What is the time complexity of Minimum Equal Sum of Two Arrays After Replacing Zeros?

The optimal solution for Minimum Equal Sum of Two Arrays After Replacing Zeros runs in O(n) time and uses O(1) space. This complexity is linear because we only traverse each array once to compute sums and counts, making it efficient for large inputs.

What is the brute force solution for Minimum Equal Sum of Two Arrays After Replacing Zeros?

The brute force approach for Minimum Equal Sum of Two Arrays After Replacing Zeros is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach tries every possible combination of replacing zeros in both arrays to find a valid equal sum. This method is straightforward but inefficient, as it explores all possibilities. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Minimum Equal Sum of Two Arrays After Replacing Zeros?

Minimum Equal Sum of Two Arrays After Replacing Zeros uses the following concepts: Array, Greedy. The recommended patterns to study are: Greedy, Array.

What are the interview tips for Minimum Equal Sum of Two Arrays After Replacing Zeros?

Always check edge cases, such as arrays with no zeros. Think about how you can minimize the values you replace zeros with. Explain your thought process clearly to the interviewer.

What are common mistakes when solving Minimum Equal Sum of Two Arrays After Replacing Zeros?

Not considering the case where zeros are insufficient to balance the sums. Overcomplicating the solution with unnecessary iterations.

#2918

Minimum Equal Sum of Two Arrays After Replacing Zeros

Medium

ArrayGreedyGreedyArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(1)

💡

Intuition

Time O(n)Space O(1)

The optimal solution leverages the fact that we can replace zeros with the smallest possible integers (1) to balance the sums of the two arrays. By calculating the difference in sums and adjusting the zeros accordingly, we can find the minimum equal sum efficiently.

⚙️

Algorithm

4 steps

1Step 1: Calculate the current sums of both arrays, ignoring zeros.
2Step 2: Count the number of zeros in both arrays.
3Step 3: Determine the minimum sum that can be achieved by replacing zeros with 1s.
4Step 4: Check if the difference between the two sums can be compensated by the zeros available in the smaller sum array.

solution.py19 lines

1# Full working Python code
2
3def minEqualSum(nums1, nums2):
4    sum1 = sum(x for x in nums1 if x > 0)
5    sum2 = sum(x for x in nums2 if x > 0)
6    zeros1 = nums1.count(0)
7    zeros2 = nums2.count(0)
8
9    min_sum = sum1 + sum2 + zeros1 + zeros2
10
11    if sum1 < sum2:
12        if (sum2 - sum1) > zeros1:
13            return -1
14    else:
15        if (sum1 - sum2) > zeros2:
16            return -1
17
18    return min_sum
19

ℹ

Complexity note: This complexity is linear because we only traverse each array once to compute sums and counts, making it efficient for large inputs.

1Replacing zeros with 1s is the minimum contribution to balance the sums.
2The array with the smaller sum must have enough zeros to cover the difference.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.