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3512. Minimum Operations to Make Array Sum Divisible by K

Description

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

Solutions

Solution 1: Sum and Modulo

The problem essentially asks for the result of the sum of the array elements modulo $k$. Therefore, we only need to iterate through the array, calculate the sum of all elements, and then take the modulo $k$. Finally, return this result.

The time complexity is $O(n)$, where $n$ is the length of the array $\textit{nums}$. The space complexity is $O(1)$.

  • class Solution {
    public int minOperations(int[] nums, int k) {
        return Arrays.stream(nums).sum() % k;
    }
}

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

  • class Solution:
    def minOperations(self, nums: List[int], k: int) -> int:
        return sum(nums) % k


  • func minOperations(nums []int, k int) (ans int) {
	for _, x := range nums {
		ans = (ans + x) % k
	}
	return
}

  • function minOperations(nums: number[], k: number): number {
    return nums.reduce((acc, x) => acc + x, 0) % k;
}

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