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This approach checks if there are conditions that confirm non-overlapping rectangles. Specifically, two rectangles do not overlap if:
If any of these conditions are true, the rectangles do not overlap. Otherwise, they do overlap.
Time Complexity: O(1), since the operation involves a constant number of comparisons.
Space Complexity: O(1), since the space used by the function is constant with respect to inputs.
1public boolean isRectangleOverlap(int[] rec1, int[] rec2) {
2 return rec1[2] > rec2[0] &&
3 rec1[0] < rec2[2] &&
4 rec1[3] > rec2[1] &&
5 rec1[1] < rec2[3];
6}The Java method checks overlap by verifying the opposite of the non-overlapping conditions simultaneously.
This approach computes the potential intersecting rectangle and verifies if its calculated dimensions form a valid rectangle with a positive area. The logic involves:
max(rec1[0], rec2[0])min(rec1[2], rec2[2])max(rec1[1], rec2[1])min(rec1[3], rec2[3])Time Complexity: O(1)
Space Complexity: O(1)
The function calculates potential overlapping parts along both axes and ensures these parts form an area (both x and y overlap must be true).