What is the time complexity of Rectangle Overlap?

The optimal solution has a time complexity of O(1).

What data structures are used to solve Rectangle Overlap?

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

What are the key insights for solving Rectangle Overlap?

Insight 1: The key is to identify the conditions under which the rectangles do *not* overlap. If we can find any of these conditions to be true, we can conclude that the rectangles don't overlap. The opposite of these non-overlapping conditions would then indicate overlap. Insight 2: The problem can be solved with simple comparisons between the coordinates of the rectangles. No complex data structures or algorithms are needed.

Which companies ask LeetCode 836. Rectangle Overlap in interviews?

This problem is frequently asked at Nvidia, Qualcomm.

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Rectangle Overlap - LeetCode 836 Solution

Rectangle Overlap - Complete Solution Guide

Rectangle Overlap is LeetCode problem 836, 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

An axis-aligned rectangle is represented as a list [x1, y1, x2, y2] , where (x1, y1) is the coordinate of its bottom-left corner, and (x2, y2) is the coordinate of its top-right corner. Its top and bottom edges are parallel to the X-axis, and its left and right edges are parallel to the Y-axis. Two rectangles overlap if the area of their intersection is positive . To be clear, two rectangles that only touch at the corner or edges do not overlap. Given two axis-aligned rectangles rec1 and rec2 ,

Detailed Explanation

The problem asks us to determine if two axis-aligned rectangles overlap. Each rectangle is defined by the coordinates of its bottom-left and top-right corners: `[x1, y1, x2, y2]`. Overlap is defined strictly; touching at edges or corners does *not* count as overlapping. The input consists of two such rectangle representations, `rec1` and `rec2`, and the output should be `true` if they overlap, and `false` otherwise.

Solution Approach

The solution approach is to check for the conditions under which the two rectangles *do not* overlap. Specifically, rectangle 1 is to the left of rectangle 2, rectangle 1 is to the right of rectangle 2, rectangle 1 is above rectangle 2, or rectangle 1 is below rectangle 2. If none of these conditions are met, then the rectangles must overlap.

Step-by-Step Algorithm

Step 1: Extract the coordinates of the first rectangle (rec1) as x1, y1, x2, y2.
Step 2: Extract the coordinates of the second rectangle (rec2) as x3, y3, x4, y4.
Step 3: Check if rec1 is to the left of rec2: x2 <= x3. If true, they don't overlap.
Step 4: Check if rec1 is to the right of rec2: x1 >= x4. If true, they don't overlap.
Step 5: Check if rec1 is below rec2: y2 <= y3. If true, they don't overlap.
Step 6: Check if rec1 is above rec2: y1 >= y4. If true, they don't overlap.
Step 7: If none of the above conditions are true, return true (they overlap). Otherwise, return false (they don't overlap).

Key Insights

Insight 1: The key is to identify the conditions under which the rectangles do *not* overlap. If we can find any of these conditions to be true, we can conclude that the rectangles don't overlap. The opposite of these non-overlapping conditions would then indicate overlap.
Insight 2: The problem can be solved with simple comparisons between the coordinates of the rectangles. No complex data structures or algorithms are needed.
Insight 3: Ensure that x1 < x2 and y1 < y2 for both rectangles, representing valid rectangles with positive area. The problem states that the given inputs represent a valid rectangle, but we should conceptually keep this in mind.

Complexity Analysis

Time Complexity: O(1)

Space Complexity: O(1)

Topics

This problem involves: Math, Geometry.

Companies

Asked at: Nvidia, Qualcomm.

Rectangle Overlap | LeetCode 836

Easy

Math Geometry

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

Solution Code

Available in:

class Solution {
    public boolean isRectangleOverlap(int[] rec1, int[] rec2) {
        return !(rec1[2] <= rec2[0] || rec1[0] >= rec2[2] || rec1[3] <= rec2[1] || rec1[1] >= rec2[3]);
    }
}

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Problem Info

Description

Problem 836: Rectangle Overlap - Detailed explanation and solution approaches.

Topics

Math Geometry

Companies

NvidiaQualcomm

Stats

Difficulty:Easy

Languages:4

Companies:2

Problem #:836

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: Extract the coordinates of the first rectangle (rec1) as x1, y1, x2, y2.
2
Step 2: Extract the coordinates of the second rectangle (rec2) as x3, y3, x4, y4.
3
Step 3: Check if rec1 is to the left of rec2: x2 <= x3. If true, they don't overlap.
4
Step 4: Check if rec1 is to the right of rec2: x1 >= x4. If true, they don't overlap.
5
Step 5: Check if rec1 is below rec2: y2 <= y3. If true, they don't overlap.
6
Step 6: Check if rec1 is above rec2: y1 >= y4. If true, they don't overlap.
7
Step 7: If none of the above conditions are true, return true (they overlap). Otherwise, return false (they don't overlap).

💡

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(1)

⚡ Constant time - Super fast! Same speed regardless of input size.

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The solution involves a fixed number of comparisons regardless of the input values. Therefore, both time and space complexities are constant.

📖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

Area of Intersection

Calculate the area of intersection between the two rectangles. If the area is positive, they overlap. This requires finding the coordinates of the intersection rectangle (if it exists) and calculating its area.

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

Trade-offs: This approach involves more calculations (finding min/max of coordinates and then area) than the main solution, but is also O(1). While logically clear, it's less efficient due to the extra calculations. It's often used for more complex geometry problems where simply checking non-overlap conditions isn't enough.

✨

Key Insights

1Insight 1: The key is to identify the conditions under which the rectangles do *not* overlap. If we can find any of these conditions to be true, we can conclude that the rectangles don't overlap. The opposite of these non-overlapping conditions would then indicate overlap.
2Insight 2: The problem can be solved with simple comparisons between the coordinates of the rectangles. No complex data structures or algorithms are needed.
3Insight 3: Ensure that x1 < x2 and y1 < y2 for both rectangles, representing valid rectangles with positive area. The problem states that the given inputs represent a valid rectangle, but we should conceptually keep this in mind.

📚

Key Terms Explained

For Students

Edge Cases

Edge case 1: Rectangles touch only at a corner. The solution correctly identifies this as not an overlap because the conditions `x2 <= x3`, `x1 >= x4`, `y2 <= y3`, or `y1 >= y4` will be met. For example, rec1 = [0, 0, 1, 1], rec2 = [1, 1, 2, 2]
Edge case 2: Rectangles touch only along an edge. Similarly, this will also be correctly identified as not an overlap. For example, rec1 = [0, 0, 1, 1], rec2 = [1, 0, 2, 1]
Edge case 3: One rectangle is completely contained within the other. The provided logic handles this case correctly. For example, rec1 = [0, 0, 4, 4], rec2 = [1, 1, 2, 2]. In this case, none of the non-overlapping conditions will be true.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrectly comparing the coordinates, for instance, confusing x1 and x2 or y1 and y2. Always ensure the correct coordinates are used in comparisons.

✗Common mistake 2: Neglecting to check all four non-overlap conditions. Missing one condition can lead to incorrect results.

✅

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 games and simulations. Determining if two objects (represented by rectangles) are colliding.

🏦Real-world scenario 2: UI design: Checking if a dragged UI element overlaps with another element to trigger an event (e.g., docking or snapping).

💡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

Don't rush. Explain your thought process clearly. Easy problems are often about communication.

⚠️ Watch Out

Always test your solution with edge cases: empty input, single element, and boundary values.

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 25, 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

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

@AlgoMasterJul 13

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

@InterviewProJul 6

❓

Frequently Asked Questions

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 "Rectangle Overlap"?

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

🤖 Need help with this problem?