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. ;-) :-) Marko