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# @lc app=leetcode.cn id=223 lang=python3
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# [223] 矩形面积
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# https://leetcode-cn.com/problems/rectangle-area/description/
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# algorithms
# Medium (46.35%)
# Likes:    133
# Dislikes: 0
# Total Accepted:    24.7K
# Total Submissions: 50.7K
# Testcase Example:  '-3\n0\n3\n4\n0\n-1\n9\n2'
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# 给你 二维 平面上两个 由直线构成的 矩形，请你计算并返回两个矩形覆盖的总面积。
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# 每个矩形由其 左下 顶点和 右上 顶点坐标表示：
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# 第一个矩形由其左下顶点 (ax1, ay1) 和右上顶点 (ax2, ay2) 定义。
# 第二个矩形由其左下顶点 (bx1, by1) 和右上顶点 (bx2, by2) 定义。
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# 示例 1：
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# 输入：ax1 = -3, ay1 = 0, ax2 = 3, ay2 = 4, bx1 = 0, by1 = -1, bx2 = 9, by2 = 2
# 输出：45
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# 示例 2：
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# 输入：ax1 = -2, ay1 = -2, ax2 = 2, ay2 = 2, bx1 = -2, by1 = -2, bx2 = 2, by2 = 2
# 输出：16
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# 提示：
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# -10^4 
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# @lc code=start
class Solution:
    def computeArea(self, ax1: int, ay1: int, ax2: int, ay2: int, bx1: int, by1: int, bx2: int, by2: int) -> int:
        area1 = (ax2-ax1)*(ay2-ay1)
        area2 = (bx2-bx1)*(by2-by1)
        overlapWidth = min(ax2, bx2) - max(ax1, bx1)
        overlapHeight = min(ay2, by2) - max(ay1, by1)
        return area1 + area2 - max(0, overlapWidth) * max(0, overlapHeight)
# @lc code=end

