main.cpp

// Source: https://leetcode.com/problems/minimum-operations-to-make-array-sum-divisible-by-k
// Title: Minimum Operations to Make Array Sum Divisible by K
// Difficulty: Easy
// Author: Mu Yang <http://muyang.pro>

////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
// You are given an integer array `nums` and an integer `k`. You can perform the following operation any number of times:
//
// - Select an index `i` and replace `nums[i]` with `nums[i] - 1`.
//
// Return the **minimum** number of operations required to make the sum of the array divisible by `k`.
//
// **Example 1:**
//
// ```
// Input: nums = [3,9,7], k = 5
// Output: 4
// Explanation:
// - Perform 4 operations on `nums[1] = 9`. Now, `nums = [3, 5, 7]`.
// - The sum is 15, which is divisible by 5.
// ```
//
// **Example 2:**
//
// ```
// Input: nums = [4,1,3], k = 4
// Output: 0
// Explanation:
// - The sum is 8, which is already divisible by 4. Hence, no operations are needed.
// ```
//
// **Example 3:**
//
// ```
// Input: nums = [3,2], k = 6
// Output: 5
// Explanation:
// - Perform 3 operations on `nums[0] = 3` and 2 operations on `nums[1] = 2`. Now, `nums = [0, 0]`.
// - The sum is 0, which is divisible by 6.
// ```
//
// **Constraints:**
//
// - `1 <= nums.length <= 1000`
// - `1 <= nums[i] <= 1000`
// - `1 <= k <= 100`
//
////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////

#include <numeric>
#include <vector>

using namespace std;

// Math
class Solution {
 public:
  int minOperations(vector<int>& nums, int k) {  //
    return accumulate(nums.cbegin(), nums.cend(), 0) % k;
  }
};