Rectangle Overlap

Rectangle Overlap - Problem

Math Geometry Easy

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, return true if they overlap, otherwise return false.

Input & Output

Example 1 — Overlapping Rectangles

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

› Output: true

💡 Note: Rectangle 1 spans from (0,0) to (2,2) and Rectangle 2 spans from (1,1) to (3,3). They overlap in the region from (1,1) to (2,2), which has positive area.

Example 2 — Non-Overlapping Rectangles

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

› Output: false

💡 Note: Rectangle 1 is at (0,0)-(1,1) and Rectangle 2 is at (2,2)-(3,3). They are completely separate with no overlapping area.

Example 3 — Touching Edges

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

› Output: false

💡 Note: The rectangles share an edge but do not overlap. Touching at edges or corners doesn't count as overlapping since the intersection area is zero.

Constraints

rec1.length == 4
rec2.length == 4
-10⁹ ≤ rec1[i], rec2[i] ≤ 10⁹
rec1[0] ≤ rec1[2] and rec1[1] ≤ rec1[3] (valid rectangle)
rec2[0] ≤ rec2[2] and rec2[1] ≤ rec2[3] (valid rectangle)

Visualization

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Asked in

G Google 15 a Amazon 12 M Microsoft 8 f Facebook 6

The key insight is to check when rectangles do NOT overlap rather than when they do. Two rectangles don't overlap if one is completely to the left, right, above, or below the other. Best approach is the optimized non-overlap condition check. Time: O(1), Space: O(1)

Common Approaches

✓ Brute Force - Corner Check

⏱️ Time: O(1) Space: O(1)

For each rectangle, check if any of its four corners lies within the boundaries of the other rectangle. This approach directly implements the definition of overlap.

Optimized - Non-Overlap Conditions

⏱️ Time: O(1) Space: O(1)

Instead of checking overlap directly, we check when rectangles DON'T overlap and negate the result. Two rectangles don't overlap if one is completely to the left, right, above, or below the other.

Brute Force - Corner Check — Algorithm Steps

Generate all 4 corners for each rectangle
Check if any corner of rec1 is inside rec2
Check if any corner of rec2 is inside rec1

Visualization

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Step-by-Step Walkthrough

Generate Corners

Create all 4 corners for each rectangle

Check Containment

Test if any corner lies inside the other rectangle

Result

Return true if any corner is contained

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <stdbool.h>

bool solution(int* rec1, int* rec2) {
    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 non-overlap conditions
    // rec1 is completely to the left of rec2
    if (x2 <= x3) return false;
    // rec1 is completely to the right of rec2
    if (x1 >= x4) return false;
    // rec1 is completely below rec2
    if (y2 <= y3) return false;
    // rec1 is completely above rec2
    if (y1 >= y4) return false;
    
    // If none of the non-overlap conditions are true, they overlap
    return true;
}

int main() {
    char line[100];
    int rec1[4], rec2[4];
    
    fgets(line, sizeof(line), stdin);
    sscanf(line, "[%d,%d,%d,%d]", &rec1[0], &rec1[1], &rec1[2], &rec1[3]);
    
    fgets(line, sizeof(line), stdin);
    sscanf(line, "[%d,%d,%d,%d]", &rec2[0], &rec2[1], &rec2[2], &rec2[3]);
    
    bool result = solution(rec1, rec2);
    printf(result ? "true\n" : "false\n");
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(1)

Fixed number of corner checks regardless of input size

✓ Linear Growth

Space Complexity

O(1)

Only uses a few variables to store coordinates

✓ Linear Space

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Input & Output

Constraints

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Related Problems

Common Approaches

Brute Force - Corner Check — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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