Generate Random Point in a Circle - Practice Coding Problems

Generate Random Point in a Circle - Problem

Imagine you're creating a game where players need to click random locations inside a circular target. You need to ensure these points are uniformly distributed across the entire circle area - not clustered toward the center!

Given a circle defined by its radius and center coordinates (x_center, y_center), your task is to implement a system that generates random points inside this circle with perfect uniform distribution.

Challenge: Simply using random radius and angle won't work - this creates more points near the center because inner rings have less area than outer rings!

Your Task:

Implement Solution(radius, x_center, y_center) constructor
Implement randPoint() that returns [x, y] coordinates
Points on the circumference are considered inside the circle
Ensure uniform distribution across the entire circle area

Input & Output

example_1.py — Basic Circle

$ Input: Solution(1.0, 0.0, 0.0)\nrandPoint() called multiple times

› Output: Possible outputs: [-0.123, 0.456], [0.789, -0.234], [0.0, 0.999]

💡 Note: For a unit circle centered at origin, all returned points should satisfy x² + y² ≤ 1.0

example_2.py — Offset Center

$ Input: Solution(2.0, 3.0, 4.0)\nrandPoint() called multiple times

› Output: Possible outputs: [4.567, 5.123], [1.234, 2.789], [3.0, 6.0]

💡 Note: For radius 2 centered at (3,4), points should satisfy (x-3)² + (y-4)² ≤ 4

example_3.py — Edge Case Small Radius

$ Input: Solution(0.01, 10.0, 10.0)\nrandPoint() called multiple times

› Output: Possible outputs: [10.005, 9.997], [9.999, 10.008], [10.001, 9.995]

💡 Note: Even with very small radius, points should be uniformly distributed within the tiny circle

Visualization

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Understanding the Visualization

The Problem

Simple random radius creates more points near center (smaller rings have less area)

The Solution

Square root transformation: r = R√(random) compensates for area differences

Perfect Distribution

Every equal-sized region now has equal probability of containing a point

Key Takeaway

🎯 Key Insight: Area scales with r², so radius must scale with √(random) for uniform distribution!

Time & Space Complexity

Time Complexity

⏱️

O(1)

Fixed number of operations: 2 random numbers, 1 sqrt, 1 cos, 1 sin

✓ Linear Growth

Space Complexity

O(1)

Only stores the circle parameters and temporary variables

✓ Linear Space

Constraints

0 < radius ≤ 10⁸
-10⁷ ≤ x_center, y_center ≤ 10⁷
At most 3 × 10⁴ calls will be made to randPoint
Points on the circumference are considered inside the circle

Asked in

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

The optimal solution uses polar coordinates with square root transformation to achieve uniform distribution. The key insight is that area grows quadratically with radius, so using r = radius × √(random) compensates for this and ensures equal probability for equal areas. This gives O(1) guaranteed performance compared to rejection sampling's variable runtime.

Common Approaches

Approach	Time	Space	Notes
✓ Polar Coordinates with Square Root (Optimal)	O(1)	O(1)	Use polar coordinates with √(random) for radius to ensure uniform area distribution
Rejection Sampling (Square Method)	O(1) average	O(1)	Generate random points in bounding square, reject those outside circle

Polar Coordinates with Square Root (Optimal) — Algorithm Steps

Generate random angle θ uniformly in [0, 2π)
Generate random radius using r = radius × √(random) for uniform area distribution
Convert to Cartesian coordinates: x = r × cos(θ), y = r × sin(θ)
Translate to actual center: (x_center + x, y_center + y)

Visualization

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Step-by-Step Walkthrough

Random Angle

Generate uniform random angle θ ∈ [0, 2π)

Square Root Radius

Use r = R√(random) for uniform area distribution

Convert to Cartesian

Apply x = r⋅cos(θ), y = r⋅sin(θ)

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <time.h>

#ifndef M_PI
#define M_PI 3.14159265358979323846
#endif

typedef struct {
    double radius;
    double x_center;
    double y_center;
} Solution;

Solution* solutionCreate(double radius, double x_center, double y_center) {
    Solution* obj = (Solution*)malloc(sizeof(Solution));
    obj->radius = radius;
    obj->x_center = x_center;
    obj->y_center = y_center;
    srand(time(NULL));
    return obj;
}

double* solutionRandPoint(Solution* obj, int* returnSize) {
    *returnSize = 2;
    double* result = (double*)malloc(2 * sizeof(double));
    
    double angle = ((double)rand() / RAND_MAX) * 2 * M_PI;
    double r = obj->radius * sqrt((double)rand() / RAND_MAX);
    
    double x = r * cos(angle);
    double y = r * sin(angle);
    
    result[0] = obj->x_center + x;
    result[1] = obj->y_center + y;
    return result;
}

Time & Space Complexity

Time Complexity

⏱️

O(1)

Fixed number of operations: 2 random numbers, 1 sqrt, 1 cos, 1 sin

✓ Linear Growth

Space Complexity

O(1)

Only stores the circle parameters and temporary variables

✓ Linear Space

Constraints

0 < radius ≤ 10⁸
-10⁷ ≤ x_center, y_center ≤ 10⁷
At most 3 × 10⁴ calls will be made to randPoint
Points on the circumference are considered inside the circle

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Input & Output

Visualization

Time & Space Complexity

Related Problems

Constraints

Common Approaches

Polar Coordinates with Square Root (Optimal) — Algorithm Steps

Visualization

Code -

Time & Space Complexity

Constraints

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