What is the time complexity of Maximum Product of First and Last Elements of a Subsequence?

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

What data structures are used to solve Maximum Product of First and Last Elements of a Subsequence?

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

What are the key insights for solving Maximum Product of First and Last Elements of a Subsequence?

Insight 1: Directly generating all subsequences is inefficient due to the potentially large number of combinations. Insight 2: Precompute maximum and minimum prefix values to efficiently find candidate first elements of a subsequence.

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Maximum Product of First and Last Elements of a Subsequence - LeetCode 3584 Solution

Q: Which companies ask LeetCode 3584. Maximum Product of First and Last Elements of a Subsequence in interviews?

This problem is frequently asked at KLA.

Maximum Product of First and Last Elements of a Subsequence - Complete Solution Guide

Maximum Product of First and Last Elements of a Subsequence is LeetCode problem 3584, 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

You are given an integer array nums and an integer m . Return the maximum product of the first and last elements of any subsequence of nums of size m . Example 1: Input: nums = [-1,-9,2,3,-2,-3,1], m = 1 Output: 81 Explanation: The subsequence [-9] has the largest product of the first and last elements: -9 * -9 = 81 . Therefore, the answer is 81. Example 2: Input: nums = [1,3,-5,5,6,-4], m = 3 Output: 20 Explanation: The subsequence [-5, 6, -4] has the largest product of the first and last eleme

Detailed Explanation

The problem requires finding a subsequence of a given array `nums` with a specific length `m`. The goal is to maximize the product of the first and last elements of all possible subsequences of length `m`. The input includes the array `nums` and the integer `m`. The output is the maximum product found.

Solution Approach

The solution uses a dynamic programming approach to precompute the maximum and minimum prefix values of the array. It then iterates through the array considering each element as the potential last element of a subsequence of length `m`. For each potential last element, it uses the precomputed maximum and minimum prefix values to efficiently determine the possible values for the first element. It then calculates the product of the first and last elements and updates the maximum product found so far.

Step-by-Step Algorithm

Step 1: Handle the base case where m is 1. In this case, return the maximum square of any element in the array.
Step 2: Create two arrays, `max_prefix` and `min_prefix`, to store the maximum and minimum prefix values, respectively.
Step 3: Initialize the first elements of `max_prefix` and `min_prefix` with the first element of `nums`.
Step 4: Iterate through `nums` from the second element onwards, updating `max_prefix[i]` to be the maximum of `max_prefix[i-1]` and `nums[i]`, and updating `min_prefix[i]` to be the minimum of `min_prefix[i-1]` and `nums[i]`.
Step 5: Initialize a variable `max_prod` to negative infinity.
Step 6: Iterate through `nums` from index `m - 1` to the end of the array. For each index `j`, consider `nums[j]` as the last element of a subsequence of length `m`.
Step 7: Calculate the index `i_limit` representing the starting index of the subsequence, which is `j - m + 1`.
Step 8: Obtain the maximum and minimum possible first elements for this subsequence from `max_prefix[i_limit]` and `min_prefix[i_limit]`, respectively.
Step 9: Calculate the product of the last element with both the maximum and minimum possible first elements, and update `max_prod` with the largest value found so far.
Step 10: Return the final `max_prod`.

Key Insights

Insight 1: Directly generating all subsequences is inefficient due to the potentially large number of combinations.
Insight 2: Precompute maximum and minimum prefix values to efficiently find candidate first elements of a subsequence.
Insight 3: Consider both the maximum and minimum prefix values when selecting the first element, as the last element could be negative, making the product of two negatives a positive.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

This problem involves: Array, Two Pointers.

Companies

Asked at: KLA.

Maximum Product of First and Last Elements of a Subsequence | LeetCode 3584

Medium

Array Two Pointers

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

Solution Code

Available in:

class Solution {
    public long maximumProduct(int[] nums, int m) {
        int n = nums.length;

        if (m == 1) {
            long maxProduct = Integer.MIN_VALUE;
            for (int x : nums) {
                maxProduct = Math.max(maxProduct, (long)x * x);
            }
            return maxProduct;
        }

        int[] max_prefix = new int[n];
        int[] min_prefix = new int[n];
        
        max_prefix[0] = nums[0];
        min_prefix[0] = nums[0];
        
        for (int i = 1; i < n; i++) {
            max_prefix[i] = Math.max(max_prefix[i-1], nums[i]);
            min_prefix[i] = Math.min(min_prefix[i-1], nums[i]);
        }

        long max_prod = Integer.MIN_VALUE;

        for (int j = m - 1; j < n; j++) {
            int i_limit = j - m + 1;
            
            int last_elem = nums[j];
            
            int first_elem_candidate_max = max_prefix[i_limit];
            int first_elem_candidate_min = min_prefix[i_limit];
            
            max_prod = Math.max(max_prod, 
                           (long)last_elem * first_elem_candidate_max);
            max_prod = Math.max(max_prod, 
                           (long)last_elem * first_elem_candidate_min);

        }

        return (int)max_prod;
    }
}

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Problem Info

Description

Problem 3584: Maximum Product of First and Last Elements of a Subsequence - Detailed explanation and solution approaches.

Topics

Array Two Pointers

Companies

KLA

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:3584

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

10 steps

1
Step 1: Handle the base case where m is 1. In this case, return the maximum square of any element in the array.
2
Step 2: Create two arrays, `max_prefix` and `min_prefix`, to store the maximum and minimum prefix values, respectively.
3
Step 3: Initialize the first elements of `max_prefix` and `min_prefix` with the first element of `nums`.
4
Step 4: Iterate through `nums` from the second element onwards, updating `max_prefix[i]` to be the maximum of `max_prefix[i-1]` and `nums[i]`, and updating `min_prefix[i]` to be the minimum of `min_prefix[i-1]` and `nums[i]`.
5
Step 5: Initialize a variable `max_prod` to negative infinity.
6
Step 6: Iterate through `nums` from index `m - 1` to the end of the array. For each index `j`, consider `nums[j]` as the last element of a subsequence of length `m`.
7
Step 7: Calculate the index `i_limit` representing the starting index of the subsequence, which is `j - m + 1`.
8
Step 8: Obtain the maximum and minimum possible first elements for this subsequence from `max_prefix[i_limit]` and `min_prefix[i_limit]`, respectively.
9
Step 9: Calculate the product of the last element with both the maximum and minimum possible first elements, and update `max_prod` with the largest value found so far.
10
Step 10: Return the final `max_prod`.

💡

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(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The time complexity is derived from the two linear scans of the input array. The space complexity is derived from the two arrays used to store the prefix maximums and minimums, each of which has size `n`.

📖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

Generate all subsequences of length `m` and calculate the product of the first and last elements for each subsequence. This would involve generating combinations of indices to select the subsequence elements.

Time: O(nCm), where nCm represents the number of combinations of choosing m elements from n, which is approximately O(n^m) in the worst case.Space: O(m), to store the subsequence.

Trade-offs: The brute force approach is much slower and more space-intensive. It's only feasible for very small values of `n` and `m`. The provided solution is significantly more efficient for larger inputs.

✨

Key Insights

1Insight 1: Directly generating all subsequences is inefficient due to the potentially large number of combinations.
2Insight 2: Precompute maximum and minimum prefix values to efficiently find candidate first elements of a subsequence.
3Insight 3: Consider both the maximum and minimum prefix values when selecting the first element, as the last element could be negative, making the product of two negatives a positive.

📚

Key Terms Explained

For Students

Array

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

Two Pointers

Using two indices to traverse data from different positions simultaneously.

Edge Cases

Edge case 1: When `m` is 1, the subsequence consists of a single element, and the problem reduces to finding the maximum value of x*x for all elements x in `nums`. The code handles this specifically in the first if statement.
Edge case 2: When all elements in the array are negative. The code correctly handles this by tracking both the maximum and minimum prefix, as multiplying two negative numbers yields a positive number which can be larger than multiplying with the prefix maximum.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Considering only the maximum prefix value as the first element candidate. This can lead to incorrect results when the last element is negative, as the minimum prefix value might yield a larger product.

✗Common mistake 2: Not handling the case when `m` is 1 correctly. This is a special case that requires considering the square of each number.

✅

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: In financial analysis, this algorithm could be adapted to find the maximum product of the first and last prices within a sequence of stock prices over a specified period.

🏦Real-world scenario 2: In signal processing, this could be useful to find specific patterns in a signal where the maximum product of the beginning and end of a short window is important. This can assist in identifying anomalies.

💡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: November 12, 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

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 14

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 7

❓

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.

When is the two-pointer technique most effective?

Two pointers work best on sorted arrays or when searching for pairs/triplets that satisfy a condition. Common patterns include: moving pointers inward from both ends (like in two-sum variants), fast/slow pointers for cycle detection, or left/right pointers for partitioning.

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 of First and Last Elements of a Subsequence"?

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 two pointers techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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