LeetCode-in-Java
1401. Circle and Rectangle Overlapping
Medium
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 <= 2000-104 <= xCenter, yCenter <= 104-104 <= x1 < x2 <= 104-104 <= y1 < y2 <= 104
Solution
public 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));
}
}