What is the time complexity of Find the Largest Area of Square Inside Two Rectangles?

The optimal solution has a time complexity of O(n^2).

What data structures are used to solve Find the Largest Area of Square Inside Two Rectangles?

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

What are the key insights for solving Find the Largest Area of Square Inside Two Rectangles?

Insight 1: The core idea is to iterate through all pairs of rectangles to check for intersections. Insight 2: To find the intersection of two rectangles, we need to find the overlapping region in both x and y coordinates.

Which companies ask LeetCode 3047. Find the Largest Area of Square Inside Two Rectangles in interviews?

This problem is frequently asked at Cisco.

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Find the Largest Area of Square Inside Two Rectangles - LeetCode 3047 Solution

Find the Largest Area of Square Inside Two Rectangles - Complete Solution Guide

Find the Largest Area of Square Inside Two Rectangles is LeetCode problem 3047, 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

There exist n rectangles in a 2D plane with edges parallel to the x and y axis. You are given two 2D integer arrays bottomLeft and topRight where bottomLeft[i] = [a_i, b_i] and topRight[i] = [c_i, d_i] represent the bottom-left and top-right coordinates of the i th rectangle, respectively. You need to find the maximum area of a square that can fit inside the intersecting region of at least two rectangles. Return 0 if such a square does not exist. Example 1: Input: bottomLeft = [[1,1],[2,2],[3,1]

Detailed Explanation

The problem asks us to find the largest possible square that can fit inside the intersection of at least two rectangles, given a list of rectangles defined by their bottom-left and top-right coordinates. We need to iterate through all possible pairs of rectangles, find the intersection region (if any), and determine the maximum side length of a square that can fit within that intersection. The final result is the area of the largest such square. If no such square exists (i.e., no intersections exist), we should return 0.

Solution Approach

The solution iterates through all possible pairs of rectangles. For each pair, it calculates the width and height of their intersection. If the intersection exists (both width and height are positive), it determines the maximum side length of a square that can fit within the intersection (which is the minimum of the intersection's width and height). The solution maintains a running maximum of the side length squared (area) across all pairs of rectangles and returns it.

Step-by-Step Algorithm

Step 1: Initialize `max_side` to 0. This variable will store the maximum side length found so far.
Step 2: Iterate through all pairs of rectangles using nested loops. The outer loop iterates from `i = 0` to `n-1`, and the inner loop iterates from `j = i+1` to `n-1`, where `n` is the number of rectangles.
Step 3: For each pair of rectangles (i, j), calculate the x-coordinates of the intersection: `intersect_width = min(x2_i, x2_j) - max(x1_i, x1_j)`. If `intersect_width` is negative, the rectangles do not intersect horizontally.
Step 4: Calculate the y-coordinates of the intersection: `intersect_height = min(y2_i, y2_j) - max(y1_i, y1_j)`. If `intersect_height` is negative, the rectangles do not intersect vertically.
Step 5: If both `intersect_width` and `intersect_height` are positive, the rectangles intersect. Calculate the maximum possible side length of a square that can fit in the intersection: `side = min(intersect_width, intersect_height)`.
Step 6: Update `max_side` with the maximum of the current `max_side` and the calculated `side`.
Step 7: After iterating through all pairs of rectangles, return the square of `max_side`, which represents the largest area of a square that can fit inside the intersection of at least two rectangles.

Key Insights

Insight 1: The core idea is to iterate through all pairs of rectangles to check for intersections.
Insight 2: To find the intersection of two rectangles, we need to find the overlapping region in both x and y coordinates.
Insight 3: The maximum side length of a square that can fit in the intersection is the minimum of the intersection's width and height.

Complexity Analysis

Time Complexity: O(n^2)

Space Complexity: O(1)

Topics

This problem involves: Array, Math, Geometry.

Companies

Asked at: Cisco.

Find the Largest Area of Square Inside Two Rectangles | LeetCode 3047

Medium

Array Math Geometry

Time: O(n^2)Space: O(1)

Solution Code

Available in:

class Solution {
    public long largestSquareArea(int[][] bottomLeft, int[][] topRight) {
        int n = bottomLeft.length;
        long max_side = 0;

        for (int i = 0; i < n; i++) {
            int x1_i = bottomLeft[i][0];
            int y1_i = bottomLeft[i][1];
            int x2_i = topRight[i][0];
            int y2_i = topRight[i][1];
            
            for (int j = i + 1; j < n; j++) {
                int x1_j = bottomLeft[j][0];
                int y1_j = bottomLeft[j][1];
                int x2_j = topRight[j][0];
                int y2_j = topRight[j][1];

                int intersect_width = Math.min(x2_i, x2_j) - Math.max(x1_i, x1_j);
                int intersect_height = Math.min(y2_i, y2_j) - Math.max(y1_i, y1_j);
                
                int side = Math.min(intersect_width, intersect_height);
                
                if (side > 0) {
                    max_side = Math.max(max_side, side);
                }
        
            }
        }
        
        return max_side * max_side;
    }
}

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Problem Info

Description

Problem 3047: Find the Largest Area of Square Inside Two Rectangles - Detailed explanation and solution approaches.

Topics

Array Math Geometry

Companies

Cisco

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:3047

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 `max_side` to 0. This variable will store the maximum side length found so far.
2
Step 2: Iterate through all pairs of rectangles using nested loops. The outer loop iterates from `i = 0` to `n-1`, and the inner loop iterates from `j = i+1` to `n-1`, where `n` is the number of rectangles.
3
Step 3: For each pair of rectangles (i, j), calculate the x-coordinates of the intersection: `intersect_width = min(x2_i, x2_j) - max(x1_i, x1_j)`. If `intersect_width` is negative, the rectangles do not intersect horizontally.
4
Step 4: Calculate the y-coordinates of the intersection: `intersect_height = min(y2_i, y2_j) - max(y1_i, y1_j)`. If `intersect_height` is negative, the rectangles do not intersect vertically.
5
Step 5: If both `intersect_width` and `intersect_height` are positive, the rectangles intersect. Calculate the maximum possible side length of a square that can fit in the intersection: `side = min(intersect_width, intersect_height)`.
6
Step 6: Update `max_side` with the maximum of the current `max_side` and the calculated `side`.
7
Step 7: After iterating through all pairs of rectangles, return the square of `max_side`, which represents the largest area of a square that can fit inside the intersection of at least two rectangles.

💡

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^2)

🐢 Quadratic - Gets slow with large inputs. Consider optimization!

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The time complexity is O(n^2) because we iterate through all possible pairs of rectangles using nested loops. The outer loop runs `n` times, and the inner loop runs approximately `n/2` times on average, resulting in O(n^2) comparisons. The space complexity is O(1) because we only use a few constant extra variables to store the maximum side length and the intersection coordinates, regardless of the input size.

📖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

Sweep Line Algorithm

A sweep line algorithm can be used, but it's more complex to implement for this specific problem. You'd sweep along the x-axis, keeping track of active rectangles and checking for intersections as you go. A similar approach could be employed for the y-axis.

Time: O(n^2 log n) or O(n^2) depending on the implementation details for intersection detection and sorting.Space: O(n)

Trade-offs: This approach could potentially improve time complexity if an efficient data structure to store active rectangles is used, but requires significantly more complex implementation. The naive version is O(n^2) like the current approach.

✨

Key Insights

1Insight 1: The core idea is to iterate through all pairs of rectangles to check for intersections.
2Insight 2: To find the intersection of two rectangles, we need to find the overlapping region in both x and y coordinates.
3Insight 3: The maximum side length of a square that can fit in the intersection is the minimum of the intersection's width and height.

📚

Key Terms Explained

For Students

Array

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

Edge Cases

Edge case 1: No intersection between any two rectangles. In this case, the `intersect_width` or `intersect_height` will be negative for all rectangle pairs, so the `if (side > 0)` condition will always be false. Consequently, `max_side` will remain 0, and the function will correctly return 0.
Edge case 2: One rectangle is completely contained within another. In this case, the intersection calculation will correctly identify the smaller rectangle as the intersection area. The `side` will be positive, and the algorithm will find the correct maximum square area.
Edge case 3: Rectangles that only intersect at an edge or a corner. In these cases, either `intersect_width` or `intersect_height` would be 0, resulting in `side = 0`, which is handled correctly.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrectly calculating the intersection coordinates. A common mistake is to mix up the min and max functions when calculating the intersection width and height, leading to incorrect intersection areas and wrong results.

✗Common mistake 2: Not considering all pairs of rectangles. Forgetting to iterate through all pairs of rectangles will lead to a suboptimal or incorrect result.

✗Common mistake 3: Integer overflow. In languages like C/C++/Java, the `side * side` calculation can overflow if the coordinates are large enough. Using `long long` or `long` as appropriate avoids this issue, as it is in the examples.

✗Common mistake 4: Not handling the case where no intersection exists, resulting in potentially returning an uninitialized value or a garbage value.

✅

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: Collision detection in video games or simulations. Determining if two objects (represented by rectangles) are colliding and finding the maximum square area of their overlap is a simplified version of real-world collision detection.

🏦Real-world scenario 2: Resource allocation and scheduling. If rectangles represent time intervals during which resources are allocated, finding the intersection represents finding time slots where two resources are being used simultaneously. Maximizing the square area in that intersection might represent maximizing the usage during that overlap.

🎮Real-world scenario 3: Map overlay analysis: Finding the maximum area where two map features overlap. For example, finding areas where two different soil types are present or where a vegetation region overlaps a specific land use type.

💡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 15, 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

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

@TechLeadJun 29

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Find the Largest Area of Square Inside Two Rectangles"?

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

🤖 Need help with this problem?