Chef and Distinct Numbers Codechef Solution: Given an array AA of size NN , check if there exist any pair of index i,ji,j such that Ai=AjAi=Aj and i≠ji≠j
Input
- The first line of the input contains a single integer TT denoting the number of test cases.
- The first line of each test case contains integer NN.
- The second line of each test case contains NN space separated integers AiAi.
Output
For each test case, print a single line containing answer “Yes” or “No” (without quotes).
Constraints
- 1≤T≤1001≤T≤100
- 2≤N≤1052≤N≤105
- 1≤Ai≤1091≤Ai≤109
- Sum of NN over all test cases doesn’t exceed 106106
Example Input
2 4 1 2 1 3 5 5 4 1 2 3
Example Output
Yes No
Explanation
Example case 1: A1A1 and A3A3 both have value 1.
Example case 2: All values are pairwise distinct.
Chef and Distinct Numbers – CodeChef Solution in JAVA
import java.util.Scanner;
import java.util.Set;
import java.util.HashSet;
public class Main {
public static void main (String[] args) {
Scanner sc = new Scanner(System.in);
int test = sc.nextInt();
while(test-- > 0) {
int n = sc.nextInt();
int[] arr = new int[n];
for(int i = 0; i < n; i++) {
arr[i] = sc.nextInt();
}
Set<Integer> set = new HashSet<>();
for (int i = 0; i < n; i++) {
set.add(arr[i]);
}
if (set.size() == n)
System.out.println("No");
else
System.out.println("Yes");
}
}
}Chef and Distinct Numbers – CodeChef Solution in CPP
#include <iostream>
#include<algorithm>
using namespace std;
int main() {
int t;
cin>>t;
while(t--){
int n;
cin>>n;
long long int arr[n];
int i=0;
while(n--){
cin>>arr[i];
i++;
}
int k=sizeof(arr)/sizeof(arr[0]);
sort(arr,arr+k);
int flag=0;
for(int i=0;i<k;i++){
if(arr[i]==arr[i-1]){
cout<<"Yes"<<endl;
flag=1;
break;
}
}
if(flag==0){
cout<<"No"<<endl;
}
}
return 0;
}
Chef and Distinct Numbers -CodeChef Solution in Python
cases=int(input())
for i in range(cases):
n=int(input())
A=list(map(int,input().split()))
A.sort()
B=set(A)
if len(A)==len(B):
print("No")
else:
print("Yes")
Disclaimer: The above Problem(Chef and Distinct Numbers ) is generated by CodeChef but the solution is provided by Chase2learn.This tutorial is only for Educational and Learning purpose.
