14 Dec
2008
14 Dec
'08
5:40 p.m.
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?