# [OT] stable algorithm with complexity O(n)

Diez B. Roggisch deets at nospam.web.de
Sat Dec 13 20:35:58 CET 2008

```Duncan Booth schrieb:
> "Diez B. Roggisch" <deets at nospam.web.de> wrote:
>
>> David HlÃ¡Äik schrieb:
>>> Hi guys,
>>>
>>> i am really sorry for making offtopic, hope you will not kill me, but
>>> this is for me life important problem which needs to be solved within
>>> next 12 hours..
>>>
>>> I have to create stable algorithm for sorting n numbers from interval
>>> [1,n^2] with time complexity O(n) .
>>>
>>> Can someone please give me a hint. Would be very very thankful!
>> Unless I grossly miss out on something in computer science 101, the
>> lower bound for sorting is O(n * log_2 n). Which makes your task
>> impossible, unless there is something to be assumed about the
>> distribution of numbers in your sequence.
>>
>> Who has given you that assignment - a professor? Or some friend who's
>> playing tricks on you?
>>
> I think you must have fallen asleep during CS101. The lower bound for
> sorting where you make a two way branch at each step is O(n * log_2 n), but
> if you can choose between k possible orderings in a single comparison you
> can get O(n * log_k n).
>
> To beat n * log_2 n just use a bucket sort: for numbers with a known
> maximum you can sort them digit by digit for O(n log_k n), and if you don't
> restrict yourself to decimal then k can be as large as you want, so for the
> problem in question I think you can set k=n and (log_n n)==1 so you get
> O(n)

Thanks. I *totally* forgot about that.

Diez

```