223

223. Rectangle Area (Medium)

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:

• -104 <= ax1 <= ax2 <= 104
• -104 <= ay1 <= ay2 <= 104
• -104 <= bx1 <= bx2 <= 104
• -104 <= by1 <= by2 <= 104

Companies: Amazon, Apple, Facebook, Zillow

Related Topics:
Math, Geometry

Similar Questions:

• Rectangle Overlap (Easy)

Solution 1.

// OJ: https://leetcode.com/problems/rectangle-area
// Author: github.com/lzl124631x
// Time: O(1)
// Space: O(1)
class Solution {
public:
    int computeArea(int ax1, int ay1, int ax2, int ay2, int bx1, int by1, int bx2, int by2) {
        if (ax1 > bx1) swap(ax1, bx1), swap(ay1, by1), swap(ax2, bx2), swap(ay2, by2);
        return (ax2 - ax1) * (ay2 - ay1) + (bx2 - bx1) * (by2 - by1) - max(0, min(ax2, bx2) - bx1) * max(0, min(ay2, by2) - max(ay1, by1));
    }
};

Solution 2.

// OJ: https://leetcode.com/problems/rectangle-area
// Author: github.com/lzl124631x
// Time: O(1)
// Space: O(1)
class Solution {
    int area(int a, int b, int c, int d) {
        return (c > a ? c - a : 0) * (d > b ? d - b : 0);
    }
public:
    int computeArea(int A, int B, int C, int D, int E, int F, int G, int H) {
        int a = max(A, E), b = max(B, F), c = min(C, G), d = min(D, H);
        return area(A, B, C, D) + area(E, F, G, H) - area(a, b, c, d);
    }
};