LeetCode 2918: Minimum Equal Sum of Two Arrays After Replacing Zeros Solution

Master LeetCode problem 2918 (Minimum Equal Sum of Two Arrays After Replacing Zeros), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

2918. Minimum Equal Sum of Two Arrays After Replacing Zeros

Medium

Array Greedy

Problem Explanation

Explanation:

To solve this problem, we need to find a way to replace the zeros in both arrays such that the sum of elements in both arrays becomes equal and the sum itself is minimized. We can approach this problem by calculating the total sum of both arrays and then distributing the increment required to make the sum equal among the zeros in both arrays.

Calculate the total sum of both arrays.
Calculate the sum of non-zero elements in both arrays.
Calculate the number of zeros in both arrays.
Calculate the increment required to make the sum of both arrays equal.
Distribute this increment among the zeros in both arrays.

If the increment required is not possible to distribute among the zeros in a way that makes the sum equal, then return -1.

Time Complexity:

Calculating the total sum, sum of non-zero elements, number of zeros: O(n) where n is the total number of elements in both arrays.
Distributing the increment among zeros: O(n) where n is the total number of zeros in both arrays.
Overall time complexity: O(n)

Space Complexity:

Additional space used for variables: O(1)

Solution Code

class Solution {
    public int minEqualSum(int[] nums1, int[] nums2) {
        int sum1 = 0, sum2 = 0, zeros1 = 0, zeros2 = 0, diff = 0;
        
        for (int num : nums1) {
            if (num == 0) zeros1++;
            else sum1 += num;
        }
        
        for (int num : nums2) {
            if (num == 0) zeros2++;
            else sum2 += num;
        }
        
        diff = Math.abs(sum1 - sum2);
        
        if ((diff + zeros1 + zeros2) % 2 != 0 || diff > zeros1 + zeros2) {
            return -1;
        }
        
        return (diff + zeros1 + zeros2) / 2;
    }
}

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