Leetcode #478: Generate Random Point in a Circle
In this guide, we solve Leetcode #478 Generate Random Point in a Circle in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
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).
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Geometry, Math, Rejection Sampling, Randomized
Intuition
There is a mathematical invariant or formula that directly leads to the result.
Using math avoids unnecessary loops and reduces complexity.
Approach
Derive the formula or update rule, then compute the answer directly.
Handle edge cases like overflow or zero carefully.
Steps:
- Identify the math relationship.
- Compute the result with a loop or formula.
- Handle edge cases.
Example
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]
Python Solution
class Solution:
def __init__(self, radius: float, x_center: float, y_center: float):
self.radius = radius
self.x_center = x_center
self.y_center = y_center
def randPoint(self) -> List[float]:
length = math.sqrt(random.uniform(0, self.radius**2))
degree = random.uniform(0, 1) * 2 * math.pi
x = self.x_center + length * math.cos(degree)
y = self.y_center + length * math.sin(degree)
return [x, y]
Complexity
The time complexity is O(n) or O(1). The space complexity is O(1).
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.