Timak <[EMAIL PROTECTED]> writes:

>> Finally, for characters of w bits, the histogram requires O(2^w log N)
>> space in the general case, not O(w) as you say.
>
> My bad.  I should be paying more careful when I post.  I meant
> O(2^w).  Why O(2^w logN)?

To answer my own stupid question, LogN is for the counters of course.

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