2022-11-17 Daily Challenge

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Posted modified:
In: DailyChallenge

Today I have done leetcode's November LeetCoding Challenge with cpp.

November LeetCoding Challenge 17

Description

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:

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

Solution

auto speedup = [](){
  cin.tie(nullptr);
  cout.tie(nullptr);
  ios::sync_with_stdio(false);
  return 0;
}();
class Solution {
public:
  int computeArea(int ax1, int ay1, int ax2, int ay2, int bx1, int by1, int bx2, int by2) {
    int x = min({ax2 - bx1, bx2 - ax1, ax2 - ax1, bx2 - bx1});
    if(x < 0) x = 0;
    int y = min({ay2 - by1, by2 - ay1, ay2 - ay1, by2 - by1});
    if(y < 0) y = 0;
    return (ax2 - ax1) * (ay2 - ay1) + (bx2 - bx1) * (by2 - by1) - x * y;
  }
};

// Accepted
// 3080/3080 cases passed (15 ms)
// Your runtime beats 57.3 % of cpp submissions
// Your memory usage beats 5.6 % of cpp submissions (6.1 MB)
Algorithm