Maximum Product Subarray Leetcode Problem 152 [Python Solution]

In this guide, we will delve into the problem of finding the maximum product subarray using Python.

This LeetCode problem, numbered 152, falls under the category of 1-D Dynamic Programming and is particularly relevant to companies like Amazon.

We will explore the problem statement, constraints, and different approaches to solving it.

Problem Overview

The problem statement is as follows:

Question: Given an integer array nums, find a subarray that has the largest product and return the product.

The test cases are designed so that the answer will fit within a 32-bit integer.

Let’s consider an example to illustrate the problem:

Example 1:
Input: nums = [2, 3, -2, 4]
Output: 6
Explanation: [2, 3] has the largest product, which is 6.

Example 2:
Input: nums = [-2, 0, -1]
Output: 0
Explanation: The result cannot be 2, because [-2, -1] is not a subarray.

Understanding Constraints

Before we delve into the solution, let’s understand the constraints of this problem:

• The length of the nums array, denoted as n, falls in the range: 1 <= n <= 2 * 10^4.
• The values of elements in nums, denoted as nums[i], range from -10 to 10.
• The product of any prefix or suffix of nums is guaranteed to fit in a 32-bit integer.

With these constraints in mind, we need to find an efficient solution to this problem.

Maximum Product Subarray LeetCode Problem Solution

1. Bruteforce Approach

One way to approach this problem is through a brute-force solution.

The idea is to consider all possible subarrays and calculate the product for each one.

We keep track of the maximum product encountered.

However, this approach has a time complexity of O(n^2), making it highly inefficient, especially for larger arrays.

def maxProduct(nums):
    n = len(nums)
    max_product = float('-inf')

    for i in range(n):
        product = 1
        for j in range(i, n):
            product *= nums[j]
            max_product = max(max_product, product)

    return max_product

While this approach works, it is not suitable for larger arrays due to its quadratic time complexity.

Let’s explore a more efficient solution.

2. An Efficient Approach with Dynamic Programming

To optimize our solution, we can make use of dynamic programming.

We will maintain two variables: current_max and current_min, which represent the maximum and minimum product subarrays ending at the current element.

We also maintain an overall result variable to store the maximum product found so far.

Here is the Python code for this efficient approach:

def maxProduct(self, nums: List[int]) -> int:
    n = len(nums)
    result = nums[0]
    current_min = current_max = 1

    for num in nums:
        # Temporary variable to hold the current max
        temp = current_max

        # Calculate the new current max
        current_max = max(num, num * current_max, num * current_min)

        # Update the current min
        current_min = min(num, num * temp, num * current_min)

        # Update the overall result
        result = max(result, current_max)

    return result

This efficient solution has a time complexity of O(n) and a space complexity of O(1), making it much faster and memory-efficient than the brute-force approach.

Time and Space Complexity

In the efficient approach, we maintain the time complexity at O(n) and the space complexity at O(1), which is highly efficient given the problem’s constraints.

This makes the solution suitable for large input arrays.

Reasoning Behind Our Approach

The dynamic programming approach relies on the observation that to find the maximum product subarray ending at a particular element, we need to consider both the maximum and minimum products ending at the previous element.

This approach ensures that we handle positive, negative, and zero values appropriately, while keeping track of the maximum product encountered so far.

By updating the current_max and current_min at each step, we maintain an efficient solution that covers all possible scenarios.

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Conclusion

In this guide, we explored the LeetCode problem Maximum Product Subarray We discussed the problem statement, examined its constraints, and presented two different approaches to solving it.

The efficient dynamic programming approach, with a time complexity of O(n) and a space complexity of O(1), stands out as the most suitable solution for this problem.

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We hope this information assists you in your coding journey and problem-solving endeavors.

Question Link: Maximum Product Subarray LeetCode Problem

Thank you for your attention, and happy coding!