Most efficient solution?

[Aahz Maruch]

In article <slrn9l68fa.o6r.alf at leo.logilab.fr>,
Alexandre Fayolle <alexandre.fayolle at logilab.fr> wrote:
>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'? 

<smacks head>
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