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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:
- zaphod October 16, 20101. 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) :)