[CentOS] [OT] stable algorithm with complexity O(n)
vvmarko at panet.co.yu
Sun Dec 14 15:24:33 UTC 2008
On Sunday 14 December 2008 03:33, 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).
Oh, you mean because the upper bound is n^2, right? Sure, of course, this
particular case is O(n^2). Your proposal in your other post with the square
roots would probably improve that in this case.
However, I was just giving the OP a hint in the general direction of a typical
O(n) algorithm, didn't have an intention to provide a full working solution
for his specific case. It's his homework, not mine. ;-)
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