On Wed, Nov 19, 2014 at 2:02 PM, Ross Paterson <R.Paterson@city.ac.uk> wrote:

On Wed, Nov 19, 2014 at 01:53:02PM -0500, David Feuer wrote:
> I meant O(m log (mn)) = O(m (log m + log n)), because there are m appends,
> building up from O(n) to O(mn), but it really doesn't matter because we can
> easily do better.

Indeed it's moot, but appending a tree of size n to one of size mn
costs O(log n).

Right, I forgot that. I got to looking at <*> just now, and it suggests the following question: is there a particularly efficient way to build a Seq when its ultimate size is known in advance, avoiding the usual incremental rebuilding?