Leetcode #1401: Circle and Rectangle Overlapping
In this guide, we solve Leetcode #1401 Circle and Rectangle Overlapping 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
You are given a circle represented as (radius, xCenter, yCenter) and 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 of the rectangle. Return true if the circle and rectangle are overlapped otherwise return false.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Geometry, Math
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: radius = 1, xCenter = 0, yCenter = 0, x1 = 1, y1 = -1, x2 = 3, y2 = 1
Output: true
Explanation: Circle and rectangle share the point (1,0).
Python Solution
class Solution:
def checkOverlap(
self,
radius: int,
xCenter: int,
yCenter: int,
x1: int,
y1: int,
x2: int,
y2: int,
) -> bool:
def f(i: int, j: int, k: int) -> int:
if i <= k <= j:
return 0
return i - k if k < i else k - j
a = f(x1, x2, xCenter)
b = f(y1, y2, yCenter)
return a * a + b * b <= radius * radius
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.