836. Rectangle Overlap

Description

An axis-aligned rectangle is represented as a list [x1, y1, x2, y2], where (x1, y1) is the coordinate of its bottom-left corner, and (x2, y2) is the coordinate of its top-right corner. Its top and bottom edges are parallel to the X-axis, and its left and right edges are parallel to the Y-axis.

Two rectangles overlap if the area of their intersection is positive. To be clear, two rectangles that only touch at the corner or edges do not overlap.

Given two axis-aligned rectangles rec1 and rec2, return true if they overlap, otherwise return false.

Example 1:

• Input: rec1 = [0,0,2,2], rec2 = [1,1,3,3]
• Output: true

Example 2:

• Input: rec1 = [0,0,1,1], rec2 = [1,0,2,1]
• Output: false

Example 3:

• Input: rec1 = [0,0,1,1], rec2 = [2,2,3,3]
• Output: false

Constraints:

• rec1.length == 4
• rec2.length == 4
• -10⁹ <= rec1[i], rec2[i] <= 10⁹
• rec1 and rec2 represent a valid rectangle with a non-zero area.

Submitted Code

class Solution(object):
    def isRectangleOverlap(self, rec1, rec2):
        """
        :type rec1: List[int]
        :type rec2: List[int]
        :rtype: bool
        """
        # 겹치지 않는 조건: rec2가 rec1보다 완전히 오른쪽or왼쪽or위쪽or아래쪽에 있는 경우
        if rec2[0] >= rec1[2] or rec2[2] <= rec1[0] or rec2[1] >= rec1[3] or rec2[3] <= rec1[1]:
            return False
        # 넷 중 어느 한 쪽이라도 겹친다면 True
        else:
            return True

Runtime: 0 ms | Beats 100.00%
Memory: 12.35 MB | Beats 79.60%

두 사각형이 겹치지 않는 조건을 먼저 파악했다.

Other Solutions

1st

class Solution:
    def isRectangleOverlap(self, rec1: 'List[int]', rec2: 'List[int]') -> 'bool':
        left = max(rec1[0], rec2[0])
        right = min(rec1[2], rec2[2])
        bottom = max(rec1[1], rec2[1])
        up = min(rec1[3], rec2[3])
        
        if left < right and up > bottom:
            return True
        else:
            return False

time complexity: 𝑂(1)
space complexity: 𝑂(1)

두 사각형의 교집합 영역을 계산해서 양수일 경우 겹쳤다고 판단할 수 있다.