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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.
1bool isRectangleOverlap(vector<int>& rec1, vector<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 C++ function achieves the same check as the Python version, evaluating non-overlapping conditions using logical operators.
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).