Is Circle and Rectangle Overlapping easy or hard?

The problem is rated Medium on LeetCode. The logic is short once you recognize the closest-point projection idea, but many candidates initially overcomplicate it by checking edges and corners separately.

Circle and Rectangle Overlapping Python/Java solution

Most implementations compute the closest point using clamp logic and then check squared distance. The same idea translates directly across Python, Java, C++, C#, JavaScript, and C because it uses only comparisons and integer arithmetic.

How to solve Circle and Rectangle Overlapping in O(1)?

Clamp the circle center coordinates to the rectangle bounds: compute closestX = max(x1, min(xCenter, x2)) and closestY = max(y1, min(yCenter, y2)). Then calculate the squared distance between (xCenter, yCenter) and (closestX, closestY). If the distance is less than or equal to radius squared, the circle overlaps the rectangle.

What is the best approach for Circle and Rectangle Overlapping?

The closest-point clamp method is the most reliable approach. Clamp the circle center coordinates to the rectangle boundaries to find the nearest point on the rectangle, then compare the squared distance to the circle radius squared. This runs in O(1) time and O(1) space and avoids handling separate edge and corner cases.

Is Circle and Rectangle Overlapping asked at Google/Amazon/Meta?

Geometry and coordinate problems like this appear in interviews at companies such as Google, Amazon, and Meta, especially in roles involving graphics, simulations, or spatial reasoning. The problem tests mathematical reasoning and careful boundary handling.

What data structure is used in Circle and Rectangle Overlapping?

No specialized data structure is required. The solution relies on coordinate arithmetic, boundary clamping, and squared distance calculations using basic math and geometry concepts.

What is the time complexity of Circle and Rectangle Overlapping?

The optimal solution runs in O(1) time and O(1) space. Only a few arithmetic operations and comparisons are required to compute the closest point and check the squared distance against the radius.

Description Hints 1 Solution Video 17:39

1401. Circle and Rectangle Overlapping

Medium67.4% acceptance2 companies

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. In other words, check if there is any point (x_i, y_i) that belongs to the circle and the rectangle at the same time.

Example 1:

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).

Example 2:

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

Example 3:

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

Constraints:

1 <= radius <= 2000
-10⁴ <= xCenter, yCenter <= 10⁴
-10⁴ <= x1 < x2 <= 10⁴
-10⁴ <= y1 < y2 <= 10⁴

Asked at 2 companies

Google Microsoft

Topics

Geometry Math

Hints open one at a time

Hint 1

Locate the closest point of the square to the circle, you can then find the distance from this point to the center of the circle and check if this is less than or equal to the radius.

17:39

Still stuck?Read 3 approaches, or watch a walkthrough.

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1class Solution:
2    def checkOverlap(self, radius: int, xCenter: int, yCenter: int, x1: int, y1: int, x2: int, y2: int) -> bool:
3        

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