# 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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