On 13 March 2013 10:43, Wolfgang Maier <wolfgang.ma...@biologie.uni-freiburg.de> wrote: > > thinking again about the question, then the min() solutions suggested so far > certainly do the job and they are easy to understand. > However, if you need to run the function repeatedly on larger lists, using > min() > is suboptimal because its performance is an O(n) one. > It's faster, though less intuitive, to sort your list first, then use bisect > on > it to find the zero position in it. Two manipulations running at O(log(n)).
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 -- http://mail.python.org/mailman/listinfo/python-list
