1401. Circle and Rectangle Overlapping
Description
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:
![](https://spcdn.pages.dev/leetcode/problems/1401.Circle%20and%20Rectangle%20Overlapping/images/sample_4_1728.png)
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:
![](https://spcdn.pages.dev/leetcode/problems/1401.Circle%20and%20Rectangle%20Overlapping/images/sample_2_1728.png)
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
Solutions
Solution 1: Mathematics
For a point $(x, y)$, its shortest distance to the center of the circle $(xCenter, yCenter)$ is $\sqrt{(x - xCenter)^2 + (y - yCenter)^2}$. If this distance is less than or equal to the radius $radius$, then this point is within the circle (including the boundary).
For points within the rectangle (including the boundary), their x-coordinates $x$ satisfy $x_1 \leq x \leq x_2$, and their y-coordinates $y$ satisfy $y_1 \leq y \leq y_2$. To determine whether the circle and rectangle overlap, we need to find a point $(x, y)$ within the rectangle such that $a = |x - xCenter|$ and $b = |y - yCenter|$ are minimized. If $a^2 + b^2 \leq radius^2$, then the circle and rectangle overlap.
Therefore, the problem is transformed into finding the minimum value of $a = |x - xCenter|$ when $x \in [x_1, x_2]$, and the minimum value of $b = |y - yCenter|$ when $y \in [y_1, y_2]$.
For $x \in [x_1, x_2]$:
- If $x_1 \leq xCenter \leq x_2$, then the minimum value of $|x - xCenter|$ is $0$;
- If $xCenter < x_1$, then the minimum value of $|x - xCenter|$ is $x_1 - xCenter$;
- If $xCenter > x_2$, then the minimum value of $|x - xCenter|$ is $xCenter - x_2$.
Similarly, we can find the minimum value of $|y - yCenter|$ when $y \in [y_1, y_2]$. We can use a function $f(i, j, k)$ to handle the above situations.
That is, $a = f(x_1, x_2, xCenter)$, $b = f(y_1, y_2, yCenter)$. If $a^2 + b^2 \leq radius^2$, then the circle and rectangle overlap.
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