O(1) base conversion can here be done as follows:-(works only when numbers
are in range 0 to n^2-1)
*From base 10 to base n
*let given base 10 no is m and we need to convert it in base n.
then base n converted no is (m/n)(m%n) ie 34 base 10 in base 6 is
(34/6)(34%6) ie 54

Now i think you can yourself write base n to base 10 conversion.

On Sun, May 6, 2012 at 11:13 AM, saurabh singh <saurab...@gmail.com> wrote:

> ^ 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 <gene.ress...@gmail.com> 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 <saurab...@gmail.com> 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 <saurab...@gmail.com>
>> 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 <cegprak...@gmail.com>
>> wrote:
>> >
>> > >> The range 1 to n^2 is already sorted
>> >
>> > >> On Sat, May 5, 2012 at 12:17 AM, Algobiz <
>> deepak.arulkan...@gmail.com>
>> > >> wrote:
>> > >> > How to sort n numbers in the range of 1 to n^2 in O(n).. Any ideas?
>> >
>> > >> > --
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-- 
Arpit Gupta
B.Tech Third Year
Computer Science And Engineering
M.N.N.I.T Allahabad
arpitgupta.211...@gmail.com

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