LeetCode in Kotlin

3512. Minimum Operations to Make Array Sum Divisible by K

Easy

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 {
    fun minOperations(nums: IntArray, k: Int): Int {
        var sum = 0
        for (num in nums) {
            sum += num
        }
        return sum % k
    }
}

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