Most efficient solution?

[Aahz Maruch]

In article <slrn9l62om.o1k.alf at leo.logilab.fr>,
Alexandre Fayolle <alexandre.fayolle at logilab.fr> wrote:
>
>I was referring to the inner loop. n is the number of elements in B 
>(or of keys in C). 'item in B' is O(n), and 'C.has_key(item)' is O(log(n)). 
>You still have to multiply by len(A) for the whole program. 

That's functionally incorrect.  Except under severely degraded
conditions, C.has_key(item) is O(n).
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