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836. Rectangle Overlap

DifficultyEasy

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

Explanation:

The rectangles overlap in the region with corners at (1,1), (2,1), (1,2), and (2,2).

This region has positive area, so the rectangles overlap.

Example 2:

Input: rec1 = [0,0,1,1], rec2 = [1,0,2,1]

Output: false

Explanation:

The rectangles share the vertical edge where x = 1, but they do not have any common interior region.

Since their intersection has zero area, the rectangles do not overlap.

Example 3:

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

Output: false

Explanation:

The rectangles are completely separate. Rectangle rec1 ends at x = 1 and y = 1, while rec2 starts at x = 2 and y = 2.

Therefore, they have no common points and do not overlap.

 

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.

Solutions

Solution 1: Determine Non-Overlap Cases

Thinking

We decide whether two axis-aligned rectangles have positive-area overlap. Describing the intersection box is messy; the complement is simpler: they miss if they are separated vertically or horizontally.

If any of the four separations holds they are disjoint; negating that test is the overlap predicate. Areas are nonzero, so degenerate edges do not arise.

Let the coordinates of rectangle \(\textit{rec1}\) be \((x_1, y_1, x_2, y_2)\), and the coordinates of rectangle \(\textit{rec2}\) be \((x_3, y_3, x_4, y_4)\).

The rectangles \(\textit{rec1}\) and \(\textit{rec2}\) do not overlap if any of the following conditions are met:

  • \(y_3 \geq y_2\): \(\textit{rec2}\) is above \(\textit{rec1}\);
  • \(y_4 \leq y_1\): \(\textit{rec2}\) is below \(\textit{rec1}\);
  • \(x_3 \geq x_2\): \(\textit{rec2}\) is to the right of \(\textit{rec1}\);
  • \(x_4 \leq x_1\): \(\textit{rec2}\) is to the left of \(\textit{rec1}\).

If none of the above conditions are met, the rectangles \(\textit{rec1}\) and \(\textit{rec2}\) overlap.

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

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class Solution:
    def isRectangleOverlap(self, rec1: List[int], rec2: List[int]) -> bool:
        x1, y1, x2, y2 = rec1
        x3, y3, x4, y4 = rec2
        return not (y3 >= y2 or y4 <= y1 or x3 >= x2 or x4 <= x1)

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class Solution {
    public boolean isRectangleOverlap(int[] rec1, int[] rec2) {
        int x1 = rec1[0], y1 = rec1[1], x2 = rec1[2], y2 = rec1[3];
        int x3 = rec2[0], y3 = rec2[1], x4 = rec2[2], y4 = rec2[3];
        return !(y3 >= y2 || y4 <= y1 || x3 >= x2 || x4 <= x1);
    }
}

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class Solution {
public:
    bool isRectangleOverlap(vector<int>& rec1, vector<int>& rec2) {
        int x1 = rec1[0], y1 = rec1[1], x2 = rec1[2], y2 = rec1[3];
        int x3 = rec2[0], y3 = rec2[1], x4 = rec2[2], y4 = rec2[3];
        return !(y3 >= y2 || y4 <= y1 || x3 >= x2 || x4 <= x1);
    }
};

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func isRectangleOverlap(rec1 []int, rec2 []int) bool {
    x1, y1, x2, y2 := rec1[0], rec1[1], rec1[2], rec1[3]
    x3, y3, x4, y4 := rec2[0], rec2[1], rec2[2], rec2[3]
    return !(y3 >= y2 || y4 <= y1 || x3 >= x2 || x4 <= x1)
}

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function isRectangleOverlap(rec1: number[], rec2: number[]): boolean {
    const [x1, y1, x2, y2] = rec1;
    const [x3, y3, x4, y4] = rec2;
    return !(y3 >= y2 || y4 <= y1 || x3 >= x2 || x4 <= x1);
}

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