) {
hm.put( i, temp);
}
}
if( n!= 1) {
hm.put( n , 1L);
}
return hm;
}
static ArrayList<Long> allfactors(long abs) {
HashMap<Long,Integer> hm = new HashMap<>();
ArrayList<Long> rtrn = new ArrayList<>();
for( long i = 2 ;i*i <= abs; i++) {
if( abs% i == 0) {
hm.put( i , 0);
hm.put(abs/i, 0);
}
}
for( long x : hm.keySet()) {
rtrn.add(x);
}
if( abs != 0) {
rtrn.add(abs);
}
return rtrn;
}
public static int[][] prefixsum( int n , int m , int arr[][] ){
int prefixsum[][] = new int[n+1][m+1];
for( int i = 1 ;i <= n ;i++) {
for( int j = 1 ; j<= m ; j++) {
int toadd = 0;
if( arr[i-1][j-1] == 1) {
toadd = 1;
}
prefixsum[i][j] = toadd + prefixsum[i][j-1] + prefixsum[i-1][j] - prefixsum[i-1][j-1];
}
}
return prefixsum;
}
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;
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given resistance value of N resistors. Find the net resistance of the system when all of these N resistors are connected in parallel.
If there are three resistors A1, A2, A3, when they are connected in parallel, the net resistance will be 1/((1/A1) + (1/A2) + (1/A3))
Since this number can also have a fraction part, you only have to print the floor of the result obtained.
For example, if value of 1/((1/A1) + (1/A2) + (1/A3)) if 7.54567, you only have to print 7.First line contains a single integer N denoting the number of resistors.
Next line contains N space separated integers containing the value of different resistors.
Constraints
1 ≤ N ≤ 10^5
1 ≤ Ai ≤ 10^9Print the integral part or floor of the value obtained from the formula 1/((1/A1) + (1/A2) + ..... + (1/AN)).Input
2
10 30
Output
7
1/((1/10) + (1/30)) = 30/4 = 7.5 and floor of 7.5 is 7, 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());
double arr[] = new double[N];
String str[] = br.readLine().trim().split(" ");
for(int i=0;i<N;i++)
arr[i]=Integer.parseInt(str[i]);
double resistance=0;
int equResistance=0;
for(int i=0;i<N;i++)
arr[i]=Integer.parseInt(str[i]);
for(int i=0;i<N;i++)
{
resistance=resistance+(1/arr[i]);
}
equResistance = (int)Math.floor((1/resistance));
System.out.println(equResistance);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given resistance value of N resistors. Find the net resistance of the system when all of these N resistors are connected in parallel.
If there are three resistors A1, A2, A3, when they are connected in parallel, the net resistance will be 1/((1/A1) + (1/A2) + (1/A3))
Since this number can also have a fraction part, you only have to print the floor of the result obtained.
For example, if value of 1/((1/A1) + (1/A2) + (1/A3)) if 7.54567, you only have to print 7.First line contains a single integer N denoting the number of resistors.
Next line contains N space separated integers containing the value of different resistors.
Constraints
1 ≤ N ≤ 10^5
1 ≤ Ai ≤ 10^9Print the integral part or floor of the value obtained from the formula 1/((1/A1) + (1/A2) + ..... + (1/AN)).Input
2
10 30
Output
7
1/((1/10) + (1/30)) = 30/4 = 7.5 and floor of 7.5 is 7, I have written this Solution Code: r = input("")
r = int(r)
n = input("").split()
resistance=0.0
for i in range(0,r):
resistor = float(n[i])
resistance = resistance + (1/resistor)
print(int(1/resistance)), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given resistance value of N resistors. Find the net resistance of the system when all of these N resistors are connected in parallel.
If there are three resistors A1, A2, A3, when they are connected in parallel, the net resistance will be 1/((1/A1) + (1/A2) + (1/A3))
Since this number can also have a fraction part, you only have to print the floor of the result obtained.
For example, if value of 1/((1/A1) + (1/A2) + (1/A3)) if 7.54567, you only have to print 7.First line contains a single integer N denoting the number of resistors.
Next line contains N space separated integers containing the value of different resistors.
Constraints
1 ≤ N ≤ 10^5
1 ≤ Ai ≤ 10^9Print the integral part or floor of the value obtained from the formula 1/((1/A1) + (1/A2) + ..... + (1/AN)).Input
2
10 30
Output
7
1/((1/10) + (1/30)) = 30/4 = 7.5 and floor of 7.5 is 7, I have written this Solution Code: #include "bits/stdc++.h"
#pragma GCC optimize "03"
using namespace std;
#define int long long int
#define ld long double
#define pi pair<int, int>
#define pb push_back
#define fi first
#define se second
#define IOS ios::sync_with_stdio(false); cin.tie(0); cout.tie(0)
#ifndef LOCAL
#define endl '\n'
#endif
const int N = 2e5 + 5;
const int mod = 1e9 + 7;
const int inf = 1e9 + 9;
signed main() {
IOS;
int n; cin >> n;
double s = 0;
for(int i = 1; i <= n; i++){
double p; cin >> p;
s = s + (1/p);
}
s = 1/s;
cout << floor(s);
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given an array A of size N. Find the number of pairs of indices (i, j) in the array A such that i < j and A<sub>i</sub> - i = A<sub>j</sub> - j.The first line of the input contains a single integer N.
The second line of the input contains N space seperated integers.
Constraints:
1 <= N <= 10<sup>5</sup>
1 <= A<sub>i</sub> <= 10<sup>5</sup>Print the number of pairs of indices (i, j) in the given array A such that i < j and A<sub>i</sub> - i = A<sub>j</sub> - j.Sample Input:
4
1 3 3 4
Sample Output:
3
Explaination:
The three pairs of indices are:
(1, 3) -> A[1] - 1 = A[3] - 3 -> 1 - 1 = 3 - 3 -> 0 = 0
(1, 4) -> A[1] - 1 = A[4] - 4 -> 1 - 1 = 4 - 4 -> 0 = 0
(3, 4) -> A[3] - 3 = A[4] - 4 -> 3 - 3 = 4 - 4 -> 0 = 0, I have written this Solution Code: #include<bits/stdc++.h>
using namespace std;
#define int long long
signed main() {
int n;
cin >> n;
map<int, int> mp;
int ans = 0;
for(int i = 1; i <= n; ++i){
int x;
cin >> x;
mp[x - i]++;
}
for(auto i:mp)
ans += i.second * (i.second - 1)/2;
cout << ans;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a simple undirected graph consisting of N nodes and M edges. Count the number of distinct ways of removing a (possibly empty) subset of edges such that the <b>remaining graph on N nodes remains edge biconnected</b>.
<ul>
<li> A graph is said to be <b>edge biconnected</b> if, for every pair of distinct nodes, there exist two edge-disjoint paths between them. Two paths are edge-disjoint if they do not share an edge.
</ul>
The number of ways may be large, so print the answer modulo 998244353.The first line of the input contains two integers, N and M, the number of nodes in the graph and the number of edges in the graph.
M lines follow, each containing 2 integers, u<sub>i</sub> and v<sub>i</sub> denoting an edge in the graph.
<b>Constraints</b>
3 <= N <= 17
1 <= M <= (N*(N-1)) / 2
1 <= u<sub>i</sub>, v<sub>i</sub> <= N for all 1 <= i <= M
It is guaranteed that there are no self-loops and multiple edges in the graph.
Print a single integer, the required number of ways modulo 998244353.Sample Input 1:
4 5
1 2
2 3
3 4
4 1
4 2
Sample Output 1:
2
<b>Explanation:</b>
For the first sample input, the two ways are to remove either the empty set of edges or to remove the edge indexed 5.
Sample Input 2:
4 6
1 2
2 3
3 4
4 1
1 3
2 4
Sample Output 2:
10
<b>Explanation:</b>
For the second sample, it can be shown that there are 6 ways to remove 1 edge, 3 ways to remove 2 edges, and 1 way to remove 0 edges., I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define debug(...) fprintf (stderr, __VA_ARGS__)
#define fi first
#define se second
#define Lep(i, l, r) for (int i = (l), i##_end = (r); i < i##_end; ++ i)
#define lep(i, l, r) for (int i = (l), i##_end = (r); i <= i##_end; ++ i)
#define rep(i, r, l) for (int i = (r), i##_end = (l); i >= i##_end; -- i)
char __c; bool __f; template <class T> inline void IN (T & x) {
__f = 0, x = 0; while (__c = getchar (), ! isdigit (__c)) if (__c == '-') __f = 1;
while (isdigit (__c)) x = x * 10 + __c - '0', __c = getchar (); if (__f) x = -x;
}
template <class T> inline void chkmin (T & x, T y) { if (x > y) x = y; }
template <class T> inline void chkmax (T & x, T y) { if (x < y) x = y; }
const int mod = 998244353;
inline int mul (int x, int y) { return 1ll * x * y % mod; }
inline void pls (int &x, int y) { x += y; if (x >= mod) x -= mod; }
inline int add (int x, int y) { x += y; if (x >= mod) x -= mod; return x; }
inline int modpow (int x, int y, int r = 1) {
for ( ; y; y >>= 1, x = mul (x, x)) if (y & 1) r = mul (r, x);
return r;
}
const int N = 20;
const int NS = (1 << 18) | 5;
int inv[N];
inline void init (int L = N - 2) {
inv[1] = 1;
lep (i, 2, L) inv[i] = mod - mul (inv[mod % i], mod / i);
}
inline void fmt (int f[NS][N], int n) {
Lep (p, 0, n) for (int k = 0; k < (1 << n); k += (1 << p + 1))
Lep (l, k, k | (1 << p)) {
lep (d, 1, n) pls (f[l | (1 << p)][d], f[l][d]);
}
}
inline void ifmt (int f[NS][N], int n) {
Lep (p, 0, n) for (int k = 0; k < (1 << n); k += (1 << p + 1))
Lep (l, k, k | (1 << p)) {
lep (d, 1, n) pls (f[l | (1 << p)][d], mod - f[l][d]);
}
}
inline void ln (int *f, int n) {
static int g[N];
lep (i, 0, n - 1) {
g[i] = mul (f[i + 1], i + 1);
lep (j, 0, i - 1) pls (g[i], mod - mul (g[j], f[i - j]));
}
lep (i, 0, n - 1) f[i + 1] = mul (g[i], inv[i + 1]);
}
inline void exp (int *f, int n) {
static int g[N];
lep (i, 0, n - 1) g[i] = mul (f[i + 1], i + 1), f[i + 1] = 0;
lep (i, 0, n - 1) {
lep (j, 0, i - 1) pls (f[i + 1], mul (g[j], f[i - j]));
f[i + 1] = mul (add (g[i], f[i + 1]), inv[i + 1]);
}
}
int n, m, G[N];
int sf[NS][N];
signed main () {
init ();
IN (n), IN (m);
for (int u, v; m; -- m) {
IN (u), IN (v), -- u, -- v;
G[u] |= 1 << v, G[v] |= 1 << u;
}
if (n == 1) return puts ("1"), 0;
Lep (S, 1, 1 << n) {
int cnt = 0;
Lep (i, 0, n) if ((S >> i) & 1) cnt += __builtin_popcount (G[i] & S);
sf[S][__builtin_popcount (S)] = modpow (2, cnt / 2);
}
fmt (sf, n);
Lep (S, 1, 1 << n) ln (sf[S], n);
Lep (i, 0, n) {
for (int k = 0; k < (1 << n); k += (1 << i + 1))
Lep (l, k, k | (1 << i)) {
lep (d, 1, n) pls (sf[l | (1 << i)][d], mod - sf[l][d]);
ln (sf[l | (1 << i)] + 1, n);
lep (d, 1, n) pls (sf[l | (1 << i)][d], sf[l][d]);
}
}
ifmt (sf, n);
Lep (S, 1, 1 << n) if (__builtin_popcount (S) <= 2) sf[S][__builtin_popcount (S)] = 0;
fmt (sf, n);
Lep (i, 0, n) {
for (int k = 0; k < (1 << n); k += (1 << i + 1))
Lep (l, k, k | (1 << i)) {
lep (d, 1, n) pls (sf[l | (1 << i)][d], mod - sf[l][d]);
exp (sf[l | (1 << i)] + 1, n);
lep (d, 1, n) pls (sf[l | (1 << i)][d], sf[l][d]);
}
}
ifmt (sf, n);
printf ("%d\n", sf[(1 << n) - 1][n]);
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Morgan was opening some boxes when she found one bearing a message “For Morgan, Dad loves you 3000". She wants to open the box but she needs to solve the question to open it. She asks Peter to do that. Help Peter and Morgan solve the question so that they can see what's inside the box.
Given two arrays Arr1 and Arr2 of size N1 and N2. Your task is to find the sum of all elements that are common to both arrays. If there are no common elements the output would be 0.
Note: The arrays may contain duplicate elements. However, you need to sum only unique elements that are common to both arrays.The first line of input contains N1 and N2 separated by a space. The second line contains N1 space separated elements of Arr1. The third line contains N2 space separated elements of Arr2.
Constraints:
1 <= N1, N2 <= 10<sup>6</sup>
1 <= Arr1[i], Arr2[i] <= 1000000000Print the sum of common elements.Sample Input:
5 6
1 2 3 4 5
2 3 4 5 6 7
Sample Output:
14
Explanation:- Common elements = 2, 3, 4 , 5
sum= 2 + 3 + 4 + 5 = 14
Sample Input:-
3 3
1 2 3
4 5 6
Sample Output:-
0, 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();
int n1 = sc.nextInt();
int n2 = sc.nextInt();
HashSet<Integer> hs = new HashSet<>();
long sum = 0l;
for(int i=0;i<n1;i++){
int curr = sc.nextInt();
hs.add(curr);
}
for(int i=0;i<n2;i++){
int sl = sc.nextInt();
if(hs.contains(sl)){
sum+=sl;
hs.remove(sl);
}
}
System.out.println(sum);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Morgan was opening some boxes when she found one bearing a message “For Morgan, Dad loves you 3000". She wants to open the box but she needs to solve the question to open it. She asks Peter to do that. Help Peter and Morgan solve the question so that they can see what's inside the box.
Given two arrays Arr1 and Arr2 of size N1 and N2. Your task is to find the sum of all elements that are common to both arrays. If there are no common elements the output would be 0.
Note: The arrays may contain duplicate elements. However, you need to sum only unique elements that are common to both arrays.The first line of input contains N1 and N2 separated by a space. The second line contains N1 space separated elements of Arr1. The third line contains N2 space separated elements of Arr2.
Constraints:
1 <= N1, N2 <= 10<sup>6</sup>
1 <= Arr1[i], Arr2[i] <= 1000000000Print the sum of common elements.Sample Input:
5 6
1 2 3 4 5
2 3 4 5 6 7
Sample Output:
14
Explanation:- Common elements = 2, 3, 4 , 5
sum= 2 + 3 + 4 + 5 = 14
Sample Input:-
3 3
1 2 3
4 5 6
Sample Output:-
0, I have written this Solution Code: N1,N2=(map(int,input().split()))
arr1=set(map(int,input().split()))
arr2=set(map(int,input().split()))
arr1=list(arr1)
arr2=list(arr2)
arr1.sort()
arr2.sort()
i=0
j=0
sum1=0
while i<len(arr1) and j<len(arr2):
if arr1[i]==arr2[j]:
sum1+=arr1[i]
i+=1
j+=1
elif arr1[i]>arr2[j]:
j+=1
else:
i+=1
print(sum1), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Morgan was opening some boxes when she found one bearing a message “For Morgan, Dad loves you 3000". She wants to open the box but she needs to solve the question to open it. She asks Peter to do that. Help Peter and Morgan solve the question so that they can see what's inside the box.
Given two arrays Arr1 and Arr2 of size N1 and N2. Your task is to find the sum of all elements that are common to both arrays. If there are no common elements the output would be 0.
Note: The arrays may contain duplicate elements. However, you need to sum only unique elements that are common to both arrays.The first line of input contains N1 and N2 separated by a space. The second line contains N1 space separated elements of Arr1. The third line contains N2 space separated elements of Arr2.
Constraints:
1 <= N1, N2 <= 10<sup>6</sup>
1 <= Arr1[i], Arr2[i] <= 1000000000Print the sum of common elements.Sample Input:
5 6
1 2 3 4 5
2 3 4 5 6 7
Sample Output:
14
Explanation:- Common elements = 2, 3, 4 , 5
sum= 2 + 3 + 4 + 5 = 14
Sample Input:-
3 3
1 2 3
4 5 6
Sample Output:-
0, I have written this Solution Code: #include<bits/stdc++.h>
using namespace std;
#define pu push_back
#define fi first
#define se second
#define mp make_pair
#define int long long
#define pii pair<int,int>
#define mm (s+e)/2
#define all(x) x.begin(), x.end()
#define For(i, st, en) for(int i=st; i<en; i++)
#define tr(x) for(auto it=x.begin(); it!=x.end(); it++)
#define fast std::ios::sync_with_stdio(false);cin.tie(NULL);
#define sz 200000
#define qw1 freopen("input1.txt", "r", stdin); freopen("output1.txt", "w", stdout);
#define qw2 freopen("input2.txt", "r", stdin); freopen("output2.txt", "w", stdout);
#define qw3 freopen("input3.txt", "r", stdin); freopen("output3.txt", "w", stdout);
#define qw4 freopen("input4.txt", "r", stdin); freopen("output4.txt", "w", stdout);
#define qw5 freopen("input5.txt", "r", stdin); freopen("output5.txt", "w", stdout);
#define qw6 freopen("input6.txt", "r", stdin); freopen("output6.txt", "w", stdout);
#define qw freopen("input.txt", "r", stdin); freopen("output.txt", "w", stdout);
signed main()
{
int n,m;
cin>>n>>m;
set<int> ss,yy;
for(int i=0;i<n;i++)
{
int a;
cin>>a;
ss.insert(a);
}
for(int i=0;i<m;i++)
{
int a;
cin>>a;
if(ss.find(a)!=ss.end()) yy.insert(a);
}
int sum=0;
for(auto it:yy)
{
sum+=it;
}
cout<<sum<<endl;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Write a program to print Five stars ('*') <b><i>vertically</i></b> and 5 <b><i>horizontally</i></b>
There will be two functions:
<ul>
<li>verticalFive(): Print stars in vertical order</li>
<li>horizontalFive(): Print stars in horizontal order</l>
</ul><b>User Task:</b>
Your task is to complete the functions <b>verticalFive()</b> and <b>horizontalFive()</b>.
Print 5 vertical stars in <b> verticalFive</b> and 5 horizontal stars(separated by whitespace) in <b>horizontalFive</b> function.
<b>Note</b>: You don't need to print the extra blank line it will be printed by the driver codeNo Sample Input:
Sample Output:
*
*
*
*
*
* * * * *, I have written this Solution Code: static void verticalFive(){
System.out.println("*");
System.out.println("*");
System.out.println("*");
System.out.println("*");
System.out.println("*");
}
static void horizontalFive(){
System.out.print("* * * * *");
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Write a program to print Five stars ('*') <b><i>vertically</i></b> and 5 <b><i>horizontally</i></b>
There will be two functions:
<ul>
<li>verticalFive(): Print stars in vertical order</li>
<li>horizontalFive(): Print stars in horizontal order</l>
</ul><b>User Task:</b>
Your task is to complete the functions <b>verticalFive()</b> and <b>horizontalFive()</b>.
Print 5 vertical stars in <b> verticalFive</b> and 5 horizontal stars(separated by whitespace) in <b>horizontalFive</b> function.
<b>Note</b>: You don't need to print the extra blank line it will be printed by the driver codeNo Sample Input:
Sample Output:
*
*
*
*
*
* * * * *, I have written this Solution Code: def vertical5():
for i in range(0,5):
print("*",end="\n")
#print()
def horizontal5():
for i in range(0,5):
print("*",end=" ")
vertical5()
print(end="\n")
horizontal5(), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: All countries by the year they first participated in the Olympics have to be ordered.
Output the National Olympics Committee (NOC) name along with the desired year.
Sort records by the year and the NOC in ascending order.DataFrame/SQL Table with the following schema -
<schema>[{'name': 'olympics_athletes_events', 'columns': [{'name': 'id', 'type': 'int64'}, {'name': 'name', 'type': 'object'}, {'name': 'sex', 'type': 'object'}, {'name': 'age', 'type': 'float64'}, {'name': 'height', 'type': 'float64'}, {'name': 'weight', 'type': 'datetime64[ns]'}, {'name': 'team', 'type': 'object'}, {'name': 'noc', 'type': 'object'}, {'name': 'games', 'type': 'object'}, {'name': 'year', 'type': 'int64'}, {'name': 'season', 'type': 'object'}, {'name': 'city', 'type': 'object'}, {'name': 'sport', 'type': 'object'}, {'name': 'event', 'type': 'object'}, {'name': 'medal', 'type': 'object'}]}]</schema>Each row in a new line and each value of a row separated by a |, i.e.,
0|1|2
1|2|3
2|3|4-, I have written this Solution Code: df = olympics_athletes_events.groupby(by=['noc'],as_index= False).min()
df2 = df.sort_values(by=['year','noc'])
for i, r in df2.iterrows():
print(f"{r['noc']}|{r['year']}"), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: All countries by the year they first participated in the Olympics have to be ordered.
Output the National Olympics Committee (NOC) name along with the desired year.
Sort records by the year and the NOC in ascending order.DataFrame/SQL Table with the following schema -
<schema>[{'name': 'olympics_athletes_events', 'columns': [{'name': 'id', 'type': 'int64'}, {'name': 'name', 'type': 'object'}, {'name': 'sex', 'type': 'object'}, {'name': 'age', 'type': 'float64'}, {'name': 'height', 'type': 'float64'}, {'name': 'weight', 'type': 'datetime64[ns]'}, {'name': 'team', 'type': 'object'}, {'name': 'noc', 'type': 'object'}, {'name': 'games', 'type': 'object'}, {'name': 'year', 'type': 'int64'}, {'name': 'season', 'type': 'object'}, {'name': 'city', 'type': 'object'}, {'name': 'sport', 'type': 'object'}, {'name': 'event', 'type': 'object'}, {'name': 'medal', 'type': 'object'}]}]</schema>Each row in a new line and each value of a row separated by a |, i.e.,
0|1|2
1|2|3
2|3|4-, I have written this Solution Code: SELECT
noc,
min(year) AS first_time_year
FROM olympics_athletes_events
GROUP BY
noc
ORDER BY
first_time_year,
noc, In this Programming Language: SQL, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a binary string S of length N. Suppose there is an integer K (1 <= K <= N). You can perform the following operation on the string any number of times (possible zero):
Select a substring in S whose size is <b>at least</b> K i. e, R − L + 1 >= K must be followed and flip this substring. Flipping here means if S<sub>i</sub> is equal to '1', make it equal to '0' and if S<sub>i</sub> is equal to '0', make it equal to '1'.
Find the maximum value of K such that you can make string S a null string. A null string is defined as a binary string in which all the characters are '0'.The first line of the input contains a single integer N.
The second line of the input contains a binary string of lenth N.
Constraints:
1 <= N <= 10<sup>5</sup>Print the maximum value of K such that you can make string S a null string.Sample Input:
3
010
Sample Output:
2
Explaination:
We can make S a null string by the following operations:
<ul>
<li>First, we pick the segment S[1,3] with length 3. S is now 101.</li>
<li>Next, we pick the segment S[1,2] with length 2. S is now 011.</li>
<li>And finally we pick the segment S[2,3] with length 2. S is now 000.</li>
</ul>, 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().split(" ");
int m=Integer.parseInt(str[0]);
String input=br.readLine();
int n=input.length();
int ans=n;
for(int i=1; i<n; i++)
{
if(input.charAt(i-1) != input.charAt(i))
{
ans=Math.min(Math.max(i,n-i), ans);
}
}
System.out.println(ans);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a binary string S of length N. Suppose there is an integer K (1 <= K <= N). You can perform the following operation on the string any number of times (possible zero):
Select a substring in S whose size is <b>at least</b> K i. e, R − L + 1 >= K must be followed and flip this substring. Flipping here means if S<sub>i</sub> is equal to '1', make it equal to '0' and if S<sub>i</sub> is equal to '0', make it equal to '1'.
Find the maximum value of K such that you can make string S a null string. A null string is defined as a binary string in which all the characters are '0'.The first line of the input contains a single integer N.
The second line of the input contains a binary string of lenth N.
Constraints:
1 <= N <= 10<sup>5</sup>Print the maximum value of K such that you can make string S a null string.Sample Input:
3
010
Sample Output:
2
Explaination:
We can make S a null string by the following operations:
<ul>
<li>First, we pick the segment S[1,3] with length 3. S is now 101.</li>
<li>Next, we pick the segment S[1,2] with length 2. S is now 011.</li>
<li>And finally we pick the segment S[2,3] with length 2. S is now 000.</li>
</ul>, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define int long long
bool solve(int k, string &s){
bool a = true, b = true;
for(int i = 0; i < s.size(); i++){
if (i >= k || i + k < s.size()) continue;
a &= (s[i] == '0');
b &= (s[i] == '1');
}
return a||b;
}
signed main(){
int n;
cin >> n;
string s;
cin >> s;
int l = 1, r = n, ans = 1;
while(r >= l){
int k = (l + r)/2;
if(solve(k, s)){
ans = k;
l = k + 1;
}
else r = k - 1;
}
cout << ans;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You have a list arr of all integers in the range [1, n] sorted in a strictly increasing order. Apply the following algorithm on arr:
:> Starting from left to right, remove the first number and every other number afterward until you reach the end of the list.
:> Repeat the previous step again, but this time from right to left, remove the rightmost number and every other number from the remaining numbers.
:> Keep repeating the steps again, alternating left to right and right to left, until a single number remains.
Given the integer n, return the last number that remains in arr.<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>lastRemaining()</b> that takes Integer "n" as parameter.
<b>Constraints:</b>
1 ≤ n ≤ 10<sup>9</sup>Return the last number that remains in arr.Sample 1:
Input:
9
Output:
6
Explanation:
arr = [1, 2, 3, 4, 5, 6, 7, 8, 9]
arr = [2, 4, 6, 8]
arr = [2, 6]
arr = [6]
Sample 2:
Input:
1
Output:
1, I have written this Solution Code:
class Solution {
public int lastRemaining(int n) {
if(n == 1) return 1;
return 2*(1 + n/2 - lastRemaining(n/2));
}
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an sorted array <b>Arr[]</b> of size <b>N</b>, containing both <b>negative</b> and <b>positive</b> integers, you need to print the squared sorted output.
<b>Note</b> Try using two pointer approachThe first line of input contains T, denoting the number of test cases. Each testcase contains 2 lines. The first line contains the N size of the array. The second line contains elements of an array separated by space.
Constraints:
1 ≤ T ≤ 100
1 ≤ N ≤ 10000
-10000 ≤ A[i] ≤ 10000
The Sum of N over all test cases does not exceed 10^6For each test case you need to print the sorted squared output in new lineInput:
1
5
-7 -2 3 4 6
Output:
4 9 16 36 49, I have written this Solution Code: import java.util.*;
import java.io.*;
class Main
{
public static void main(String[] args)throws IOException {
BufferedReader read = new BufferedReader(new InputStreamReader(System.in));
int t = Integer.parseInt(read.readLine());
while (t-- > 0) {
int n = Integer.parseInt(read.readLine());
int[] arr = new int[n];
String str[] = read.readLine().trim().split(" ");
for(int i = 0; i < n; i++)
arr[i] = Integer.parseInt(str[i]);
arr = sortedSquares(arr);
for(int i = 0; i < n; i++)
System.out.print(arr[i] + " ");
System.out.println();
}
}
public static int[] sortedSquares(int[] A) {
int[] nums = new int[A.length];
int k=A.length-1;
int i=0, j=A.length-1;
while(i<=j){
if(Math.abs(A[i]) <= Math.abs(A[j])){
nums[k--] = A[j]*A[j];
j--;
}
else{
nums[k--] = A[i]*A[i];
i++;
}
}
return nums;
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an sorted array <b>Arr[]</b> of size <b>N</b>, containing both <b>negative</b> and <b>positive</b> integers, you need to print the squared sorted output.
<b>Note</b> Try using two pointer approachThe first line of input contains T, denoting the number of test cases. Each testcase contains 2 lines. The first line contains the N size of the array. The second line contains elements of an array separated by space.
Constraints:
1 ≤ T ≤ 100
1 ≤ N ≤ 10000
-10000 ≤ A[i] ≤ 10000
The Sum of N over all test cases does not exceed 10^6For each test case you need to print the sorted squared output in new lineInput:
1
5
-7 -2 3 4 6
Output:
4 9 16 36 49, I have written this Solution Code: t = int(input())
for i in range(t):
n = int(input())
for i in sorted(map(lambda j:int(j)**2,input().split())):
print(i,end=' ')
print(), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer Q representing the number of queries and an integer K. There are two types of queries:
(i) 1 x : Add the number x to the stream
(ii) 2 : Print the sum of last K numbers of the stream. If there are less than K numbers in the stream, print the sum of current stream.
Process all the queries.First line contains two integers Q and K.
Next Q lines contains the queries.
Constraints
1 <= Q <= 10^5
1 <= x <= 10^5
1 <= K <= Q
There is atleast one query of 2nd type. For each query of type 2, print the sum of last K numbers of the stream on a new line.Sample Input 1:
5 2
1 4
2
1 1
1 3
2
Output
4
4
Explanation:
Initial Stream = {}
Add 4. Stream = {4}
Sum of last two elements = 4
Add 1. Stream = {4, 1}
Add 3. Stream = {4, 1, 3}
Sum of last two elements = 4
Sample Input 2:
3 1
1 1
2
2
Output
1
1
Explanation
Initial Stream = {}
Add 1. Stream = {1}
Sum of last element = 1
Sum of last element = 1, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main{
public static void main (String[] args)throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int n,k;
String line = br.readLine();
String[] strs = line.trim().split("\\s+");
n=Integer.parseInt(strs[0]);
k=Integer.parseInt(strs[1]);
Long sum=0L;
Queue<Integer>x=new LinkedList<Integer>();
while(n>0){
n--;
line = br.readLine();
strs = line.trim().split("\\s+");
int y=Integer.parseInt(strs[0]);
if(y==1){
y=Integer.parseInt(strs[1]);
x.add(y);
sum += y;
if (x.size() > k) {
sum -= x.peek();
x.remove();
}
}
else{
System.out.print(sum+"\n");
}
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer Q representing the number of queries and an integer K. There are two types of queries:
(i) 1 x : Add the number x to the stream
(ii) 2 : Print the sum of last K numbers of the stream. If there are less than K numbers in the stream, print the sum of current stream.
Process all the queries.First line contains two integers Q and K.
Next Q lines contains the queries.
Constraints
1 <= Q <= 10^5
1 <= x <= 10^5
1 <= K <= Q
There is atleast one query of 2nd type. For each query of type 2, print the sum of last K numbers of the stream on a new line.Sample Input 1:
5 2
1 4
2
1 1
1 3
2
Output
4
4
Explanation:
Initial Stream = {}
Add 4. Stream = {4}
Sum of last two elements = 4
Add 1. Stream = {4, 1}
Add 3. Stream = {4, 1, 3}
Sum of last two elements = 4
Sample Input 2:
3 1
1 1
2
2
Output
1
1
Explanation
Initial Stream = {}
Add 1. Stream = {1}
Sum of last element = 1
Sum of last element = 1, I have written this Solution Code: /**
* author: tourist1256
* created: 2022-06-14 14:26:47
**/
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace std;
using namespace __gnu_pbds;
template <class T>
using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
template <class key, class value, class cmp = std::less<key>>
using ordered_map = tree<key, value, cmp, rb_tree_tag, tree_order_statistics_node_update>;
// find_by_order(k) returns iterator to kth element starting from 0;
// order_of_key(k) returns count of elements strictly smaller than k;
#ifdef LOCAL
#define debug(...) cerr << "[" << #__VA_ARGS__ << "]:", debug_out(__VA_ARGS__)
#else
#define debug(...) 2351
#endif
#define int long long
mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
inline int64_t random_long(long long l = LLONG_MIN, long long r = LLONG_MAX) {
uniform_int_distribution<int64_t> generator(l, r);
return generator(rng);
}
int32_t main() {
int Q, K;
cin >> Q >> K;
deque<int> st;
int sum = 0;
while (Q--) {
int x;
cin >> x;
if (x == 1) {
int y;
cin >> y;
if (st.size() == K) {
sum -= st.back();
st.pop_back();
st.push_front(y);
sum += y;
} else {
st.push_front(y);
sum += y;
}
} else {
cout << sum << "\n";
}
}
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A[], of size N containing positive integers. You need to print the elements of an array in increasing order.The first line of the input denotes the number of test cases 'T'. The first line of the test case is the size of the array and the second line consists of array elements.
For Python Users just complete the given function.
<b>Constraints:</b>
1 ≤ T ≤ 100
1 ≤ N ≤ 1000
1 ≤ A[i] ≤ 1000For each testcase print the sorted array in a new line.Input:
2
5
4 1 3 9 7
10
10 9 8 7 6 5 4 3 2 1
Output:
1 3 4 7 9
1 2 3 4 5 6 7 8 9 10
<b>Explanation:</b>
Testcase 1: 1 3 4 7 9 are in sorted form.
Testcase 2: For the given input, 1 2 3 4 5 6 7 8 9 10 are in sorted form., I have written this Solution Code: #include "bits/stdc++.h"
#pragma GCC optimize "03"
using namespace std;
#define int long long int
#define ld long double
#define pi pair<int, int>
#define pb push_back
#define fi first
#define se second
#define IOS ios::sync_with_stdio(false); cin.tie(0); cout.tie(0)
#ifndef LOCAL
#define endl '\n'
#endif
const int N = 2e5 + 5;
const int mod = 1e9 + 7;
const int inf = 1e9 + 9;
int a[N];
signed main() {
IOS;
int t; cin >> t;
while(t--){
int n; cin >> n;
for(int i = 1; i <= n; i++)
cin >> a[i];
sort(a + 1, a + n + 1);
for(int i = 1; i <= n; i++)
cout << a[i] << " ";
cout << endl;
}
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A[], of size N containing positive integers. You need to print the elements of an array in increasing order.The first line of the input denotes the number of test cases 'T'. The first line of the test case is the size of the array and the second line consists of array elements.
For Python Users just complete the given function.
<b>Constraints:</b>
1 ≤ T ≤ 100
1 ≤ N ≤ 1000
1 ≤ A[i] ≤ 1000For each testcase print the sorted array in a new line.Input:
2
5
4 1 3 9 7
10
10 9 8 7 6 5 4 3 2 1
Output:
1 3 4 7 9
1 2 3 4 5 6 7 8 9 10
<b>Explanation:</b>
Testcase 1: 1 3 4 7 9 are in sorted form.
Testcase 2: For the given input, 1 2 3 4 5 6 7 8 9 10 are in sorted form., I have written this Solution Code: // arr is unsorted array
// n is the number of elements in the array
function bubbleSort(arr, n) {
// write code here
// do not console.log the answer
// return sorted array
return arr.sort((a, b) => a - b)
}, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A[], of size N containing positive integers. You need to print the elements of an array in increasing order.The first line of the input denotes the number of test cases 'T'. The first line of the test case is the size of the array and the second line consists of array elements.
For Python Users just complete the given function.
<b>Constraints:</b>
1 ≤ T ≤ 100
1 ≤ N ≤ 1000
1 ≤ A[i] ≤ 1000For each testcase print the sorted array in a new line.Input:
2
5
4 1 3 9 7
10
10 9 8 7 6 5 4 3 2 1
Output:
1 3 4 7 9
1 2 3 4 5 6 7 8 9 10
<b>Explanation:</b>
Testcase 1: 1 3 4 7 9 are in sorted form.
Testcase 2: For the given input, 1 2 3 4 5 6 7 8 9 10 are in sorted form., I have written this Solution Code:
import java.util.*;
import java.lang.*;
import java.io.*;
class Main
{
public static void main (String[] args) throws java.lang.Exception
{
Scanner sc = new Scanner(System.in);
int t = sc.nextInt();
while(t-->0){
int n = sc.nextInt();
int a[] = new int[n];
for(int i=0;i<n;i++){
a[i]=sc.nextInt();
}
int temp;
for(int i=1;i<n;i++){
if(a[i]<a[i-1]){
for(int j=i;j>0;j--){
if(a[j]<a[j-1]){
temp=a[j];
a[j]=a[j-1];
a[j-1]=temp;
}
else{
break;
}
}
}
}
for(int i=0;i<n;i++){
System.out.print(a[i]+" ");
}
System.out.println();
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A[], of size N containing positive integers. You need to print the elements of an array in increasing order.The first line of the input denotes the number of test cases 'T'. The first line of the test case is the size of the array and the second line consists of array elements.
For Python Users just complete the given function.
<b>Constraints:</b>
1 ≤ T ≤ 100
1 ≤ N ≤ 1000
1 ≤ A[i] ≤ 1000For each testcase print the sorted array in a new line.Input:
2
5
4 1 3 9 7
10
10 9 8 7 6 5 4 3 2 1
Output:
1 3 4 7 9
1 2 3 4 5 6 7 8 9 10
<b>Explanation:</b>
Testcase 1: 1 3 4 7 9 are in sorted form.
Testcase 2: For the given input, 1 2 3 4 5 6 7 8 9 10 are in sorted form., I have written this Solution Code: def bubbleSort(arr):
arr.sort()
return arr
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a circular linked list consisting of N nodes and an integer K, your task is to add the integer K at the end of the list.
<b>Note:
Sample Input and Output just show how a linked list will look depending on the questions. Do not copy-paste as it is in custom input</b><b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>Insertion()</b> that takes head node of circular linked list and the integer K as parameter.
Constraints:
1 <=N <= 1000
1 <= Node.data, K<= 1000Return the head node of the modified circular linked list.Sample Input 1:-
3
1- >2- >3
4
Sample Output 1:-
1- >2- >3- >4
Sample Input 2:-
3
1- >3- >2
1
Sample Output 2:-
1- >3- >2- >1, I have written this Solution Code: public static Node Insertion(Node head, int K){
Node node=head;
while ( node.next != head)
{node = node.next; }
Node temp = new Node(K);
node.next=temp;
temp.next=head;
return head;}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a boolean Matrix of size N*M, A cell of the matrix is called "Good" if it is completely surrounded by the cells containing '1' i.e. each adjacent cell (which shares a common edge) contains '1'. Your task is to find the number of such cells.
See the below example for a better understandingFirst line of input contains two space- separated integers N and M. Next N lines of input contain M space- separated integers depicting the values of the matrix.
Constraints:-
3 <= N, M <= 500
0 <= Matrix[][] <= 1Print the number of good cells.Sample Input:-
3 3
1 1 0
1 1 1
1 1 1
Sample Output:-
1
Explanation:-
Only cell at position 1, 1 is good
Sample Input:-
5 4
1 0 1 0
0 1 0 1
1 0 1 0
0 1 0 1
1 0 1 0
Sample Output:-
3
Explanation:-
(1, 2), (2, 1) and (3, 2) are good cells, I have written this Solution Code: // mat is the matrix/ 2d array
// n,m are dimensions
function goodCell(mat, n, m) {
// write code here
// do not console.log
// return the answer as a number
let cnt = 0;
for (let i = 1; i < n - 1; i++) {
for (let j = 1; j < m - 1; j++) {
if (mat[i - 1][j] == 1 && mat[i + 1][j] == 1 && mat[i][j - 1] == 1 && mat[i][j + 1] == 1) {
cnt++;
}
}
}
return cnt
}
, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a boolean Matrix of size N*M, A cell of the matrix is called "Good" if it is completely surrounded by the cells containing '1' i.e. each adjacent cell (which shares a common edge) contains '1'. Your task is to find the number of such cells.
See the below example for a better understandingFirst line of input contains two space- separated integers N and M. Next N lines of input contain M space- separated integers depicting the values of the matrix.
Constraints:-
3 <= N, M <= 500
0 <= Matrix[][] <= 1Print the number of good cells.Sample Input:-
3 3
1 1 0
1 1 1
1 1 1
Sample Output:-
1
Explanation:-
Only cell at position 1, 1 is good
Sample Input:-
5 4
1 0 1 0
0 1 0 1
1 0 1 0
0 1 0 1
1 0 1 0
Sample Output:-
3
Explanation:-
(1, 2), (2, 1) and (3, 2) are good cells, I have written this Solution Code: N, M= list(map(int,input().split()))
mat =[]
for i in range(N):
List =list(map(int,input().split()))[:M]
mat.append(List)
count =0
for i in range(1,N-1):
for j in range(1,M-1):
if (mat[i][j-1] == 1 and mat[i][j+1] == 1 and mat[i-1][j] == 1 and mat[i+1][j] == 1):
count +=1
print(count), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a boolean Matrix of size N*M, A cell of the matrix is called "Good" if it is completely surrounded by the cells containing '1' i.e. each adjacent cell (which shares a common edge) contains '1'. Your task is to find the number of such cells.
See the below example for a better understandingFirst line of input contains two space- separated integers N and M. Next N lines of input contain M space- separated integers depicting the values of the matrix.
Constraints:-
3 <= N, M <= 500
0 <= Matrix[][] <= 1Print the number of good cells.Sample Input:-
3 3
1 1 0
1 1 1
1 1 1
Sample Output:-
1
Explanation:-
Only cell at position 1, 1 is good
Sample Input:-
5 4
1 0 1 0
0 1 0 1
1 0 1 0
0 1 0 1
1 0 1 0
Sample Output:-
3
Explanation:-
(1, 2), (2, 1) and (3, 2) are good cells, I have written this Solution Code:
#include <bits/stdc++.h>
using namespace std;
#define MEM(a, b) memset(a, (b), sizeof(a))
#define FOREACH(it, l) for (auto it = l.begin(); it != l.end(); it++)
#define IN(A, B, C) assert( B <= A && A <= C)
#define MP make_pair
#define FOR(i,a) for(int i=0;i<a;i++)
#define FOR1(i,j,a) for(int i=j;i<a;i++)
#define EB emplace_back
#define INF (int)1e9
#define EPS 1e-9
#define PI 3.1415926535897932384626433832795
#define max1 1000001
#define MOD 1000000007
#define read(type) readInt<type>()
#define out(x) cout<<x<<'\n'
#define out1(x) cout<<x<<" "
#define END cout<<'\n'
#define int long long
#define sz(v) ((int)(v).size())
#define all(v) (v).begin(), (v).end()
void fast(){
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
}
signed main(){
int n,m;
cin>>n>>m;
int a[n][m];
FOR(i,n){
FOR(j,m){
cin>>a[i][j];}}
int sum=0,sum1=0;;
FOR1(i,1,n-1){
FOR1(j,1,m-1){
if(a[i-1][j]==1 && a[i+1][j]==1 && a[i][j-1]==1 && a[i][j+1]==1){
sum++;
}
}
}
out1(sum);
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a boolean Matrix of size N*M, A cell of the matrix is called "Good" if it is completely surrounded by the cells containing '1' i.e. each adjacent cell (which shares a common edge) contains '1'. Your task is to find the number of such cells.
See the below example for a better understandingFirst line of input contains two space- separated integers N and M. Next N lines of input contain M space- separated integers depicting the values of the matrix.
Constraints:-
3 <= N, M <= 500
0 <= Matrix[][] <= 1Print the number of good cells.Sample Input:-
3 3
1 1 0
1 1 1
1 1 1
Sample Output:-
1
Explanation:-
Only cell at position 1, 1 is good
Sample Input:-
5 4
1 0 1 0
0 1 0 1
1 0 1 0
0 1 0 1
1 0 1 0
Sample Output:-
3
Explanation:-
(1, 2), (2, 1) and (3, 2) are good cells, 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 m= sc.nextInt();
int a[][]= new int[n][m];
for(int i=0;i<n;i++){
for(int j=0;j<m;j++){
a[i][j]=sc.nextInt();}}
int cnt=0;
for(int i=1;i<n-1;i++){
for(int j=1;j<m-1;j++){
if(a[i-1][j]==1 && a[i+1][j]==1 && a[i][j-1]==1 && a[i][j+1]==1){
cnt++;
}
}
}
System.out.print(cnt);
}
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of N integers, an integer X, and Q queries. For each query, you will be given an integer K. Your task is to find the Kth occurrence of the number X in the array. If the Kth occurrence does not exist print -1.The first line of input contains three integers N, X, and Q. The second line of input contains N space- separated integers. The Next Q lines of input contain a single integer each representing K.
Constraints:-
1 <= Q, K, N <= 100000
1 <= X, Arr[] <= 100000For each query in a new line print the index of the Kth occurrence. If the Kth occurrence does not exist print -1.Sample Input:-
10 3 4
1 2 3 4 2 3 4 5 2 3
1
2
3
4
Sample Output:_
3
6
10
-1, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws IOException {
BufferedReader inp = new BufferedReader(new InputStreamReader(System.in));
String[] NXQ = inp.readLine().split("\\s+");
int N = Integer.parseInt(NXQ[0]);
int X = Integer.parseInt(NXQ[1]);
int Q = Integer.parseInt(NXQ[2]);
String[] StrArr = inp.readLine().split("\\s+");
int[] arr = new int[N];
ArrayList<Integer> posX = new ArrayList<>();
int j = 0;
for (int i = 0; i < N; i++) {
arr[i] = Integer.parseInt(StrArr[i]);
if (arr[i] == X) {
posX.add(i + 1);
}
}
for (int i = 0; i < Q; i++) {
int query = Integer.parseInt(inp.readLine());
if (query > posX.size()) {
System.out.println("-1");
}
else {
System.out.println(posX.get(query - 1));
}
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of N integers, an integer X, and Q queries. For each query, you will be given an integer K. Your task is to find the Kth occurrence of the number X in the array. If the Kth occurrence does not exist print -1.The first line of input contains three integers N, X, and Q. The second line of input contains N space- separated integers. The Next Q lines of input contain a single integer each representing K.
Constraints:-
1 <= Q, K, N <= 100000
1 <= X, Arr[] <= 100000For each query in a new line print the index of the Kth occurrence. If the Kth occurrence does not exist print -1.Sample Input:-
10 3 4
1 2 3 4 2 3 4 5 2 3
1
2
3
4
Sample Output:_
3
6
10
-1, I have written this Solution Code: N, X , Q = map(int, input().split())
L = list(map(int, input().split()))
Queries = []
while Q > 0:
Queries.append(int(input()))
Q -= 1
indices = [i for i,x in enumerate(L) if x == X]
for query in Queries:
if len(indices)< query:
print(-1)
else:
print(indices[query-1]+1), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of N integers, an integer X, and Q queries. For each query, you will be given an integer K. Your task is to find the Kth occurrence of the number X in the array. If the Kth occurrence does not exist print -1.The first line of input contains three integers N, X, and Q. The second line of input contains N space- separated integers. The Next Q lines of input contain a single integer each representing K.
Constraints:-
1 <= Q, K, N <= 100000
1 <= X, Arr[] <= 100000For each query in a new line print the index of the Kth occurrence. If the Kth occurrence does not exist print -1.Sample Input:-
10 3 4
1 2 3 4 2 3 4 5 2 3
1
2
3
4
Sample Output:_
3
6
10
-1, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define int long long
signed main(){
int n,x,q;
cin>>n>>x>>q;
vector<int> a(n),v;
for(int i=0;i<n;i++){
cin>>a[i];
if(a[i]==x){
v.push_back(i+1);
}
}
while(q--){
cin>>x;
if(x<=v.size()){
cout<<v[x-1]<<'\n';
}
else{
cout<<-1<<'\n';
}
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given N flags, initially set to 0. Now you have to perform two operations on them:
1. Increase(F) by 1: flag F is increased by 1.
2. max_flag: all flags are set to a maximum value of any flag.
A non-empty array arr[] will be given of size M. This array represents consecutive operations:
a) If arr[K] = F, such that 1 <= F <= N then operation K is Increase(F).
b) If arr[K] = N+1 then operation K is max_flag.
The goal is to calculate the value of every flag after all operations.The input line contains T, denoting the number of test cases.
Each test case contains two lines.
The first line contains an integer N, the number of flags, and an integer M, the size of the array 'arr'.The second line contains elements of the array 'arr' separated by space.
<b>Constraints:</b>
1 <= T <= 100
1 <= N, M <= 10^5
1 <= arr[i] <= N+1
Sum of N and M for each test case is less than or equal to 10^5For each testcase you need to print the updated array after all operations in new line.Sample Input:
1
5 7
3 4 4 6 1 4 4
Sample Output:
3 2 2 4 2
<b>Explanation:</b>
Testcase 1:
the values of the flags after each consecutive operation will be:
(0, 0, 1, 0, 0)
(0, 0, 1, 1, 0)
(0, 0, 1, 2, 0)
(2, 2, 2, 2, 2)
(3, 2, 2, 2, 2)
(3, 2, 2, 3, 2)
(3, 2, 2, 4, 2), I have written this Solution Code: t=int(input())
while t>0:
t-=1
n,m=map(int,input().split())
a=map(int,input().split())
b=[0]*(n+1)
for i in a:
if i==n+1:
v=max(b)
for i in range(1,n+1):
b[i]=v
else:b[i]+=1
for i in range(1,n+1):
print(b[i],end=' ')
print(), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given N flags, initially set to 0. Now you have to perform two operations on them:
1. Increase(F) by 1: flag F is increased by 1.
2. max_flag: all flags are set to a maximum value of any flag.
A non-empty array arr[] will be given of size M. This array represents consecutive operations:
a) If arr[K] = F, such that 1 <= F <= N then operation K is Increase(F).
b) If arr[K] = N+1 then operation K is max_flag.
The goal is to calculate the value of every flag after all operations.The input line contains T, denoting the number of test cases.
Each test case contains two lines.
The first line contains an integer N, the number of flags, and an integer M, the size of the array 'arr'.The second line contains elements of the array 'arr' separated by space.
<b>Constraints:</b>
1 <= T <= 100
1 <= N, M <= 10^5
1 <= arr[i] <= N+1
Sum of N and M for each test case is less than or equal to 10^5For each testcase you need to print the updated array after all operations in new line.Sample Input:
1
5 7
3 4 4 6 1 4 4
Sample Output:
3 2 2 4 2
<b>Explanation:</b>
Testcase 1:
the values of the flags after each consecutive operation will be:
(0, 0, 1, 0, 0)
(0, 0, 1, 1, 0)
(0, 0, 1, 2, 0)
(2, 2, 2, 2, 2)
(3, 2, 2, 2, 2)
(3, 2, 2, 3, 2)
(3, 2, 2, 4, 2), I have written this Solution Code: #include "bits/stdc++.h"
#pragma GCC optimize "03"
using namespace std;
#define int long long int
#define ld long double
#define pi pair<int, int>
#define pb push_back
#define fi first
#define se second
#define IOS ios::sync_with_stdio(false); cin.tie(0); cout.tie(0)
#ifndef LOCAL
#define endl '\n'
#endif
const int N = 1e5 + 5;
const int mod = 1e9 + 7;
const int inf = 1e9 + 9;
int a[N];
signed main() {
IOS;
int t; cin >> t;
while(t--){
memset(a, 0, sizeof a);
int n, m;
cin >> n >> m;
int mx = 0, flag = 0;
for(int i = 1; i <= m; i++){
int p; cin >> p;
if(p == n+1){
flag = mx;
}
else{
a[p] = max(a[p], flag) + 1;
mx = max(mx, a[p]);
}
}
for(int i = 1; i <= n; i++){
a[i] = max(a[i], flag);
cout << a[i] << " ";
}
cout << endl;
}
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Bob is at the origin of a number line. He wants to reach a goal at coordinate X.
There is a wall at coordinate Y, which Bob cannot go beyond at first. However, after picking up a hammer at coordinate Z, he can destroy that wall and pass through.
Determine whether Bob can reach the goal. If he can, find the minimum total distance he needs to travel to do so.The input is given from Standard Input in the following format:
X Y Z
<b>Constraints</b>
−1000 ≤ X, Y, Z ≤ 1000
X, Y, and Z are distinct, and none of them is 0.
All values in the input are integers.If Bob can reach the goal, print the minimum total distance he needs to travel to do so. If he cannot, print -1 instead.<b>Sample Input 1</b>
10 -10 1
<b>Sample Output 1</b>
10
<b>Sample Input 2</b>
20 10 -10
<b>Sample Output 2</b>
40, I have written this Solution Code: #include<bits/stdc++.h>
#define L(i, j, k) for(int i = (j); i <= (k); ++i)
#define R(i, j, k) for(int i = (j); i >= (k); --i)
#define ll long long
#define sz(a) ((int) (a).size())
#define vi vector < int >
#define me(a, x) memset(a, x, sizeof(a))
#define ull unsigned long long
#define ld __float128
using namespace std;
const int N = 1e6 + 7;
int n, x, y, z;
int main() {
ios :: sync_with_stdio(false);
cin.tie(0); cout.tie(0);
cin >> x >> y >> z;
if(abs(y) + abs(x - y) == abs(x)) {
if(abs(y - z) + abs(y) == abs(z)) {
cout << -1 << '\n';
} else {
cout << abs(z) + abs(z - x) << '\n';
}
} else {
cout << abs(x) << '\n';
}
if(y > 0 && x < y) {
}
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Create a class Teacher with name, age, and salary attributes, where salary must be a private attribute that cannot be accessed outside the class.
Print the details of the teacher including salary.The first line contains the name, the second line contains the age and the third line contains the salary of the teacher.
<b>Constraints:</b>
20<=age<=80Prints the details of the teacher in the following format:
Name:
Age:
Salary:Sample Input:
raj
22
20000
Sample Output:
Name: raj
Age: 22
Salary: 20000, I have written this Solution Code: class Teacher():
def __init__(self, name, age, salary):
self.name = name
self.age = age
# private variable
self.__salary = salary
def show_details(self):
print("Name:", self.name)
print("Age:", self.age)
#access private attribute inside the class
print("Salary:", self.__salary)
name=input()
age=int(input())
salary=int(input())
teacher = Teacher(name,age,salary)
teacher.show_details(), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Create a class Teacher with name, age, and salary attributes, where salary must be a private attribute that cannot be accessed outside the class.
Print the details of the teacher including salary.The first line contains the name, the second line contains the age and the third line contains the salary of the teacher.
<b>Constraints:</b>
20<=age<=80Prints the details of the teacher in the following format:
Name:
Age:
Salary:Sample Input:
raj
22
20000
Sample Output:
Name: raj
Age: 22
Salary: 20000, I have written this Solution Code: import java.io.*;
import java.util.*;
class Teacher{
String name;
int age;
private int salary;
Teacher(String name , int age , int salary){
this.name = name;
this.age = age;
this.salary = salary;
}
public void details(){
System.out.println("Name: "+this.name);
System.out.println("Age: "+ this.age);
System.out.print("Salary: "+this.salary);
}
}
class Main {
public static void main (String[] args) {
Scanner sc = new Scanner(System.in);
String name = sc.nextLine();
int age = sc.nextInt();
int salary = sc.nextInt();
Teacher obj = new Teacher (name , age , salary);
obj.details();
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer N, find the closest prime number to N. If there are multiple print the smaller one.The input contains a single integer N.
Constraints:
1 <= N <= 1000000000Print the closest prime to N.Sample Input 1
12
Sample Output 1
11
Explanation: Closest prime to 12 is 11 and 13 smaller of which is 11.
Sample Input 2
17
Sample Output 2
17
Explanation: Closest prime to 17 is 17., 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());
if(n==1){
System.out.println(2);
}else{
int after = afterPrime(n);
int before = beforePrime(n);
if(before>after){
System.out.println(n+after);
}
else{System.out.println(n-before);}
}
}
public static boolean isPrime(int n)
{
int count=0;
for(int i=2;i*i<n;i++)
{
if(n%i==0)
count++;
}
if(count==0)
return true;
else
return false;
}
public static int beforePrime(int n)
{
int c=0;
while(true)
{
if(isPrime(n))
return c;
else
{
n=n-1;
c++;
}
}
}
public static int afterPrime(int n)
{
int c=0;
while(true)
{
if(isPrime(n))
return c;
else
{
n=n+1;
c++;
}
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer N, find the closest prime number to N. If there are multiple print the smaller one.The input contains a single integer N.
Constraints:
1 <= N <= 1000000000Print the closest prime to N.Sample Input 1
12
Sample Output 1
11
Explanation: Closest prime to 12 is 11 and 13 smaller of which is 11.
Sample Input 2
17
Sample Output 2
17
Explanation: Closest prime to 17 is 17., I have written this Solution Code: from math import sqrt
def NearPrime(N):
if N >1:
for i in range(2,int(sqrt(N))+1):
if N%i ==0:
return False
break
else: return True
else: return False
N=int(input())
i =0
while NearPrime(N-i)==False and NearPrime(N+i)==False:
i+=1
if NearPrime(N-i) and NearPrime(N+i):print(N-i)
elif NearPrime(N-i):print(N-i)
elif NearPrime(N+i): print(N+i), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer N, find the closest prime number to N. If there are multiple print the smaller one.The input contains a single integer N.
Constraints:
1 <= N <= 1000000000Print the closest prime to N.Sample Input 1
12
Sample Output 1
11
Explanation: Closest prime to 12 is 11 and 13 smaller of which is 11.
Sample Input 2
17
Sample Output 2
17
Explanation: Closest prime to 17 is 17., 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>
/////////////
bool isPrime(int n){
if(n<=1)
return false;
for(int i=2;i*i<=n;++i)
if(n%i==0)
return false;
return true;
}
signed main()
{
fastio;
#ifdef ANIKET_GOYAL
freopen("inputf.in","r",stdin);
freopen("outputf.in","w",stdout);
#endif
int n;
cin>>n;
if(n==1)
cout<<"2";
else{
int v1=n,v2=n;
while(isPrime(v1)==false)
--v1;
while(isPrime(v2)==false)
++v2;
if(v2-n==n-v1)
cout<<v1;
else{
if(v2-n<n-v1)
cout<<v2;
else
cout<<v1;
}
}
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a String of length S, reverse the whole string without reversing the individual words in it. Words are separated by dots.First line of input contains T, number of testcases. Each testcase contains a string S containing characters.
Constraints:
1 <= T <= 100
1 <= |S| <= 10000For each test case, in a new line, output a single line containing the reversed String.Example:
Input:
2
i.like.this.program.very.much
pqr.mno
Output:
much.very.program.this.like.i
mno.pqr, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws IOException
{
InputStreamReader ip=new InputStreamReader(System.in);
BufferedReader br=new BufferedReader(ip);
int T=Integer.parseInt(br.readLine());
while(T-->0)
{
String S[]=br.readLine().split("\\.");
String res[]=new String[S.length];
int j=0;
for(int i=S.length-1;i>=0;i--)
{
res[j]=S[i];
j++;
}
System.out.print(res[0]);
for(int i=1;i<res.length;i++)
{
System.out.print("."+res[i]);
}
System.out.println();
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a String of length S, reverse the whole string without reversing the individual words in it. Words are separated by dots.First line of input contains T, number of testcases. Each testcase contains a string S containing characters.
Constraints:
1 <= T <= 100
1 <= |S| <= 10000For each test case, in a new line, output a single line containing the reversed String.Example:
Input:
2
i.like.this.program.very.much
pqr.mno
Output:
much.very.program.this.like.i
mno.pqr, I have written this Solution Code: #include<bits/stdc++.h>
using namespace std;
#define pu push_back
#define fi first
#define se second
#define mp make_pair
#define int long long
#define pii pair<int,int>
#define mm (s+e)/2
#define all(x) x.begin(), x.end()
#define For(i, st, en) for(int i=st; i<en; i++)
#define tr(x) for(auto it=x.begin(); it!=x.end(); it++)
#define fast std::ios::sync_with_stdio(false);cin.tie(NULL);
#define sz 200000
int A[2000][2000];
int vis[200][200];
int dist[200][200];
int xa[8]={1,1,-1,-1,2,2,-2,-2};
int ya[8]={2,-2,2,-2,1,-1,1,-1};
int n,m;
int val,hh;
int check(int x,int y)
{
if (x>=0 && y>=0 && x<n && y<m )return 1;
else return 0;
}
#define qw1 freopen("input1.txt", "r", stdin); freopen("output1.txt", "w", stdout);
#define qw2 freopen("input2.txt", "r", stdin); freopen("output2.txt", "w", stdout);
#define qw3 freopen("input3.txt", "r", stdin); freopen("output3.txt", "w", stdout);
#define qw4 freopen("input4.txt", "r", stdin); freopen("output4.txt", "w", stdout);
#define qw5 freopen("input5.txt", "r", stdin); freopen("output5.txt", "w", stdout);
#define qw6 freopen("input6.txt", "r", stdin); freopen("output6.txt", "w", stdout);
#define qw freopen("input.txt", "r", stdin); freopen("output.txt", "w", stdout);
signed main()
{
int t;
cin>>t;
while(t>0)
{
t--;
string s,p="";
cin>>s;
vector<string> ss;
int n=s.size();
int cnt=0,ans=0;
for(int i=0;i<n;i++)
{
if(s[i]!='.') {
p+=s[i];
}
else {
ss.pu(p);
p="";
}
}
if(p!="") ss.pu(p);
reverse(all(ss));
int nn=ss.size();
for(int i=0;i<nn;i++)
{
cout<<ss[i];
if(i!=nn-1) cout<<".";
}
cout<<endl;
}
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Numbers are awesome, the xor of them is more.
You are given an array A of N natural numbers. For each A[i], you need to find the Xor Value, A[i]^A[i+1]^A[i+2]^...^A[n].
Find the sum of Xor values for each element A[i].The first line of the input contains an integer N denoting the number of elements in the array.
The next line contains N single spaced integers.
<b>Constraints</b>
1 <= N <= 100000
1 <= A[i] <= 1000000000Output the single integer, the sum of Xor values for each element in array.
Note:-you may need to use long longSample Input
5
1 2 3 4 5
Sample Output
9
<b>Explanation</b>
Xor values are as follows: 1, 0, 2, 1, 5 and their sum is 9, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main(){
int n,k;
cin>>n;
int a[n];
long long sum=0;
for(int i=0;i<n;i++){
cin>>a[i];sum^=a[i];
}
long long res=0;
for(int i=0;i<n;i++){
res+=sum;
sum^=a[i];
}
cout<<res<<endl;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Numbers are awesome, the xor of them is more.
You are given an array A of N natural numbers. For each A[i], you need to find the Xor Value, A[i]^A[i+1]^A[i+2]^...^A[n].
Find the sum of Xor values for each element A[i].The first line of the input contains an integer N denoting the number of elements in the array.
The next line contains N single spaced integers.
<b>Constraints</b>
1 <= N <= 100000
1 <= A[i] <= 1000000000Output the single integer, the sum of Xor values for each element in array.
Note:-you may need to use long longSample Input
5
1 2 3 4 5
Sample Output
9
<b>Explanation</b>
Xor values are as follows: 1, 0, 2, 1, 5 and their sum is 9, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args)throws IOException {
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
int n=Integer.parseInt(br.readLine());
int a[]=new int[n];
int b=0;
int i=0;
long c=0;
String s[]=br.readLine().split(" ");
for(int k=0;k<n;k++){
a[k]=Integer.parseInt(s[k]);
}
while(i<n){
int j=i;
while(j<n){
b=b ^a[j];
j++;
}
c +=b;
b=0;
i++;
}
System.out.print(c);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Numbers are awesome, the xor of them is more.
You are given an array A of N natural numbers. For each A[i], you need to find the Xor Value, A[i]^A[i+1]^A[i+2]^...^A[n].
Find the sum of Xor values for each element A[i].The first line of the input contains an integer N denoting the number of elements in the array.
The next line contains N single spaced integers.
<b>Constraints</b>
1 <= N <= 100000
1 <= A[i] <= 1000000000Output the single integer, the sum of Xor values for each element in array.
Note:-you may need to use long longSample Input
5
1 2 3 4 5
Sample Output
9
<b>Explanation</b>
Xor values are as follows: 1, 0, 2, 1, 5 and their sum is 9, I have written this Solution Code: n = int(input())
arr = list(map(int,input().split()))
xor = 0
result = 0
for i in range(n-1,-1,-1):
xor = xor^arr[i]
# print(xor)
result += xor
print(result), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Emily was playing with triplets. She was excited to find out how many triples of non-negative integers (a, b, c) satisfy a+b+c≤S and a×b×c≤T, where S & T are non-negative integers.The input line contains S, and T separated by space.
<b>Constraints</b>
0≤S≤100
0≤T≤10000
S and T are integers.Print the number of triples of non-negative integers (a, b, c) satisfying the conditions.<b>Sample Input 1</b>
1 0
<b>Sample Output 1 </b>
4
<b>Sample Input 2</b>
2 5
<b>Sample Output 2 </b>
10
<b>Sample Input 3</b>
10 10
<b>Sample Output 3 </b>
213
(In example 1,the triplets are(0,0,0),(0,0,1),(0,1,0) and (1,0,0), I have written this Solution Code: #include<bits/stdc++.h>
using namespace std;
int main(){
int S, T; cin >> S >> T;
int cnt = 0;
for(int a = 0; a <= S; a++){
for(int b = 0; a+b <= S; b++){
for(int c = 0; a+b+c <= S; c++){
if(a*b*c <= T) cnt++;
}
}
}
cout << cnt << endl;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
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