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Rectangle Overlap

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.

Solution Explanation for Rectangle Overlap

This problem asks to determine if two axis-aligned rectangles overlap. Axis-aligned means their sides are parallel to the x and y axes. We're given the bottom-left and top-right coordinates of each rectangle.

Approach:

Instead of calculating the intersection area directly (which is more complex), we can efficiently determine overlap by checking for non-overlap conditions. If none of the non-overlap conditions are met, then the rectangles must overlap.

The non-overlap conditions are:

  1. One rectangle is completely above the other: rec2[1] (y3) >= rec1[3] (y2) or rec1[1] (y1) >= rec2[3] (y4)
  2. One rectangle is completely to the right of the other: rec2[0] (x3) >= rec1[2] (x2) or rec1[0] (x1) >= rec2[2] (x4)

If none of these conditions are true, it implies the rectangles must intersect.

Code Explanation (Python):

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)

The Python code directly implements the logic above. It unpacks the coordinates of both rectangles, then uses not to invert the boolean result of the non-overlap conditions. If any non-overlap condition is true, the expression (y3 >= y2 or y4 <= y1 or x3 >= x2 or x4 <= x1) evaluates to True, and not makes the overall result False (no overlap). Otherwise, the result is True (overlap).

Time and Space Complexity:

  • Time Complexity: O(1). The algorithm performs a constant number of comparisons, regardless of the size of the input rectangles.
  • Space Complexity: O(1). The algorithm uses a constant amount of extra space to store variables.

Code in other languages:

The logic remains the same across different programming languages; only syntax changes. The examples below demonstrate this:

Java:

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);
    }
}

C++:

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);
    }
};

Go:

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)
}

TypeScript:

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);
}

All these implementations share the same core logic and have the same time and space complexity.