Rectangle Overlap - Practice Coding Problems

Rectangle Overlap - Problem

Math Geometry Easy

Imagine you're designing a collision detection system for a 2D game! 🎮

You have two axis-aligned rectangles represented as lists [x1, y1, x2, y2], where:

(x1, y1) is the bottom-left corner
(x2, y2) is the top-right corner
All edges are parallel to the X and Y axes

Your task is to determine if these rectangles overlap - meaning they share some positive area in common. Two rectangles that only touch at corners or edges do not count as overlapping!

Goal: Return true if the rectangles overlap, false otherwise.

Example: Rectangle A: [0,0,2,2] and Rectangle B: [1,1,3,3] → They overlap in the area from (1,1) to (2,2) → Return true

Input & Output

example_1.py — Basic Overlap

$ Input: rec1 = [0,0,2,2], rec2 = [1,1,3,3]

› Output: true

💡 Note: The rectangles overlap in the area from (1,1) to (2,2), which has positive area.

example_2.py — No Overlap (Separated)

$ Input: rec1 = [0,0,1,1], rec2 = [2,2,3,3]

› Output: false

💡 Note: Rectangle 1 is completely separated from Rectangle 2 (both horizontally and vertically), so no overlap.

example_3.py — Edge Touch (No Overlap)

$ Input: rec1 = [0,0,1,1], rec2 = [1,0,2,1]

› Output: false

💡 Note: The rectangles only touch along the edge x=1, but don't share any positive area, so no overlap.

Constraints

Each rectangle rec[i] = [xi1, yi1, xi2, yi2] where (xi1, yi1) is the bottom-left corner and (xi2, yi2) is the top-right corner
-10⁹ ≤ xi1 < xi2 ≤ 10⁹
-10⁹ ≤ yi1 < yi2 ≤ 10⁹
Both rectangles are guaranteed to be valid (bottom-left coordinates are less than top-right coordinates)

Visualization

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Understanding the Visualization

Parse Coordinates

Extract x1,y1,x2,y2 for both rectangles

Check Left Separation

Is rect1 completely left of rect2? (x2 <= x3)

Check Right Separation

Is rect1 completely right of rect2? (x1 >= x4)

Check Vertical Separation

Is rect1 completely above/below rect2?

Determine Overlap

If no separation exists, rectangles must overlap!

Key Takeaway

🎯 Key Insight: Instead of checking all possible overlap scenarios, simply verify that the rectangles are NOT completely separated in any direction (left, right, above, below). If they're not separated, they must overlap!

Asked in

G Google 35 a Amazon 28 ⊞ Microsoft 22 f Meta 18

The optimal approach uses the key insight that two rectangles overlap if and only if they are NOT separated in any direction. Instead of checking all overlap cases, we simply verify that neither rectangle is completely to the left, right, above, or below the other. This reduces the problem to just 4 simple comparisons in O(1) time.

Common Approaches

Approach	Time	Space	Notes
✓ Brute Force (Boundary Checking)	O(1)	O(1)	Check all possible overlap conditions explicitly
Optimal Solution (Non-Separation Check)	O(1)	O(1)	Check if rectangles are NOT separated in both X and Y directions

Brute Force (Boundary Checking) — Algorithm Steps

Check if rectangle 1's corners are inside rectangle 2
Check if rectangle 2's corners are inside rectangle 1
Check if rectangles intersect along edges
Handle all possible intersection cases
Return true if any overlap condition is met

Visualization

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Step-by-Step Walkthrough

Check Corner Inside

Test if any corner of rectangle 1 falls inside rectangle 2

Reverse Check

Test if any corner of rectangle 2 falls inside rectangle 1

Cross Intersections

Check if rectangles intersect along edges without corners inside

Code -

solution.c — C

#include <stdbool.h>

bool rectangleOverlap(int* rec1, int rec1Size, int* rec2, int rec2Size) {
    int x1 = rec1[0], y1 = rec1[1], x2 = rec1[2], y2 = rec1[3];
    int x3 = rec2[0], y3 = rec2[1], x4 = rec2[2], y4 = rec2[3];
    
    // Check if rec1 corners are inside rec2
    if ((x3 < x1 && x1 < x4 && y3 < y1 && y1 < y4) || 
        (x3 < x2 && x2 < x4 && y3 < y2 && y2 < y4)) {
        return true;
    }
    if ((x3 < x1 && x1 < x4 && y3 < y2 && y2 < y4) || 
        (x3 < x2 && x2 < x4 && y3 < y1 && y1 < y4)) {
        return true;
    }
    
    // Check if rec2 corners are inside rec1
    if ((x1 < x3 && x3 < x2 && y1 < y3 && y3 < y2) || 
        (x1 < x4 && x4 < x2 && y1 < y4 && y4 < y2)) {
        return true;
    }
    if ((x1 < x3 && x3 < x2 && y1 < y4 && y4 < y2) || 
        (x1 < x4 && x4 < x2 && y1 < y3 && y3 < y2)) {
        return true;
    }
    
    // Check cross intersections
    if (((x1 < x3 && x3 < x2) || (x1 < x4 && x4 < x2)) && 
        ((y3 < y1 && y1 < y4) || (y3 < y2 && y2 < y4))) {
        return true;
    }
    if (((x3 < x1 && x1 < x4) || (x3 < x2 && x2 < x4)) && 
        ((y1 < y3 && y3 < y2) || (y1 < y4 && y4 < y2))) {
        return true;
    }
        
    return false;
}

Time & Space Complexity

Time Complexity

⏱️

O(1)

Constant time as we only check fixed number of conditions

✓ Linear Growth

Space Complexity

O(1)

No extra space needed, only variables for comparisons

✓ Linear Space

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Smart Actions

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force (Boundary Checking) — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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