Circle and Rectangle Overlapping - Practice Coding Problems

Circle and Rectangle Overlapping - Problem

Math Geometry Hard

Imagine you're designing a collision detection system for a 2D game! You need to determine if a circular object (like a ball) and a rectangular object (like a platform) are overlapping.

You are given:

A circle represented as (radius, xCenter, yCenter)
An axis-aligned rectangle represented as (x1, y1, x2, y2), where (x1, y1) are the coordinates of the bottom-left corner, and (x2, y2) are the coordinates of the top-right corner

Goal: Return true if the circle and rectangle overlap (share at least one point), false otherwise.

Key insight: Two shapes overlap if there exists any point (xi, yi) that belongs to both the circle and the rectangle simultaneously.

Input & Output

example_1.py — Basic Overlap

$ Input: radius = 1, xCenter = 0, yCenter = 0, x1 = 1, y1 = -1, x2 = 3, y2 = 1

› Output: true

💡 Note: The circle centered at (0,0) with radius 1 overlaps with the rectangle. The closest point on rectangle to circle center is (1,0), and distance from (0,0) to (1,0) is 1, which equals the radius.

example_2.py — No Overlap

$ Input: radius = 1, xCenter = 1, yCenter = 1, x1 = 1, y1 = -3, x2 = 2, y2 = -1

› Output: false

💡 Note: The circle centered at (1,1) with radius 1 does not overlap with the rectangle. The closest point on rectangle to circle center is (1,-1), and distance from (1,1) to (1,-1) is 2, which is greater than the radius of 1.

example_3.py — Circle Inside Rectangle

$ Input: radius = 1, xCenter = 0, yCenter = 0, x1 = -1, y1 = -1, x2 = 1, y2 = 1

› Output: true

💡 Note: The circle is completely inside the rectangle. The closest point on rectangle to circle center is the center itself (0,0), so distance is 0, which is less than radius 1.

Visualization

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Understanding the Visualization

Setup Shapes

Position the circle and rectangle in coordinate space

Find Closest Point

Identify the point on rectangle closest to circle center

Measure Distance

Calculate distance from circle center to closest point

Compare with Radius

If distance ≤ radius, shapes overlap

Key Takeaway

🎯 Key Insight: Geometric optimization reduces an area-checking problem to a single point-distance calculation!

Time & Space Complexity

Time Complexity

⏱️

O(1)

Only performs constant number of min/max operations and one distance calculation

✓ Linear Growth

Space Complexity

O(1)

Uses only a few variables to store coordinates and distance

✓ Linear Space

Constraints

1 ≤ radius ≤ 2000
-1000 ≤ xCenter, yCenter ≤ 1000
-1000 ≤ x1 < x2 ≤ 1000
-1000 ≤ y1 < y2 ≤ 1000
All values are integers

Asked in

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

The optimal solution uses the closest point method in O(1) time. Find the closest point on the rectangle to the circle's center using coordinate clamping, then check if the distance is within the radius. This elegant geometric approach avoids checking every point and handles all edge cases perfectly.

Common Approaches

Approach	Time	Space	Notes
✓ Closest Point Method (Optimal)	O(1)	O(1)	Find the closest point on rectangle to circle center and check if it's within radius

Closest Point Method (Optimal) — Algorithm Steps

Find the closest point on rectangle to circle center by clamping coordinates
Clamp x-coordinate: closest_x = max(x1, min(xCenter, x2))
Clamp y-coordinate: closest_y = max(y1, min(yCenter, y2))
Calculate distance from circle center to this closest point
Return true if distance ≤ radius, false otherwise

Visualization

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Step-by-Step Walkthrough

Identify Circle Center

Start with the circle's center coordinates

Clamp to Rectangle

Find closest point on rectangle using min/max clamping

Distance Check

Calculate distance and compare with radius

Code -

solution.c — C

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

int min(int a, int b) {
    return a < b ? a : b;
}

int max(int a, int b) {
    return a > b ? a : b;
}

bool checkOverlap(int radius, int xCenter, int yCenter, int x1, int y1, int x2, int y2) {
    // Find closest point on rectangle to circle center
    int closestX = max(x1, min(xCenter, x2));
    int closestY = max(y1, min(yCenter, y2));
    
    // Calculate distance from circle center to closest point
    int dx = xCenter - closestX;
    int dy = yCenter - closestY;
    int distanceSquared = dx * dx + dy * dy;
    
    // Check if distance is within radius
    return distanceSquared <= radius * radius;
}

int main() {
    int radius, xCenter, yCenter, x1, y1, x2, y2;
    scanf("%d %d %d %d %d %d %d", &radius, &xCenter, &yCenter, &x1, &y1, &x2, &y2);
    printf("%s\n", checkOverlap(radius, xCenter, yCenter, x1, y1, x2, y2) ? "true" : "false");
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(1)

Only performs constant number of min/max operations and one distance calculation

✓ Linear Growth

Space Complexity

O(1)

Uses only a few variables to store coordinates and distance

✓ Linear Space

Constraints

1 ≤ radius ≤ 2000
-1000 ≤ xCenter, yCenter ≤ 1000
-1000 ≤ x1 < x2 ≤ 1000
-1000 ≤ y1 < y2 ≤ 1000
All values are integers

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Input & Output

Visualization

Time & Space Complexity

Related Problems

Constraints

Common Approaches

Closest Point Method (Optimal) — Algorithm Steps

Visualization

Code -

Time & Space Complexity

Constraints

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