HackerEarth A bitwise pair problem solution

In this HackerEarth A bitwise pair problem solution You are given an array A of N positive integers. Find the number of pairs (i,j) such that:

(A[i] | A[j]) - (A[i] & A[j]) = (A[i] - A[j]), where | is a bitwise OR operator and & is a bitwise AND operator.

HackerEarth A bitwise pair problem solution.

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

void solve(){
    int n;
    cin >> n;
    assert(1 <= n and n <= 100000);

    int cnt[1001] = {0};
    for(int i = 1 ; i <= n ; i++){
        int val;
        cin >> val;
        assert(1 <= val and val <= 1000);
        cnt[val]++;
    }

    int answer = 0;
    for(int i = 1 ; i <= 1000 ; i++)
    {
        for(int j = 1 ; j <= i ; j++)
        {
            if(i == j)
            {
                answer += (cnt[i]*cnt[i]);
            }
            else if((i^j) == (i - j))
            {
                answer += (cnt[i]*cnt[j]);
            }
        }
    }

    cout << answer << endl;
}

signed main(){
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    int t;
    cin >> t;
    assert(1 <= t and t <= 10);
    while(t--){
        solve();
    }
}

Second solution

MAXN = 1001
t = int(input())
while t > 0:
    t -= 1
    n = int(input())
    a = list(map(int, input().split()))
    cnt = [0] * MAXN
    for x in a:
        cnt[x] += 1
    ans = 0
    for i in range(0, MAXN):
        for j in range(0, MAXN):
            ans += (i ^ j == i - j) * cnt[i] * cnt[j]
    print(ans)