What is the time complexity of Maximum Area of Longest Diagonal Rectangle?

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

What data structures are used to solve Maximum Area of Longest Diagonal Rectangle?

This problem uses Array. Understanding these concepts is essential for solving this Easy problem efficiently.

What are the key insights for solving Maximum Area of Longest Diagonal Rectangle?

The problem requires calculating the diagonal length of each rectangle using the Pythagorean theorem (diagonal = sqrt(length^2 + width^2)). A straightforward iterative approach is sufficient to find the maximum diagonal and corresponding area. No sophisticated data structures are needed.

Which companies ask LeetCode 3000. Maximum Area of Longest Diagonal Rectangle in interviews?

This problem is frequently asked at Accenture, Atlassian.

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Maximum Area of Longest Diagonal Rectangle - LeetCode 3000 Solution

Q: How to solve Maximum Area of Longest Diagonal Rectangle?

The solution iterates through each rectangle in the `dimensions` array. For each rectangle, it calculates the diagonal length and the area. It keeps track of the maximum diagonal length encountered so far and the area of the rectangle with that diagonal. If a rectangle has a longer diagonal, it updates the maximum diagonal and its corresponding area. If a rectangle has the same maximum diagonal, it compares its area with the current maximum area and updates the maximum area if necessary.

Maximum Area of Longest Diagonal Rectangle - Complete Solution Guide

Maximum Area of Longest Diagonal Rectangle is LeetCode problem 3000, a Easy 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 a 2D 0-indexed integer array dimensions . For all indices i , 0 <= i < dimensions.length , dimensions[i][0] represents the length and dimensions[i][1] represents the width of the rectangle i . Return the area of the rectangle having the longest diagonal. If there are multiple rectangles with the longest diagonal, return the area of the rectangle having the maximum area. Example 1: Input: dimensions = [[9,3],[8,6]] Output: 48 Explanation: For index = 0, length = 9 and width = 3. Dia

Detailed Explanation

The problem asks you to find the area of the rectangle with the longest diagonal from a given list of rectangles. Each rectangle is represented by its length and width in a 2D array. If multiple rectangles have the same longest diagonal, the one with the largest area is selected. The input is a 2D integer array `dimensions` where each inner array `[length, width]` represents a rectangle. The output is a single integer representing the area of the selected rectangle.

Solution Approach

The solution iterates through each rectangle in the `dimensions` array. For each rectangle, it calculates the diagonal length and the area. It keeps track of the maximum diagonal length encountered so far and the area of the rectangle with that diagonal. If a rectangle has a longer diagonal, it updates the maximum diagonal and its corresponding area. If a rectangle has the same maximum diagonal, it compares its area with the current maximum area and updates the maximum area if necessary.

Step-by-Step Algorithm

Initialize `maxDiagonal` to 0 and `maxArea` to 0.
Iterate through each rectangle in the `dimensions` array.
Calculate the diagonal length using the Pythagorean theorem: `diagonal = sqrt(length^2 + width^2)`.
Calculate the area: `area = length * width`.
If `diagonal > maxDiagonal`, update `maxDiagonal` to `diagonal` and `maxArea` to `area`.
If `diagonal == maxDiagonal`, update `maxArea` to `max(maxArea, area)`.
After iterating through all rectangles, return `maxArea`.

Key Insights

The problem requires calculating the diagonal length of each rectangle using the Pythagorean theorem (diagonal = sqrt(length^2 + width^2)).
A straightforward iterative approach is sufficient to find the maximum diagonal and corresponding area. No sophisticated data structures are needed.
Special handling is required to manage cases where multiple rectangles share the same maximum diagonal length; in this case, we need to choose the one with the maximum area.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: Array.

Companies

Asked at: Accenture, Atlassian.

Maximum Area of Longest Diagonal Rectangle | LeetCode 3000

Easy

Array

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

Solution Code

Available in:

class Solution {
    public int areaOfMaxDiagonal(int[][] dimensions) {
        double maxDiagonal = 0;
        int maxArea = 0;
        for (int[] dimension : dimensions) {
            int length = dimension[0];
            int width = dimension[1];
            double diagonal = Math.sqrt(length * length + width * width);
            int area = length * width;
            if (diagonal > maxDiagonal) {
                maxDiagonal = diagonal;
                maxArea = area;
            } else if (diagonal == maxDiagonal) {
                maxArea = Math.max(maxArea, area);
            }
        }
        return maxArea;
    }
}

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Problem Info

Description

Problem 3000: Maximum Area of Longest Diagonal Rectangle - Detailed explanation and solution approaches.

Topics

Array

Companies

AccentureAtlassian

Stats

Difficulty:Easy

Languages:4

Companies:2

Problem #:3000

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
Initialize `maxDiagonal` to 0 and `maxArea` to 0.
2
Iterate through each rectangle in the `dimensions` array.
3
Calculate the diagonal length using the Pythagorean theorem: `diagonal = sqrt(length^2 + width^2)`.
4
Calculate the area: `area = length * width`.
5
If `diagonal > maxDiagonal`, update `maxDiagonal` to `diagonal` and `maxArea` to `area`.
6
If `diagonal == maxDiagonal`, update `maxArea` to `max(maxArea, area)`.
7
After iterating through all rectangles, return `maxArea`.

💡

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 iterates through the input array once, performing constant-time operations for each element. No additional data structures proportional to the input size are used. Therefore, the time complexity is linear (O(n)) and the space complexity is constant (O(1)).

📖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

Using a custom data structure

Create a class or struct to store both diagonal length and area together, and then sort the array of these objects based on diagonal length (descending), then area (descending). The first element would contain the answer.

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

Trade-offs: This approach is less efficient in terms of time complexity than the linear approach but might be more readable for some. The increased space complexity is due to the creation of a new array of objects.

✨

Key Insights

1The problem requires calculating the diagonal length of each rectangle using the Pythagorean theorem (diagonal = sqrt(length^2 + width^2)).
2A straightforward iterative approach is sufficient to find the maximum diagonal and corresponding area. No sophisticated data structures are needed.
3Special handling is required to manage cases where multiple rectangles share the same maximum diagonal length; in this case, we need to choose the one with the maximum area.

📚

Key Terms Explained

For Students

Array

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

Edge Cases

Empty input array: The solution correctly handles an empty input array by returning 0.
Rectangles with equal diagonals: The solution correctly identifies the rectangle with the largest area among those with the longest diagonal.

⚠️

Common Mistakes to Avoid

✗Forgetting to handle the case where multiple rectangles have the same maximum diagonal length, leading to incorrect area selection.

✗Incorrectly implementing the Pythagorean theorem or calculating the diagonal length, resulting in inaccurate results.

✗Using integer division instead of floating-point division when calculating the diagonal which could lead to inaccurate results in the comparison of diagonal lengths.

✅

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?

💼Optimizing warehouse layout: Finding the largest rectangular area with the longest diagonal can be useful in determining the optimal placement of storage units to minimize travel time.

🏦Choosing building dimensions: In construction, selecting building dimensions (length and width) might involve maximizing the diagonal length while maintaining the greatest floor space, which would map directly to this problem.

💡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

Even for easy problems, discuss time and space complexity. Interviewers want to see you think analytically.

⚠️ Watch Out

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

👨‍💻 From Experience

This is one of the first problems I solved when learning Array. It's a great foundation for more complex problems.

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: September 29, 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

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

@DevExpert92Jul 13

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 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 should I approach easy-level algorithm problems?

Easy problems build foundational skills. Focus on: 1) Understanding the problem completely before coding, 2) Writing clean, readable code, 3) Handling all edge cases (empty input, single element, etc.), 4) Analyzing time and space complexity. Don't skip complexity analysis even for simple solutions.

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 Area of Longest Diagonal Rectangle"?

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

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