How do you solve Maximum Length of Subarray With Positive Product on LeetCode?

Maximum Length of Subarray With Positive Product (LeetCode #1567) can be solved using Brute Force. The optimal approach is Brute Force with O(n²) time complexity and O(1) space complexity. Key insight: The brute force approach involves checking every possible subarray and calculating the product of its elements. If the product is positive, we keep track of the maximum length found.

What is the time complexity of Maximum Length of Subarray With Positive Product?

The optimal solution for Maximum Length of Subarray With Positive Product runs in O(n²) time and uses O(1) space. This complexity arises because we are using nested loops to check all subarrays, leading to a quadratic number of operations.

What algorithm or data structure is used to solve Maximum Length of Subarray With Positive Product?

Maximum Length of Subarray With Positive Product uses the following concepts: Array, Dynamic Programming, Greedy. The recommended patterns to study are: Two Pointers, Sliding Window, Dynamic Programming.

What are the interview tips for Maximum Length of Subarray With Positive Product?

Always think about edge cases, like zeros in the array Keep track of the state of your variables carefully Practice explaining your thought process as you code

What are common mistakes when solving Maximum Length of Subarray With Positive Product?

Forgetting to reset counters when encountering zeros Not considering the case of even and odd negative counts properly

Maximum Length of Subarray With Positive Product — LeetCode #1567 (Medium)

Tags: Array, Dynamic Programming, Greedy

Related patterns: Two Pointers, Sliding Window, Dynamic Programming

Brute Force approach

Time complexity: O(n²). Space complexity: O(1).

The brute force approach involves checking every possible subarray and calculating the product of its elements. If the product is positive, we keep track of the maximum length found.

This complexity arises because we are using nested loops to check all subarrays, leading to a quadratic number of operations.

Step 1: Iterate through each element in the array as a starting point for subarrays.
Step 2: For each starting point, iterate through all possible ending points to form subarrays.
Step 3: Calculate the product of the current subarray and check if it's positive. If it is, update the maximum length.

Input: nums = [1,-2,-3,4] 1. Start at index 0, product = 1. Subarray [1] is positive, length = 1. 2. Extend to index 1, product = -2. Not positive. 3. Extend to index 2, product = 6. Subarray [1,-2,-3] is positive, length = 3. 4. Extend to index 3, product = 24. Subarray [1,-2,-3,4] is positive, length = 4. 5. Max length found: 4.

Common Mistakes

Forgetting to reset counters when encountering zeros
Not considering the case of even and odd negative counts properly

Interview Tips

Always think about edge cases, like zeros in the array
Keep track of the state of your variables carefully
Practice explaining your thought process as you code

#1567

Maximum Length of Subarray With Positive Product

Medium

Array↗Dynamic Programming↗Greedy↗Two Pointers↗Sliding Window↗Dynamic Programming↗

LeetCode ↗

Approaches

💡

Intuition

Time O(n²)Space O(1)

The brute force approach involves checking every possible subarray and calculating the product of its elements. If the product is positive, we keep track of the maximum length found.

⚙️

Algorithm

3 steps

1Step 1: Iterate through each element in the array as a starting point for subarrays.
2Step 2: For each starting point, iterate through all possible ending points to form subarrays.
3Step 3: Calculate the product of the current subarray and check if it's positive. If it is, update the maximum length.

solution.py10 lines

1def maxLength(nums):
2    max_length = 0
3    n = len(nums)
4    for i in range(n):
5        product = 1
6        for j in range(i, n):
7            product *= nums[j]
8            if product > 0:
9                max_length = max(max_length, j - i + 1)
10    return max_length

ℹ

Complexity note: This complexity arises because we are using nested loops to check all subarrays, leading to a quadratic number of operations.

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