LeetCode #478 — MEDIUM

Generate Random Point in a Circle

Move from brute-force thinking to an efficient approach using math strategy.

Solve on LeetCode
The Problem

Problem Statement

Given the radius and the position of the center of a circle, implement the function randPoint which generates a uniform random point inside the circle.

Implement the Solution class:

  • Solution(double radius, double x_center, double y_center) initializes the object with the radius of the circle radius and the position of the center (x_center, y_center).
  • randPoint() returns a random point inside the circle. A point on the circumference of the circle is considered to be in the circle. The answer is returned as an array [x, y].

Example 1:

Input
["Solution", "randPoint", "randPoint", "randPoint"]
[[1.0, 0.0, 0.0], [], [], []]
Output
[null, [-0.02493, -0.38077], [0.82314, 0.38945], [0.36572, 0.17248]]

Explanation
Solution solution = new Solution(1.0, 0.0, 0.0);
solution.randPoint(); // return [-0.02493, -0.38077]
solution.randPoint(); // return [0.82314, 0.38945]
solution.randPoint(); // return [0.36572, 0.17248]

Constraints:

  • 0 < radius <= 108
  • -107 <= x_center, y_center <= 107
  • At most 3 * 104 calls will be made to randPoint.

Roadmap

  1. Brute Force Baseline
  2. Core Insight
  3. Algorithm Walkthrough
  4. Edge Cases
  5. Full Annotated Code
  6. Interactive Study Demo
  7. Complexity Analysis
Step 01

Brute Force Baseline

Problem summary: Given the radius and the position of the center of a circle, implement the function randPoint which generates a uniform random point inside the circle. Implement the Solution class: Solution(double radius, double x_center, double y_center) initializes the object with the radius of the circle radius and the position of the center (x_center, y_center). randPoint() returns a random point inside the circle. A point on the circumference of the circle is considered to be in the circle. The answer is returned as an array [x, y].

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Math

Example 1

["Solution","randPoint","randPoint","randPoint"]
[[1.0,0.0,0.0],[],[],[]]

Related Problems

  • Random Point in Non-overlapping Rectangles (random-point-in-non-overlapping-rectangles)
Step 02

Core Insight

What unlocks the optimal approach

  • No official hints in dataset. Start from constraints and look for a monotonic or reusable state.
Interview move: turn each hint into an invariant you can check after every iteration/recursion step.
Step 03

Algorithm Walkthrough

Iteration Checklist

  1. Define state (indices, window, stack, map, DP cell, or recursion frame).
  2. Apply one transition step and update the invariant.
  3. Record answer candidate when condition is met.
  4. Continue until all input is consumed.
Use the first example testcase as your mental trace to verify each transition.
Step 04

Edge Cases

Minimum Input
Single element / shortest valid input
Validate boundary behavior before entering the main loop or recursion.
Duplicates & Repeats
Repeated values / repeated states
Decide whether duplicates should be merged, skipped, or counted explicitly.
Extreme Constraints
Upper-end input sizes
Re-check complexity target against constraints to avoid time-limit issues.
Invalid / Corner Shape
Empty collections, zeros, or disconnected structures
Handle special-case structure before the core algorithm path.
Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #478: Generate Random Point in a Circle
class Solution {
  public Solution(double radius, double x_center, double y_center) {
    this.radius = radius;
    this.x_center = x_center;
    this.y_center = y_center;
  }

  public double[] randPoint() {
    final double length = Math.sqrt(Math.random()) * radius;
    final double degree = Math.random() * 2 * Math.PI;
    final double x = x_center + length * Math.cos(degree);
    final double y = y_center + length * Math.sin(degree);
    return new double[] {x, y};
  }

  private double radius;
  private double x_center;
  private double y_center;
}
Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Press Step or Run All to begin.
Step 07

Complexity Analysis

Time
O(log n)
Space
O(1)

Approach Breakdown

ITERATIVE
O(n) time
O(1) space

Simulate the process step by step — multiply n times, check each number up to n, or iterate through all possibilities. Each step is O(1), but doing it n times gives O(n). No extra space needed since we just track running state.

MATH INSIGHT
O(log n) time
O(1) space

Math problems often have a closed-form or O(log n) solution hidden behind an O(n) simulation. Modular arithmetic, fast exponentiation (repeated squaring), GCD (Euclidean algorithm), and number theory properties can dramatically reduce complexity.

Shortcut: Look for mathematical properties that eliminate iteration. Repeated squaring → O(log n). Modular arithmetic avoids overflow.
Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.

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