Rectangle Area
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).
Examples​
Example 1:
Input: rax1 = -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 <= 104104 <= ay1 <= ay2 <= 104104 <= bx1 <= bx2 <= 104104 <= by1 <= by2 <= 104
Approach​
Find area of both rectangle and subtract the common region.
Solution​
Java​
class Solution {
public int computeArea(int ax1, int ay1, int ax2, int ay2, int bx1, int by1, int bx2, int by2) {
int areaA = Math.abs(ax1 - ax2) * Math.abs(ay1 - ay2);
int areaB = Math.abs(bx1 - bx2) * Math.abs(by1 - by2);
int intersectionWidth = Math.max(0, Math.min(ax2, bx2) - Math.max(ax1, bx1));
int intersectionHeight = Math.max(0, Math.min(ay2, by2) - Math.max(ay1, by1));
int intersectionArea = intersectionWidth * intersectionHeight;
return areaA + areaB - intersectionArea;
}
}
C++​
class Solution {
public:
int computeArea(int ax1, int ay1, int ax2, int ay2, int bx1, int by1, int bx2, int by2) {
int ans=(ax2-ax1)*(ay2-ay1)+(bx2-bx1)*(by2-by1);
int rem=(min(ax2,bx2)-max(ax1,bx1))*(min(ay2,by2)-max(ay1,by1));
if(bx1>=ax2||by1>=ay2||bx2<=ax1||by2<=ay1)rem=0;
if(rem<0)rem=0;
return ans-rem;
}
};
Python​
class Solution:
def computeArea(self, ax1: int, ay1: int, ax2: int, ay2: int, bx1: int, by1: int, bx2: int, by2: int) -> int:
a1=(ax2-ax1)*(ay2-ay1)
a2=(bx2-bx1)*(by2-by1)
x1=max(ax1,bx1)
x2=min(ax2,bx2)
y1=max(ay1,by1)
y2=min(ay2,by2)
if x2-x1<0 or y2-y1<0: #No intersection will occur if one of the side is negative
return a1+a2
a3=(x2-x1)*(y2-y1)
return a1+a2-a3