223. Rectangle Area Leetcode Solution
Rectangle Area Leetcode 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).
Rectangle Area Leetcode Solution :
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
Example 1:
- Input: ax1 = -2, ay1 = -2, ax2 = 2, ay2 = 2, bx1 = -2, by1 = -2, bx2 = 2, by2 = 2
- Output: 16
Intuition :
Intuition of this problem is simple. To calculate the common area between two rectangles,we use (area_1 + area_2 – common_area(between two rectangles)).
Approach :
Firstly,we will calculate the area1 and area2 by simple mathematics .i.e. length * width .Now our main task is compute the common area.There are few cases that can exist between two rectangles that are as follows:
Case 1: Rectangle 1 is enclosed in rectangle 2.
- Condition: ax1>=bx1 && ax2<=bx2 && ay1>=by1 && ay2<=by2
- Common Area:Area 1
Case 2: Rectangle 2 is enclosed in rectangle 1.
- Condition: bx1>=ax1 && bx2<=ax2 && by1>=ay1 && by2<=ay2
- Common Area:Area 2
Case 3: Rectangle 1 and rectangle 2 share some common area.
- Condition: max(ax1,bx1)<= min(ax2,bx2) && max(ay1,by1)<= min(ay2,by2)
- common_area= (max(ax1,bx1)- min(ax2,bx2))*(max(ay1,by1)- min(ay2,by2))
Case 4: Rectangle 1 and rectangle 2 has no common area.
- Common area=0
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Code :
class Solution {
public:
int computeArea(int ax1, int ay1, int ax2, int ay2, int bx1, int by1, int bx2, int by2) {
long long area1=(ax2-ax1)*(ay2-ay1);
long long area2=(bx2-bx1)*(by2-by1);
long long common_area;
if( ax1>=bx1 && ax2<=bx2 && ay1>=by1 && ay2<=by2){
common_area= area1;
}
else if( bx1>=ax1 && bx2<=ax2 && by1>=ay1 && by2<=ay2){
common_area= area2;
}
else if( max(ax1,bx1)<= min(ax2,bx2) && max(ay1,by1)<= min(ay2,by2)){
common_area= (max(ax1,bx1)- min(ax2,bx2))*(max(ay1,by1)- min(ay2,by2));
}
else common_area=0;
return area1+area2-common_area;
}
};
class Solution {
public int computeArea(int ax1, int ay1, int ax2, int ay2, int bx1, int by1, int bx2, int by2) {
if(ax2< bx1||bx2< ax1 )
return (bx2-bx1)*(by2-by1) + (ax2-ax1)*(ay2-ay1);
if(ay2< by1 || by2< ay1)
return (bx2-bx1)*(by2-by1) + (ax2-ax1)*(ay2-ay1);
int right = Math.min(ax2,bx2);
int left = Math.max(ax1,bx1);
int top = Math.min(by2,ay2);
int bottom = Math.max(by1,ay1);
return (bx2-bx1)*(by2-by1) + (ax2-ax1)*(ay2-ay1) - (right-left)*(top-bottom);
}
}class Solution:
def computeArea(self, ax1: int, ay1: int, ax2: int, ay2: int, bx1: int, by1: int, bx2: int, by2: int) -> int:
# Function to caculate area of rectangle (Length * width)
def area(x1,y1,x2,y2):
return (x2-x1)*(y2-y1)
# Finding the overlap rectangle length and width
overlapX = max(min(ax2,bx2)-max(ax1,bx1), 0)
overlapY = max(min(ay2,by2)-max(ay1,by1), 0)
# Area1 + Area2 - Overlap Rectangle Area
return area(ax1,ay1,ax2,ay2) + area(bx1,by1,bx2,by2) - overlapX * overlapY
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