LeetCode 0223. Rectangle Area Solution in Java, C++, Python & More | Explanation + Code

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0223. Rectangle Area

Description

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:

Rectangle Area
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

Solutions

Solution 1: Calculate Overlapping Area

First, we calculate the area of the two rectangles separately, denoted as a and b. Then we calculate the overlapping width width and height height. The overlapping area is max(width, 0) × max(height, 0). Finally, we subtract the overlapping area from a and b.

The time complexity is O(1), and the space complexity is O(1).

PythonJavaC++GoTypeScriptC#
class Solution: def computeArea( self, ax1: int, ay1: int, ax2: int, ay2: int, bx1: int, by1: int, bx2: int, by2: int, ) -> int: a = (ax2 - ax1) * (ay2 - ay1) b = (bx2 - bx1) * (by2 - by1) width = min(ax2, bx2) - max(ax1, bx1) height = min(ay2, by2) - max(ay1, by1) return a + b - max(height, 0) * max(width, 0)(code-box)

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