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

Jerry Franz jfranz at freerun.com
Sun Dec 14 17:17:16 UTC 2008



Nicolas Thierry-Mieg wrote:
> 
> Jerry Franz wrote:
>> Marko Vojinovic wrote:
>> [...]
>>> Basically, count the number of appearances of every number in your set. If you 
>>> have a set a priori bounded from above and below --- which you do,
>>> [1, n^2] --- you first allocate an array of integers of length n^2. 
>> By definition, your proposed algorithm is O(n^2), not O(n).
> 
> No it isn't, it's O(n) in time.
> O(n^2) in memory but that wasn't the question, right?

Look closer at it.

"[...]
you first allocate an array of integers of length n^2. Set all
elements to zero,"
[...]
go through the whole set from 1 to n^2, and if the value of k-th element 
is nonzero, print number k appropriate number of times.
[...]"

O(n^2) operations are required. It is O(n^2) both in time and memory as 
described.

-- 
Benjamin Franz


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