@L: It was asked if we could take advantage of the ranges of the
integers between 1-N^2..
I doubt if its possible.

On Jun 26, 5:33 pm, ross <[email protected]> wrote:
> @radhakrishnan: Counting sort in this case, will be O(n2).. as it
> involves traversing the entire array!
>
> On Jun 26, 5:03 pm, L <[email protected]> wrote:
>
>
>
>
>
>
>
> > Count sort is O(n+k), since k~n^2 here, it will be O(N^2).
> > Radix sort has complexity O(r.n) which is nearly O(n logn).
> > Are you sure that the person asking this question wanted O(n) ?
>
> > On Jun 26, 1:31 pm, radha krishnan <[email protected]>
> > wrote:
>
> > > Yes ! Count Sort !!
>
> > > On Sun, Jun 26, 2011 at 1:44 PM, ross <[email protected]> wrote:
> > > > Given a sequence of numbers in the range of 1-N^2, what is the most
> > > > efficient way to sort the numbers (better than NlgN)..
> > > > Can counting sort be used here? Is an O(N) solution possible..
>
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