[Solved] XOR Pair Contest Problem

XOR Pair: Given an array of positive element having size N and an integer C. Check if there exists a pair (A,B) such that A xor B = C.

Input : 
First line of input contains number of testcases T. The 1^st line of each testcase contains a two integers N and C. The 2^nd line of each testcase, contains N space separated integers denoting the elements of the array A.

Output: 
Print “Yes” is the pair exists else print “No” without quotes.(Change line after every answer).

Constraints:
1 <= T <= 100
1 <= N <= 10⁵
1 <= C <= 10⁵
0 <= arr[i] <= 10⁵

Example:
Input:
2
7 7
2 1 10 3 4 9 5 
5 1
9 9 10 10 3 

Output:
Yes
No

Explanation :
In first case, pair (2,5) give 7. Hence answer is “Yes”. In second case no pair exist such that satisfies the condition hance the answer is “No”.

Solution:

import java.util.*;
import java.lang.*;
import java.io.*;

public class XORPair {
    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());
        }
        String nextLine(){
            String str = "";
            try {
                str = br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }
    }
    public static void main(String[] args) {
        FastReader fr = new FastReader();
        int t = fr.nextInt();
        while (t-->0){
            int n = fr.nextInt();
            int c = fr.nextInt();
            int[] arr = new int[n];
            for (int i = 0; i < n; i++) {
                arr[i] = fr.nextInt();
            }
            HashSet<Integer> set = new HashSet<>();
            int r = 0;
            for (int i = 0; i < n; i++) {
                if (set.contains(arr[i]^c)){
                    r+=1;
                }
                set.add(arr[i]);
            }
            if (r>0){
                System.out.println("Yes");
            }else {
                System.out.println("No");
            }
        }
    }
}

#include <bits/stdc++. h>
using namespace std;

int main() {
	//
	int t;
	cin>>t;
	
	while(t--){
	    
	    int n,c;
	    cin>>n>>c;
	    
	    int arr[n];
	    
	    for(int i=0; i<n; i++){
	        cin>>arr[i];
	    }
	    
	    unordered_set<int> s;
	    
	    bool flag = false;
	    
	    for(int i=0; i<n; i++){
	        
	        if(s.find(c^arr[i]) != s.end()){
	           flag = true;
	           break;
	        }
	        
	        s.insert(arr[i]);
	    }
	    
	    if(flag){
	        cout<<"Yes"<<endl;
	    }else{
	        cout<<"No"<<endl;
	    }
	}
	return 0;
}

Happy Learning – If you require any further information, feel free to contact me.