^ And this completes the solution....
Saurabh Singh
B.Tech (Computer Science)
MNNIT
blog:geekinessthecoolway.blogspot.com



On Sun, May 6, 2012 at 11:12 AM, Gene <[email protected]> wrote:

> Ah, but you can pick the radix to be n.  Then at most 3 passes will
> always sort the array. O(3n) = O(n), so you are done.
>
> This topic has come up before.  There is code at
> http://groups.google.com/group/algogeeks/msg/90ce2df194aba2d2 .
>
> It is true this code assumes math including mod takes constant time,
> but that's normal for RAM computation models used for most algorithms.
>
> Gene
>
> On May 5, 4:32 am, saurabh singh <[email protected]> wrote:
> > After giving some thought,I think even radix sort may not be sufficient.
> > Complexity of radix sort is O(k*n) where k is the number of buckets
> > required to sort the given range.
> > The number of buckets is proportional to the number of bits required to
> > represent the *maximum number in the given range.*For our case the
> maximum
> > number is O(n^2).Hence *the number of buckets required would be
> > proportional to log(n^2) in the worst case.*
> > Hence the worst case complexity for the given constraints using radix
> sort
> > would be *O(n*(log n^2)) = O(n*logn).*
> > This is no better than comparision sort.A slight optimization that we can
> > make is to use a higher base which would reduce the number of buckets
> > required but would add the cost of converting each number into  the
> higher
> > base.
> > Somehow I am getting convinced worst case O(n) algorithm may not be
> > possible.Working on the mathematical proof.
> > Saurabh Singh
> > B.Tech (Computer Science)
> > MNNIT
> > blog:geekinessthecoolway.blogspot.com
> >
> > On Sat, May 5, 2012 at 8:37 AM, saurabh singh <[email protected]>
> wrote:
> > > @cegprakash They are n numbers lying in the range 1 to n^2.Not
> necessarily
> > > sorted.
> > > eg 3 4 1 2 5 8 (6 numbers satisfying the conditions given in the
> problem)
> > > Saurabh Singh
> > > B.Tech (Computer Science)
> > > MNNIT
> > > blog:geekinessthecoolway.blogspot.com
> >
> > > On Sat, May 5, 2012 at 5:17 AM, Prakash D <[email protected]>
> wrote:
> >
> > >> The range 1 to n^2 is already sorted
> >
> > >> On Sat, May 5, 2012 at 12:17 AM, Algobiz <[email protected]
> >
> > >> wrote:
> > >> > How to sort n numbers in the range of 1 to n^2 in O(n).. Any ideas?
> >
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