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    <p>These kind of questions are largely answered by some standard <code>Computational Geometry</code> algorithms. I can think of a <code>vertical sweep line</code> algorithm for this particular problem.</p> <p>Assuming a rectangle is represented by a pair of points <code>(p1, p2)</code>, where <code>p1</code> is the upper left corner and <code>p2</code> is the bottom right corner. And a point has two attributes (can be accessed as <code>p.x</code> and <code>p.y</code>).</p> <p>Here is the algorithm.</p> <ol> <li>Sort all the pairs of points - <code>O(n log n)</code></li> <li>Initialize a list called <code>sweep line status</code>. This will hold all the rectangles that are encountered till now, that are <code>alive</code>. Also initialize another list called <code>event queue</code> that holds upcoming events. This <code>event queue</code> currently holds starting points of all the rectagles.</li> <li>Process the events, start from the first element in the <code>event queue</code>. <ul> <li>If the event is a <code>start point</code>, then add that rectangle to <code>sweep line status</code> (in sorted order by y-coordinate) (in <code>O(log n)</code> time) and add its bottom-right point to the <code>event queue</code> at the appropriate position (sorted by the points) (again in <code>O(log n)</code> time). When you add it to <code>sweep line status</code>, you just need to check if this point lies in the rectangle alive just above it in the <code>sweep line status</code>. If it does lie inside, this is not your rectangle, otherwise, add this to your list of required rectangles.</li> <li>If the event is an end point, just remove the correspoinding rectangle from the <code>sweep line status</code>.</li> </ul></li> </ol> <p>Running time (for n rectangles):</p> <ul> <li>Sorting takes <code>O(n log n)</code>.</li> <li>Number of events = 2*n = <code>O(n)</code></li> <li>Each event takes <code>O(log n)</code> time (for insertions in <code>event queue</code> as well as <code>sweep line status</code>. So total is <code>O(n log n)</code>.</li> </ul> <p>Therefore, <code>O(n log n)</code>.</p> <p>For more details, please refer to <a href="http://en.wikipedia.org/wiki/Bentley%E2%80%93Ottmann_algorithm" rel="nofollow">Bentley–Ottmann algorithm</a>. The above just a simple modification of this.</p> <p>EDIT:</p> <p>Just realized that input is in terms of line segments, but since they always form rectangles (according to question), a linear traversal for a pre-process can convert them into the rectangle (pair of points) form.</p>
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    1. POGiven many horizontal and vertical lines, how to find all the rectangles that do have any sub-rectangle inside them?
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    1. POGiven many horizontal and vertical lines, how to find all the rectangles that do have any sub-rectangle inside them?
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