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I think we can do it in O(log n) using interval trees. Lets say the given point is (x,y). Lets say, the positions of lower-left (xL, yL) & upper-right (xH, yH) corners of each rectangle are given. We will create two interval trees. Each node of first interval tree (say treeX) will have pair (xL, xH) corresponding to each rectangle. Similarly, each node of second interval tree (say treeY) will have pair (yL, hH) corresponding to each rectangle. In first pass (which is O(log n)), we will use treeX to get all the rectangles for which (xL < x < xH). In second pass (which is also order O(log n)), we will use treeY to get the minimum rectangle that contains (x,y).
- abhishekseth8887 April 10, 2016