LeetCode in Kotlin
223. Rectangle Area
Medium
Given the coordinates of two rectilinear rectangles in a 2D plane, return the total area covered by the two rectangles.
The first rectangle is defined by its bottom-left corner (ax1, ay1) and its top-right corner (ax2, ay2).
The second rectangle is defined by its bottom-left corner (bx1, by1) and its top-right corner (bx2, by2).
Example 1:
Input: ax1 = -3, ay1 = 0, ax2 = 3, ay2 = 4, bx1 = 0, by1 = -1, bx2 = 9, by2 = 2
Output: 45
Example 2:
Input: ax1 = -2, ay1 = -2, ax2 = 2, ay2 = 2, bx1 = -2, by1 = -2, bx2 = 2, by2 = 2
Output: 16
Constraints:
-104 <= ax1 <= ax2 <= 104-104 <= ay1 <= ay2 <= 104-104 <= bx1 <= bx2 <= 104-104 <= by1 <= by2 <= 104
Solution
@Suppress("kotlin:S107")
class Solution {
fun computeArea(ax1: Int, ay1: Int, ax2: Int, ay2: Int, bx1: Int, by1: Int, bx2: Int, by2: Int): Int {
val left = Math.max(ax1, bx1).toLong()
val right = Math.min(ax2, bx2).toLong()
val top = Math.min(ay2, by2).toLong()
val bottom = Math.max(ay1, by1).toLong()
var area = (right - left) * (top - bottom)
// if not overlaping, either of these two will be non-posittive
// if right - left = 0, are will automtically be 0 as well
if (right - left < 0 || top - bottom < 0) {
area = 0
}
return ((ax2 - ax1) * (ay2 - ay1) + (bx2 - bx1) * (by2 - by1) - area).toInt()
}
}