@Dave:

Sorry my bad. Yes in that case, I cant think of a solution in O(n). I guess
your solution of sorting O(nlogn) is the best one in that case.

Thanks,
Immanuel

On Sat, May 21, 2011 at 2:20 AM, Dave <dave_and_da...@juno.com> wrote:

> @Immanuel: It still looks like you are finding the nearest neighbors
> of only one point, while the problem was to find the neighbors of
> _each_ of the given points.
>
> Dave
>
> On May 20, 3:07 pm, immanuel kingston <kingston.imman...@gmail.com>
> wrote:
> > I guess a <O(nk),O(k)> solution exists.
> >
> > Have a maxHeap of k elements in our case its 3.
> >
> > Iterate through the array, compare the (difference between the position
> > along a number
> > line between ) and the top element of the maxHeap. It it happens to be
> > lesser than the top element, pop off the top element and push the current
> > element into the maxHeap. Proceeding till the end of the array we will be
> > getting the 3 friends of a given person.
> >
> > Hope I am not wrong.
> >
> > Thanks,
> > Immanuel
> >
> >
> >
> > On Thu, May 19, 2011 at 7:33 PM, Dave <dave_and_da...@juno.com> wrote:
> > > @Sravanreddy001: You are to find _each_ person's friends. Can you do
> > > that in O(n)?
> >
> > > Dave
> >
> > > On May 19, 8:59 am, sravanreddy001 <sravanreddy...@gmail.com> wrote:
> > > > Also, I think there is no need for sorting the number, its still okey
> if
> > > the
> > > > 3rd person is standing 1st and has the lowest number line value.
> >
> > > > And, finding the closest 3 number takes, 3*n time.. so.. its O(n)
> running
> > > > time..
> >
> > > > @Dave.. good catch.. :)
> >
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