Return True if the circle and rectangle are overlapped otherwise return False.

In other words, check if there are any point (xi, yi) such that belongs to the circle and the rectangle at the same time.

Example 1:

Input: radius = 1, x_center = 0, y_center = 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, x_center = 0, y_center = 0, x1 = -1, y1 = 0, x2 = 0, y2 = 1
Output: true

Example 3:

Input: radius = 1, x_center = 1, y_center = 1, x1 = -3, y1 = -3, x2 = 3, y2 = 3
Output: true

Example 4:

Input: radius = 1, x_center = 1, y_center = 1, x1 = 1, y1 = -3, x2 = 2, y2 = -1
Output: false

Constraints:

• 1 <= radius <= 2000
• -10^4 <= x_center, y_center, x1, y1, x2, y2 <= 10^4
• x1 < x2
• y1 < y2

Solution: Geometry

Find the shortest distance from the center to the rectangle, return dist <= radius.

Time complexity: O(1)
Space complexity: O(1)

// Author: Huahua
class Solution {
public:
  bool checkOverlap(int radius, int x_center, int y_center, int x1, int y1, int x2, int y2) {
    int dx = x_center - max(x1, min(x2, x_center));
    int dy = y_center - max(y1, min(y2, y_center));
    return dx * dx + dy * dy <= radius * radius;
  }
};

The post 花花酱 LeetCode 1401. Circle and Rectangle Overlapping appeared first on Huahua's Tech Road.