Maximum Product Subarray – Think Before You Multiply!

 Maximum Product Subarray – Finding the Largest Product

Subarray problems are a classic challenge in Data Structures and Algorithms (DSA). Today’s problem, Maximum Product Subarray, involves finding a contiguous subarray within an integer array that yields the highest possible product. This problem requires careful handling of negative numbers and zeros, making it an interesting edge-case-heavy problem.

The Best Data Structure for Solving It (and Why!)

For this problem, no additional data structures are needed! Since we only need to track maximum product values, simple integer variables work best. Using an iterative approach ensures an efficient solution in O(N) time without requiring extra memory.

Different Approaches – Brute Force to Optimized Solutions

Brute Force Approach – Check All Subarrays

Idea: Compute the product of every possible subarray and track the maximum.

  • Time Complexity: O(N²) – Nested loops make this inefficient for large arrays.

  • Space Complexity: O(1) – Only a few extra variables are used.

  • Why It Fails: Too slow for large inputs (N ≤ 20,000).

Optimized Approach – Prefix & Suffix Products

Instead of trying every subarray, we can iterate through the array once while maintaining the maximum possible product.

Key Idea:

  • A subarray product can be negatively impacted by a negative number.

  • A zero resets the product, so we start fresh when encountering one.

  • To handle both cases, compute prefix and suffix products in a single pass.

Implementation (Java):

class Solution {
    public int maxProduct(int[] nums) {
        int n = nums.length; 
        int pre = 1, suff = 1;
        int ans = Integer.MIN_VALUE;
        for (int i = 0; i < n; i++) {
            if (pre == 0) pre = 1;
            if (suff == 0) suff = 1;
            pre *= nums[i];
            suff *= nums[n - i - 1];
            ans = Math.max(ans, Math.max(pre, suff));
        }
        return ans;
    }
}

⚠️ Common Mistakes & Pitfalls to Avoid

  • Ignoring negative numbers – A negative number can turn a small product into a large one if multiplied correctly.

  • Forgetting to reset on zeroes – A zero wipes out the product, so handling resets is crucial.

  • Using only forward iteration – Some maximum products exist in the suffix of the array, requiring both prefix and suffix tracking.

📊 Time & Space Complexity Analysis

Brute Force: O(N²) – Too slow.
Optimized (Prefix-Suffix): O(N) – Fast and efficient.
Space Complexity: O(1) – No extra arrays needed.

🌍 Real-World Applications

  • Financial Market Analysis – Detecting maximum profit trends in stock data.

  • Signal Processing – Identifying high-energy segments in waveforms.

  • AI & Machine Learning – Feature extraction for time-series data.


I hope you found this article insightful! Keep exploring, keep learning, and don’t forget to check back daily for more exciting problem breakdowns. If you know someone who would benefit from this, share it with them and help them grow too! That’s it for today—see you in the next post!

                                                                                                                                         Signing off!! 

                                                                                                Master DSA One Problem at a Time :)

Comments

Post a Comment

Popular posts from this blog

Trailing Zeros in Factorial – How Many and Why?

Trailing Zeros – Counting Zeros in Factorial Factorial-related problems are a fundamental part of Data Structures and Algorithms (DSA). One such classic problem involves counting the number of trailing zeros in N! . In this article, we will explore different approaches to solving this problem, optimize our solution, and examine its real-world applications. Optimal Data Structure for Solving the Problem For this problem, no additional data structures are required. Since factorials grow exponentially, storing the actual factorial value is impractical. Instead, we leverage mathematical properties and division operations to efficiently determine the number of trailing zeros. This approach ensures an optimal solution in terms of both time and space complexity. Approach – From Brute Force to an Optimized Solution Brute Force Approach – Compute and Count A naive approach involves computing the factorial of N and counting the number of trailing zeros. However, this method is highly ineff...
Read more

Divisible Drama — Who’s In, Who’s Out?

 Divisible Drama — Who’s In, Who’s Out? Problem Statement You’re given two positive integers n and m . num1 = sum of all numbers from 1 to n not divisible by m num2 = sum of all numbers from 1 to n divisible by m Your mission, should you choose to accept it: Return num1 - num2 Example Input: n = 10 , m = 3 Output: 19 Breakdown: Not divisible by 3 → [ 1 , 2 , 4 , 5 , 7 , 8 , 10 ] → num1 = 37 Divisible by 3 → [ 3 , 6 , 9 ] → num2 = 18 Answer = 37 - 18 = 19 Best Data Structure for Solving the Problem This one’s a math.  Why overthink it when Gauss already gave us the formula for sum of 1 to n? Different Approaches (from Brute Force to Optimized) Approach 1: Brute Force (aka List & Sum Vibes) Loop from 1 to n , split the nums based on whether they’re divisible by m , and sum them up. class Solution { public int differenceOfSums ( int n, int m) { int num1 = 0 , num2 = 0 ; for ( int i = 1 ; i <= n; i++)...
Read more

Zero Array After Queries

Zero Array After Queries You are given an integer array nums and a list of queries, where each query is a pair [li, ri] representing a range. Each query allows you to decrement any subset of elements in the range [li, ri] by 1. Your task is to determine if it is possible to make every element in nums equal to zero after applying all queries in order . Example Input: nums = [ 1 , 0 , 1 ] queries = [[0, 2]] Output: true Explanation: We can select indices 0 and 2 and decrement both by 1 to obtain [0, 0, 0] . Best Data Structure for the Job We are working with multiple range updates and need to evaluate coverage over indices. The best choice here is the difference array combined with a prefix sum . It allows efficient range additions and helps track how many times each index is affected by the queries. Different Approaches Approach 1: Brute Force This naive method iterates over every index for every query and performs the decrement operations directly. class Solution {...
Read more