I think that we can do it by first passing through an array once and convert
all the interger to the base N in O(N). Since all elements are in the range
1 to N^2 they can have atmost 3 digits.Now apply radix sort which will be O(
dn )  where d=3 in this case .=> O(n) when done convert them back to decimal
notation in O(N) .
But here N must be maximum no of single digit characters representable in ur
system.

Check if possible !!!!


On Tue, Aug 2, 2011 at 1:02 PM, payel roy <smithpa...@gmail.com> wrote:

> Can you please elaborate ??
>
>
> On 2 August 2011 12:38, sunny agrawal <sunny816.i...@gmail.com> wrote:
>
>> Radix sort is one of the solution.
>> because this Question is in the section Radix sort in CLRS. :)
>>
>>
>> On Tue, Aug 2, 2011 at 11:10 AM, Ravinder Kumar <ravinde...@gmail.com>wrote:
>>
>>> I have little thought on this :
>>>
>>> divide the whole array by n and store their remainder also in an array
>>> now number in original array are in range 1-n
>>> sort the array and when two number with same break the tie using
>>> remainder array
>>> recreate the array using remainder array .
>>> --
>>> *With Regards :*
>>>
>>> Ravinder Kumar
>>> B.Tech Final Year
>>> Computer Science and Engineering
>>> MNNIT Allahabad
>>>
>>>
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>>
>>
>>
>> --
>> Sunny Aggrawal
>> B-Tech IV year,CSI
>> Indian Institute Of Technology,Roorkee
>>
>>
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