Math, Implementation

We are given an array of size n with total sum S.
If S is divisible by n, the array is already balanced and the answer is "YES".
Otherwise, we can try deleting one element a[i]. After deletion, the new sum becomes S − a[i] and the size becomes n − 1.
To be balanced, (S − a[i]) must be divisible by (n − 1), which means a[i] % (n − 1) == S % (n − 1).
We just need to check if such an element exists.

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

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    int t; cin >> t;
    while (t--) {
        int n; cin >> n;
        vector<int> a(n);
        long long sum = 0;
        for (int &x : a) cin >> x, sum += x;

        if (sum % n == 0) {
            cout << "YES\n";
            continue;
        }

        long long rem = sum % (n - 1);
        bool ok = false;
        for (int x : a)
            if (x % (n - 1) == rem)
                ok = true;

        cout << (ok ? "YES\n" : "NO\n");
    }
}