@jeevitesh yeah that may be right but it requires extra space as lot
of space will be wasted...

On May 5, 1:44 pm, Jeevitesh <[email protected]> wrote:
> Hi all,
>
> I am new to this group.
>
> My last post was deleted i do not know the reason behind it.
>
> I will explain my logic here:-
>
> as the range is 1 to n^2 we have a input array like input[n^2].
> We can take a auxillary array of size n^2 like aux[n^2].
> Scan the input array.
> For each input input[i] increment by one corresponding aux[input[i]].
> After this just iterate through the aux array printing the index aux[i]
> times.
>
> This way we can sort it in O(n) time.
>
>
>
>
>
>
>
>
>
> On Sat, May 5, 2012 at 2:02 PM, 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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> --
> *Thanks,
> Jeevitesh Shekhar Singh.*

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