Flipkart Interview Question for Software Engineer / Developers






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3
of 3 vote

You need to keep track of the 100 closest points. Use a binary max heap with the comparator being distance from (x,y) (you can optimize this by comparing distance squared rather than calculating square-root but it complicates things so to keep it simple just use distance). Now add the first 100 points. Starting with the 101st point:

1. add the point to the heap
2. delete the max element from the heap

The heap will always contain the 100 smallest elements of those looked at so far. Once you've compared all the points the heap will contain the 100 closest points to (x,y). Since the heap is always the same size the heap operations are kinda-sorta "constant" so if you squint just right it runs in O(n) :)

- zaphod October 16, 2010 | Flag Reply
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0
of 0 votes

awesome :)

- Anonymous November 14, 2010 | Flag
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0
of 0 votes

The cost of calculating (y^2 - x^2) can be reduced to (y+x)*(y-x).

- Anonymous December 23, 2010 | Flag
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0
of 0 vote

Look at the wiki page for R-Tree.

- Anonymous October 19, 2010 | Flag Reply
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0
of 0 vote

KD-Tree is appropriate data structure for this problem.

- dongre.avinash August 03, 2011 | Flag Reply


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