Easy
Minimum Operations to Make Array Sum Divisible by K — C++
Full explanation · Time O(n) · Space O(1)
// Time: O(n)
// Space: O(1)
// math
class Solution {
public:
int minOperations(vector<int>& nums, int k) {
return accumulate(cbegin(nums), cend(nums), 0) % k;
}
};