2918. Minimum Equal Sum of Two Arrays After Replacing Zeros

Problem Description

In this problem, we are provided with two arrays, nums1 and nums2, each containing positive integers. However, both arrays may contain zeros, and we need to find a way to replace these zeros with positive integers. Our goal is to ensure that after the replacements, the sum of the integers in both arrays is the same.

The question then is: What is the minimum equal sum we can achieve after replacing all zeros with strictly positive integers? If it is not possible to make the sums of both arrays equal by replacing zeros, the function should return -1.

It's a problem that demands a careful examination of the sums of arrays and the zeros they contain to find a method to balance the sums with minimum changes.

Intuition

To approach the problem, we first handle the zeros in a straightforward way. We consider replacing each zero with 1, as this is the smallest strictly positive integer. This gives us a baseline sum for each array.

Let s1 and s2 be the sums of nums1 and nums2 respectively when each zero is treated as 1. Based on these sums, we can reason about the possible scenarios:

1. If s1 equals s2, we already have equality and do not need to replace any zeros with larger numbers, so the baseline sum is the answer.
2. If s1 is less than s2, and there are zeros in nums1, we can increase the sum of nums1 by replacing zeros with numbers larger than 1 to match s2. In this case, since s2 is larger, it is our target, and we return s2 because it represents the minimum equal sum we can achieve.
3. Conversely, if s1 is more than s2, we swap the roles of nums1 and nums2 to apply the above reasoning.
4. If s1 is less than s2, but nums1 has no zeros, we cannot increase the sum of nums1 to match s2. Thus, it is impossible to make the sums equal and we return -1.

This way, we identify the minimum sum after replacements that makes both arrays' sums equal, or determine the impossibility of such an operation.

Solution Approach

The solution approach is to first calculate the sums of the arrays while treating all zeros as ones. This calculation is straightforward and can be done using the built-in sum function in Python, along with the count method to account for the zeros. The following code calculates these sums for nums1 and nums2:

1s1 = sum(nums1) + nums1.count(0)
2s2 = sum(nums2) + nums2.count(0)

Once the sums are calculated, we compare them to decide the next steps:

• If s1 is greater than s2, we have a helper function, self.minSum, that we call with the arrays swapped. This is an example of recursion that helps us reduce the problem to a single direction where s1 is less than or equal to s2, ensuring a more straightforward decision process:

1if s1 > s2:
2    return self.minSum(nums2, nums1)

• If the sums are already equal (s1 == s2), we have achieved the goal of equal sums with minimum effort, merely by treating zeros as ones. The baseline sum s1 (or s2, since they are equal now) is the minimum sum we can achieve, so we return it:

1if s1 == s2:
2    return s1

• In the case that s1 is less than s2, we have two scenarios. If there are zeros in nums1, we can replace them with numbers greater than one to match the sum s2. Therefore, we return s2 which is the target sum:

1if nums1.count(0) != 0:
2    return s2

• Finally, if there are no zeros in nums1, it implies we cannot adjust the sum of nums1 to match s2, making it impossible for both array sums to be equal, and we return -1:

1else:
2    return -1

By considering these conditions and applying the recursive step, the algorithm ensures minimum modifications are made to achieve equal sums, or it correctly identifies when the task is impossible. The use of simple built-in functions and conditional logic, along with recursion, forms the core of this solution, eliminating the need for complex data structures or patterns.

Example Walkthrough

Consider the arrays nums1 = [0, 2, 3] and nums2 = [1, 0, 4, 0].

1. We start by treating all zeros in nums1 and nums2 as ones and calculate their sums.

   • For nums1: The sum is 1 (replacing 0) + 2 + 3 = 6.
   • For nums2: The sum is 1 + 1 (replacing 0) + 4 + 1 (replacing 0) = 7.
   • s1 becomes 6 and s2 becomes 7.

2. We compare s1 and s2. Since s1 is less than s2, we proceed without swapping the arrays.

3. Because s1 (sum of nums1) is less than s2 (sum of nums2), we check if there are zeros present in nums1 that we can replace to match the sum s2. Indeed, nums1 has one zero.

4. We can replace the zero in nums1 with a positive integer that will raise the sum of nums1 to match s2. Since s2 is 7, and the current sum of nums1 is 6, we can replace the zero with any number greater than 1 to meet or exceed the sum of nums2. For minimal increment, we can choose to replace zero with the smallest number greater than 1, which is 2. It is now nums1 = [2, 2, 3].

5. After replacement, the sum of nums1 is now 2 + 2 + 3 = 7, which matches s2.

6. Since we can match the sums by replacing zeros with strictly positive integers, we return s2. Here, the minimum equal sum we achieve is 7.

According to this logic, if nums1 had no zeros but still had a smaller sum than nums2, we would return -1, indicating it's not possible to match the sums.

Now, let's break down the steps according to the code snippets provided in the solution approach:

• Step 1 corresponds to the initial calculations:

1s1 = sum(nums1) + nums1.count(0)  # s1 = 6
2s2 = sum(nums2) + nums2.count(0)  # s2 = 7

• Since s1 < s2, we do not swap nums1 and nums2, and the first recursive call is skipped.
• Checking s1 == s2 would skip the second condition in this case since they are not equal.
• In the third condition, we evaluate if nums1.count(0) != 0:. Because nums1 contains a zero, we return s2 (7), which represents the minimum equal sum we can achieve after replacements.
• There is no need to execute the last condition else: return -1 since we did not return -1.

1if nums1.count(0) != 0:
2    return s2  # returns 7

The given example demonstrates how the solution approach is applied to find the minimum equal sum after replacing all zeros with strictly positive integers or determining when it's not possible to match the sums.

Solution Implementation

Time and Space Complexity

Time Complexity

The time complexity of the code can be broken down as follows:

• Calculating sum of nums1 and nums2 requires traversing both arrays, which costs O(n) and O(m) respectively, where n is the length of nums1 and m is the length of nums2.
• Counting the zeroes in nums1 and nums2 with nums1.count(0) and nums2.count(0) also takes O(n) and O(m) time respectively.
• The recursive call self.minSum(nums2, nums1) only happens if s1 > s2. If multiple invocations occur, the process will still not exceed O(n + m) because the sum and count operations are repeated once for each list.
• The if comparisons are constant time operations, O(1).

Aggregating all these costs, the total time complexity is O(n + m) because we are summing separate linear terms associated with each list.

Space Complexity

• The space complexity is O(1) as no additional space that grows with the input size is used. Variables s1 and s2 use constant space, and the recursive call does not exceed a single level of stack space since it's a direct swap with no further recursion.