How do you solve Maximum Product Subarray on LeetCode?

Maximum Product Subarray (LeetCode #152) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: The product of negative numbers can become positive, so we need to track both maximum and minimum products.

What is the brute force solution for Maximum Product Subarray?

The brute force approach for Maximum Product Subarray is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking every possible subarray and calculating its product. This is simple to understand but inefficient for larger arrays. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Maximum Product Subarray?

Maximum Product Subarray uses the following concepts: Array, Dynamic Programming. The recommended patterns to study are: Dynamic Programming, Array.

What are the interview tips for Maximum Product Subarray?

Always think about edge cases, especially with negative numbers. Explain your thought process clearly while coding. Practice writing both brute force and optimal solutions to understand the trade-offs.

What are common mistakes when solving Maximum Product Subarray?

Forgetting to handle the case of negative numbers correctly. Not considering the possibility of a single element being the maximum product.

#152

Maximum Product Subarray

Medium

ArrayDynamic ProgrammingDynamic ProgrammingArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(n)Space O(1)

The optimal solution uses a single pass through the array, keeping track of both the maximum and minimum products at each position. This is crucial because a negative number can turn a small product into a large one when multiplied.

⚙️

Algorithm

3 steps

1Step 1: Initialize variables for max_product, min_product, and result.
2Step 2: Iterate through the array, updating max_product and min_product based on the current number.
3Step 3: Update the result with the maximum of itself and max_product.

solution.py9 lines

1def maxProduct(nums):
2    max_product = min_product = result = nums[0]
3    for num in nums[1:]:
4        if num < 0:
5            max_product, min_product = min_product, max_product
6        max_product = max(num, max_product * num)
7        min_product = min(num, min_product * num)
8        result = max(result, max_product)
9    return result

ℹ

Complexity note: The time complexity is O(n) because we only go through the array once. The space complexity is O(1) since we are using a constant amount of extra space.

1The product of negative numbers can become positive, so we need to track both maximum and minimum products.
2A single pass through the array is sufficient to find the maximum product subarray.

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