Hi Dave,

I don't think ur logic will cover all cases like   (1,1)(-3,-3),      (1,1)
(2,2)  a line connecting these points passes through origin,

i think the solution is, we need to compute the slope of the point at index
i with origin and build a binary tree with theses slopes.

but worst cases of this algo is N*N , if we try balancing the tree while
inserting I guess it can be done in NlogN

Thanks
Vinay


On Fri, Feb 25, 2011 at 9:20 AM, Gene <[email protected]> wrote:

> Dave's solution is best if numerical error is possible.
>
> If the points are precise, you can also do it in linear time.  Just hash
> the points on abs(y/x).
>
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