LeetCode //C - 223. Rectangle Area

223. Rectangle Area

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:

• $-10^4<=ax1<=ax2<=10^4$
• $-10^4<=ay1<=ay2<=10^4$
• $-10^4<=bx1<=bx2<=10^4$
• $-10^4<=by1<=by2<=10^4$

From: LeetCode
Link: 223. Rectangle Area

Solution:

Ideas:

1. Compute the area of each rectangle individually.
2. Calculate the overlapping area, if any.
3. Subtract the overlapping area from the sum of the areas of the two rectangles.

Code:

int computeArea(int ax1, int ay1, int ax2, int ay2, int bx1, int by1, int bx2, int by2) {
    // Calculate the area of the first rectangle
    int areaA = (ax2 - ax1) * (ay2 - ay1);
    // Calculate the area of the second rectangle
    int areaB = (bx2 - bx1) * (by2 - by1);

    // Find the overlapping area
    int overlapWidth = (ax2 > bx1 && bx2 > ax1) ? (ax2 < bx2 ? ax2 : bx2) - (ax1 > bx1 ? ax1 : bx1) : 0;
    int overlapHeight = (ay2 > by1 && by2 > ay1) ? (ay2 < by2 ? ay2 : by2) - (ay1 > by1 ? ay1 : by1) : 0;
    int overlapArea = overlapWidth * overlapHeight;

    // Total area is the sum of the areas minus the overlapping area
    return areaA + areaB - overlapArea;
}