Meta Data#
Difficulty: medium First Attempt: 2026-01-05
- Submissions: Wrong Answer ×3 → Accepted
Approach#
The circle and the rectangle intersect ⇔ the rectangle’s nearest point to the center lies in the circle.
That point is easy: x and y are independent, so clamp the center’s x into [x1, x2] and y into [y1, y2]:
nx = min(max(xc, x1), x2)
ny = min(max(yc, y1), y2)
With the center inside the rectangle, the nearest point is the center itself at distance 0, which also counts as overlapping.
Compare squared distances to avoid floating-point square roots.
My Attempts#
My version split into four directions (the center left of, below, right of, or above the rectangle) and found the nearest point in each. Two of the four ifs were wrong:
| Version | Problem |
|---|---|
| Attempt 1 | Missed the center-inside case; the last direction (above) was yCenter < y2 instead of yCenter > y2. With radius 2, center (6, 4) and rectangle (5, 0)–(6, 2), the top edge point (6, 2) is exactly 2 away, so the answer is True; it returned False |
| Attempt 2 | Added center-inside, but the “above” condition was still wrong; the same example returned False |
| Attempt 3 | Changed it to <=, still the wrong direction. With radius 3, center (3, 4) and rectangle (2, 2)–(5, 3), the answer is True; it returned False |
| AC | Changed it to yCenter >= y2 |
Clamping needs no cases, so there is no direction to get backwards.
Solution#
class Solution:
def checkOverlap(self, radius: int, xCenter: int, yCenter: int, x1: int, y1: int, x2: int, y2: int) -> bool:
nx = min(max(xCenter, x1), x2) # the rectangle's nearest point to the center
ny = min(max(yCenter, y1), y2)
return (nx - xCenter) ** 2 + (ny - yCenter) ** 2 <= radius ** 2
Complexity#
Time O(1), space O(1).
Mistakes to Avoid#
- Writing a comparison backwards in a case split.
- Missing the center-inside case.
- Comparing floating-point square roots: squared comparisons are more robust.
Interview Explanation#
I find the rectangle’s nearest point to the center by clamping each coordinate of the center into the rectangle; they overlap when its squared distance is at most the squared radius. With the center inside, the nearest point is the center itself, which works naturally. Time O(1).