Problem
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:
-10^4 <= ax1 <= ax2 <= 10^4-10^4 <= ay1 <= ay2 <= 10^4-10^4 <= bx1 <= bx2 <= 10^4-10^4 <= by1 <= by2 <= 10^4
Solution
class Solution {
public int computeArea(int ax1, int ay1, int ax2, int ay2, int bx1, int by1, int bx2, int by2) {
long left = Math.max(ax1, bx1);
long right = Math.min(ax2, bx2);
long top = Math.min(ay2, by2);
long bottom = Math.max(ay1, by1);
long 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 (int) ((ax2 - ax1) * (ay2 - ay1) + (bx2 - bx1) * (by2 - by1) - area);
}
}
Explain:
nope.
Complexity:
- Time complexity : O(n).
- Space complexity : O(n).