Most efficient solution?

[Alexandre Fayolle]

On 16 Jul 2001 09:18:40 -0700, Aahz Maruch <aahz at panix.com> wrote:
>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).

I'm pretty sure I was wrong, but are you sure you are not too? ;o)
Should it not read: 'C.has_key(item) is constant'? 

Alexandre Fayolle
-- 
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