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2010-07-24T12:32:15.363
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2014-12-03T21:24:05.473
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2014-12-03T21:24:05.473
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Here is a brief description of the intersection algorithm presented in DuduAlul's link. The solution requires the usage of a search tree capable of performing range queries. A range query asks for all items with values between K1 and K2, and it should be an O(R+log N) operation, where N is the total number of tree items, and R is the number of results. The algorithm uses the sweep approach: 1) Sort all left and right rectangle edges, according to their X value, into list L. 2) Create a new empty range search tree T, for Y ordering of rectangle tops/bottoms 3) Create a new empty result set RS of unique rectangle pairs 4) Traverse L in ascending order. For all V in L:    If V.isRightEdge()       T.remove(V.rectangle.top)       T.remove(V.rectangle.bottom)    else        For all U in T.getRange(V.rectangle.top, V.rectangle.bottom)          RS.add(<V.rectangle, U.rectangle>)       T.add(V.rectangle.top)       T.add(V.rectangle.bottom) 5) return RS <hr> The time complexity is O(R + N log N) where R is the actual number of intersections. -- EDIT -- I just figured out that the solution is not fully correct - the intersection test in tree T misses some cases. The tree should maintain Y intervals rather than Y values, and it should ideally be an <a href="http://en.wikipedia.org/wiki/Interval_tree" rel="nofollow noreferrer">Interval Tree</a>.
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1. POaxis‐aligned rectangles intersection
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1. USEyal Schneider
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1. USEyal Schneider
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1. COThis is a good answer. I also found it at the handout of the class _Hacking a Google Interview_ from csail.mit.edu
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Querying!

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