Finding the Min for positive and negative in python 3.3 list

[Chris Angelico]

On Wed, Mar 13, 2013 at 10:34 PM, Wolfgang Maier
<wolfgang.maier at biologie.uni-freiburg.de> wrote:
> Oscar Benjamin <oscar.j.benjamin <at> gmail.com> writes:
>
>>
>> Sort cannot be O(log(n)) and it cannot be faster than a standard O(n)
>> minimum finding algorithm. No valid sorting algorithm can have even a
>> best case performance that is better than O(n). This is because it
>> takes O(n) just to verify that a list is sorted.
>>
>> Oscar
>>
>
> Oops, you're right of course.
> Wrote this in a hurry before and got confused a bit.
> So, the two min()s take O(n) each, the sort takes O(n),
> but the bisect takes O(log n),
> which means that sorting and bisecting together should still be faster
> than 2xmin(), although it's a bit less striking than what I wrote first.
> Thanks for the correction,
> Wolfgang

Your sort is usually going to be O(n log n), not O(log n).

ChrisA