Problem
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 (xi, yi) 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 <= 2000-10^4 <= xCenter, yCenter <= 10^4-10^4 <= x1 < x2 <= 10^4-10^4 <= y1 < y2 <= 10^4
Solution (Java)
class Solution {
public boolean checkOverlap(
int radius, int xCenter, int yCenter, int x1, int y1, int x2, int y2) {
// Find the closest point to the circle within the rectangle
int closestX = clamp(xCenter, x1, x2);
int closestY = clamp(yCenter, y1, y2);
// Calculate the distance between the circle's center and this closest point
int distanceX = xCenter - closestX;
int distanceY = yCenter - closestY;
// If the distance is less than the circle's radius, an intersection occurs
int distanceSquared = distanceX * distanceX + distanceY * distanceY;
return distanceSquared <= radius * radius;
}
private int clamp(int val, int min, int max) {
return Math.max(min, Math.min(max, val));
}
}
Explain:
nope.
Complexity:
- Time complexity : O(n).
- Space complexity : O(n).