LeetCode 3512: Minimum Operations to Make Array Sum Divisible by K Solution

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3512. Minimum Operations to Make Array Sum Divisible by K

Easy

Array Math

Problem Explanation

Explanation

To solve this problem, we need to find the minimum number of operations required to make the sum of the array divisible by k. We can achieve this by iterating through the array and calculating the sum modulo k. If the sum modulo k is greater than 0, we need to adjust the array elements to reduce the sum modulo k to 0. We can do this by decrementing the elements in the array until the sum becomes divisible by k. The minimum number of operations required will be the sum modulo k.

Algorithm:

Calculate the sum of all elements in the array.
Calculate the sum modulo k.
If the sum modulo k is greater than 0, decrement the elements in the array until the sum becomes divisible by k.
The minimum number of operations required will be the sum modulo k.

Time Complexity: O(n) Space Complexity: O(1)

Solution Code

class Solution {
    public int minOperations(int[] nums, int k) {
        int sum = 0;
        for (int num : nums) {
            sum += num;
        }
        int mod = sum % k;
        if (mod == 0) {
            return 0;
        }
        
        int operations = 0;
        int target = sum - (sum / k) * k;
        
        for (int i = 0; i < nums.length; i++) {
            if (mod >= target) {
                return operations;
            }
            if (nums[i] > mod) {
                int diff = Math.min(nums[i] - mod, mod);
                nums[i] -= diff;
                mod -= diff;
                operations += diff;
            }
        }
        
        return operations;
    }
}

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