> > A lazy sort can be done in n time, easily:
> >
> > @highest;
> > foreach element {
> > if (element > lowest in @highest) {
> ^^^^^^^^^^^^^^^^^^
> You haven't kept @highest sorted, so there is another
> factor of n.
Nope, not so. @highest is of a fixed size, and hence
requires a fixed amount of time to sort. That's O(1).
O(N)*O(1) = O(N)
you can probably get me on how I expressed this, but not
on the fact it's O(N).
Jonathan Paton
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