2025-11-29 Daily Challenge

Posted on:
Posted modified:
In: DailyChallenge

Today I have done leetcode's November LeetCoding Challenge with cpp.

November LeetCoding Challenge 29

Description

Minimum Operations to Make Array Sum Divisible by K

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

Solution

class Solution {
public:
  int minOperations(vector<int>& nums, int k) {
    int sum = accumulate(nums.begin(), nums.end(), 0);
    return sum % k;
  }
};

// Accepted
// 855/855 cases passed (0 ms)
// Your runtime beats 100 % of cpp submissions
// Your memory usage beats 3.18 % of cpp submissions (45.4 MB)
Algorithm