Most efficient solution?

[Alexandre Fayolle]

On 16 Jul 2001 11:02:12 -0400, Alex <new_name at mit.edu> wrote:
>
>> You may get some speedup by making B a dictionary, and using has_key()
>> to see if the word is there. This should get you a O(log(n)) instead
>> of O(n) inside the loop. To gain further performance, use filter to
>> skim A.
>
>What's n, here?  This looks linear in the length of A, and constant time
>in the number of unique elements in B, to me.

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. 

Alexandre Fayolle
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