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.