What is the time complexity of Maximum Product Subarray?

The optimal solution has a time complexity of O(n).

What data structures are used to solve Maximum Product Subarray?

This problem uses Array, Dynamic Programming. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Maximum Product Subarray?

Insight 1: The presence of negative numbers is crucial. Two negative numbers multiplied together result in a positive number, which could potentially lead to a larger product than any subarray containing only positive numbers. Insight 2: We need to keep track of both the maximum and minimum product ending at each index. The minimum product is important because multiplying it by a negative number could yield a new maximum product.

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Maximum Product Subarray - LeetCode 152 Solution

Q: How to solve Maximum Product Subarray?

The solution uses dynamic programming principles to efficiently track the maximum and minimum products ending at each index of the array. For each element, it updates the current maximum and minimum product by considering the element itself, the element multiplied by the previous maximum product, and the element multiplied by the previous minimum product. The overall maximum product is then updated accordingly.

Q: Which companies ask LeetCode 152. Maximum Product Subarray in interviews?

This problem is frequently asked at Arcesium, DE Shaw, Goldman Sachs, Google, HashedIn and 10 more companies.

Maximum Product Subarray - Complete Solution Guide

Maximum Product Subarray is LeetCode problem 152, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

Given an integer array nums , find a subarray that has the largest product, and return the product . The test cases are generated so that the answer will fit in a 32-bit integer. Example 1: Input: nums = [2,3,-2,4] Output: 6 Explanation: [2,3] has the largest product 6. Example 2: Input: nums = [-2,0,-1] Output: 0 Explanation: The result cannot be 2, because [-2,-1] is not a subarray. Constraints: 1 <= nums.length <= 2 * 10 4 -10 <= nums[i] <= 10 The product of any subarray of nums is guaranteed

Detailed Explanation

The problem asks us to find the maximum product of any non-empty subarray within a given integer array `nums`. The key constraint is that the product of any subarray will fit within a 32-bit integer. This means we don't need to worry about integer overflow causing incorrect results. A subarray is a contiguous sequence of elements within the array. The goal is to find the subarray with the largest possible product.

Solution Approach

The solution uses dynamic programming principles to efficiently track the maximum and minimum products ending at each index of the array. For each element, it updates the current maximum and minimum product by considering the element itself, the element multiplied by the previous maximum product, and the element multiplied by the previous minimum product. The overall maximum product is then updated accordingly.

Step-by-Step Algorithm

Step 1: Initialize `res`, `cur_max`, and `cur_min` to the first element of the array. `res` will store the overall maximum product, `cur_max` will store the maximum product ending at the current index, and `cur_min` will store the minimum product ending at the current index.
Step 2: Iterate through the array from the second element (index 1).
Step 3: For each element `num`, calculate the potential new maximum product: `max(num, num * cur_max, num * cur_min)`. This considers the case where the current number is greater than the product of the current number multiplied by the previous maximum or minimum.
Step 4: Similarly, calculate the potential new minimum product: `min(num, num * cur_max, num * cur_min)`. The `cur_max` is used in calculation because multiplying a negative number by a larger number produces a smaller result (min).
Step 5: Update `cur_max` and `cur_min` with the calculated maximum and minimum products. Importantly, we need to save the previous `cur_max` value before updating it since it's used in the `cur_min` calculation.
Step 6: Update `res` with the maximum of `res` and `cur_max` to keep track of the overall maximum product encountered so far.
Step 7: After iterating through the entire array, return `res`.

Key Insights

Insight 1: The presence of negative numbers is crucial. Two negative numbers multiplied together result in a positive number, which could potentially lead to a larger product than any subarray containing only positive numbers.
Insight 2: We need to keep track of both the maximum and minimum product ending at each index. The minimum product is important because multiplying it by a negative number could yield a new maximum product.
Insight 3: Zeroes can reset the current product. If we encounter a zero, the current maximum and minimum products become 0. This effectively splits the array into separate subproblems.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: Array, Dynamic Programming.

Companies

Asked at: Arcesium, DE Shaw, Goldman Sachs, Google, HashedIn, Hiver, LinkedIn, Nvidia, Oracle, PayPal, ServiceNow, TikTok, Wayfair, Yahoo, tcs.

Maximum Product Subarray | LeetCode 152

Medium

Array Dynamic Programming

Time: O(n)Space: O(1)

Solution Code

Available in:

class Solution {
    public int maxProduct(int[] nums) {
        if (nums == null || nums.length == 0) {
            return 0;
        }

        int res = nums[0];
        int cur_min = nums[0];
        int cur_max = nums[0];

        for (int i = 1; i < nums.length; i++) {
            int num = nums[i];
            
            // We need to store the previous cur_max because it gets updated
            // but is needed for the cur_min calculation.
            int temp_max = cur_max;
            
            cur_max = Math.max(num, Math.max(num * temp_max, num * cur_min));
            cur_min = Math.min(num, Math.min(num * temp_max, num * cur_min));
            
            res = Math.max(res, cur_max);
        }

        return res;
    }
}

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Problem Info

Description

Problem 152: Maximum Product Subarray - Detailed explanation and solution approaches.

Topics

Array Dynamic Programming

Companies

ArcesiumDE ShawGoldman SachsGoogleHashedInHiverLinkedInNvidia

Stats

Difficulty:Medium

Languages:4

Companies:15

Problem #:152

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

7 steps

1
Step 1: Initialize `res`, `cur_max`, and `cur_min` to the first element of the array. `res` will store the overall maximum product, `cur_max` will store the maximum product ending at the current index, and `cur_min` will store the minimum product ending at the current index.
2
Step 2: Iterate through the array from the second element (index 1).
3
Step 3: For each element `num`, calculate the potential new maximum product: `max(num, num * cur_max, num * cur_min)`. This considers the case where the current number is greater than the product of the current number multiplied by the previous maximum or minimum.
4
Step 4: Similarly, calculate the potential new minimum product: `min(num, num * cur_max, num * cur_min)`. The `cur_max` is used in calculation because multiplying a negative number by a larger number produces a smaller result (min).
5
Step 5: Update `cur_max` and `cur_min` with the calculated maximum and minimum products. Importantly, we need to save the previous `cur_max` value before updating it since it's used in the `cur_min` calculation.
6
Step 6: Update `res` with the maximum of `res` and `cur_max` to keep track of the overall maximum product encountered so far.
7
Step 7: After iterating through the entire array, return `res`.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The algorithm performs a single pass through the input array. Inside the loop, it performs a fixed number of arithmetic operations. Therefore, the time complexity is directly proportional to the size of the input array, resulting in O(n) time complexity. The space used is independent of the input size, so it's constant, resulting in O(1) space complexity.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Brute Force

Iterate through all possible subarrays and calculate their products. Keep track of the maximum product encountered.

Time: O(n^2), where n is the length of the array, to generate all subarrays and then O(n) to calculate the product in each subarray. Overall, O(n^3)Space: O(1). Constant extra space is used.

Trade-offs: This approach is simpler to understand but much less efficient than the dynamic programming solution. It's suitable for small input sizes but becomes impractical for larger arrays.

✨

Key Insights

1Insight 1: The presence of negative numbers is crucial. Two negative numbers multiplied together result in a positive number, which could potentially lead to a larger product than any subarray containing only positive numbers.
2Insight 2: We need to keep track of both the maximum and minimum product ending at each index. The minimum product is important because multiplying it by a negative number could yield a new maximum product.
3Insight 3: Zeroes can reset the current product. If we encounter a zero, the current maximum and minimum products become 0. This effectively splits the array into separate subproblems.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

Edge Cases

Edge case 1: Empty array. The Python and C++ solutions handle an empty array implicitly by returning the first element, which would cause an exception. The Java and C solutions explicitly check for this and return 0.
Edge case 2: Array with a single element. The algorithm correctly handles this by initializing `res`, `cur_max`, and `cur_min` to the first element, which will be the result.
Edge case 3: Array with only negative numbers. The algorithm correctly finds the largest product, which will be either a single number (if odd # of elements) or product of two smallest magnitude numbers. For example, nums = [-2, -3, -4] will give result 12, and nums = [-2] will result in -2.
Edge case 4: Array containing zero. The algorithm correctly handles zeros by using `max` and `min`, where if zero is one of the options, it gets properly set. So the subarray before and after zero is calculated separately.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Only tracking the maximum product and not considering the minimum. This will fail in cases where a negative number multiplied by a negative number results in a larger positive product.

✗Common mistake 2: Not correctly handling the case when the current element is zero. This can lead to incorrect results if the zero is part of the subarray with the maximum product. Failing to re-initialize the current min/max when a 0 is encountered.

✗Common mistake 3: Integer overflow. The problem states the output fits in 32-bits, but intermediate calculations might exceed this limit if not handled carefully. This is generally avoided in this case because the input range is limited to [-10, 10]

✗Common mistake 4: Not initializing res properly. Initially, it's important to ensure the res value is not 0 and it's properly initialized with first element of the array.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Financial analysis. This algorithm can be used to find the period of maximum revenue growth in a company's financial data (e.g., daily stock prices), where negative numbers represent losses.

🏦Real-world scenario 2: Signal processing. It can be applied to find the strongest signal within a given time frame, even if the signal fluctuates between positive and negative values.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Consider multiple approaches before coding. Mentioning alternatives shows breadth of knowledge.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: October 31, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 13

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 6

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

How do I identify if a problem needs dynamic programming?

DP problems typically have: 1) Overlapping subproblems (same calculation repeated), 2) Optimal substructure (optimal solution uses optimal solutions to subproblems). Look for keywords like "minimum/maximum," "count ways," "is it possible," or "longest/shortest."

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Maximum Product Subarray"?

For this medium-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether dynamic programming techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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