Introduction

In this blog post, we will explore how to solve the LeetCode problem "2918" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

You are given two arrays nums1 and nums2 consisting of positive integers. You have to replace all the 0's in both arrays with strictly positive integers such that the sum of elements of both arrays becomes equal. Return the minimum equal sum you can obtain, or -1 if it is impossible. Example 1: Input: nums1 = [3,2,0,1,0], nums2 = [6,5,0] Output: 12 Explanation: We can replace 0's in the following way: - Replace the two 0's in nums1 with the values 2 and 4. The resulting array is nums1 = [3,2,2,1,4]. - Replace the 0 in nums2 with the value 1. The resulting array is nums2 = [6,5,1]. Both arrays have an equal sum of 12. It can be shown that it is the minimum sum we can obtain. Example 2: Input: nums1 = [2,0,2,0], nums2 = [1,4] Output: -1 Explanation: It is impossible to make the sum of both arrays equal. Constraints: 1 <= nums1.length, nums2.length <= 105 0 <= nums1[i], nums2[i] <= 106

Explanation

Here's the approach:

• Calculate the sum of each array and the number of zeros in each.
• Check if it's possible to make the sums equal. The minimum value to replace a 0 with is 1. If sum1 + count1 < sum2 or sum2 + count2 < sum1, it's impossible.

• Determine the minimum equal sum. If sum1 > sum2, we must ensure sum2 can reach at least sum1 by filling its zeros with ones. Similarly if sum2 > sum1. The minimum equal sum is the maximum of sum1 and sum2. We must also handle cases where one array has zeros and the other does not.

• Runtime Complexity: O(n), Storage Complexity: O(1)

Code

    def solve():
    nums1 = [int(x) for x in input().split(",")]
    nums2 = [int(x) for x in input().split(",")]

    sum1 = sum(nums1)
    sum2 = sum(nums2)
    count1 = nums1.count(0)
    count2 = nums2.count(0)

    if sum1 + count1 < sum2 or sum2 + count2 < sum1:
        print(-1)
        return

    if count1 == 0 and count2 == 0:
        if sum1 == sum2:
            print(sum1)
        else:
            print(-1)
        return

    print(max(sum1 + count1 if count1 > 0 else sum1, sum2 + count2 if count2 > 0 else sum2))

solve()