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

Jerry Franz jfranz at freerun.com
Sun Dec 14 02:33:52 UTC 2008

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).


Benjamin Franz

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