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 <stdio.h>

int main() {
    int x, y, n;
    scanf("%d%d", &x, &y);
    scanf("%d", &n);

    if((x + y) >= n) {   // if total chocolates are greater than or equal to number of children, then print YES
        printf("YES\n");
    } else {   // less chocolates than children, print NO
        printf("NO\n");
    }
    return 0;
}

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 

int main() {
    int t;
    scanf("%d", &t); // taking the number of test cases

    while (t--) { // loop runs for each test case
        long long n;
        scanf("%lld", &n); // input number for the current test case

        int maxi = 0;
        while (n > 0) {
            int rem = n % 10; // extract last digit
            if (rem > maxi) {
                maxi = rem; // update maximum digit
            }
            n /= 10; // remove last digit
        }
        printf("%d\n", maxi); // print maximum digit
    }
    return 0;
}