stable algorithm with complexity O(n)
David Hláčik
david at hlacik.eu
Sun Dec 14 21:01:33 UTC 2008
Thank you guys for help and support! My homework is done and waiting
for grading.
Here it comes - bucket sort with time complexity O(n) == linear complexity
#! /usr/bin/python
def sort(numbers):
"sort n positive integers in O(n) provided that they are all from
interval [1, n^2]"
N = len(numbers) # get size of test numbers
buckets_mod = [[] for i in xrange(N)]
buckets_sorted = [[] for i in xrange(N+1)]
# group numbers to buckets (list of numbers) with common modulus
for n in numbers:
buckets_mod[n % N].append(n)
print "buckets_mod: %s" % buckets_mod
# check numbers in buckets
for l in buckets_mod:
for n in l:
# place number into bucket number grouped by result of division
buckets_sorted[n / N].append(n)
print "buckets_sorted: %s" % buckets_sorted
# search through sorted buckets and return list of sorted numbers
return [n for l in buckets_sorted for n in l]
Regards,
David
On Sun, Dec 14, 2008 at 4:08 PM, Ed Greshko <Ed.Greshko at greshko.com> wrote:
> Aaron Konstam wrote:
>> On Sun, 2008-12-14 at 12:10 +1100, Cameron Simpson wrote:
>>
>>> | There is the generation counter algorithm. It is O(n), and relies on
>>> the fact
>>> | that one knows the boundaries of the possible elements of the
>>> dataset (and
>>> | it's domain). Basically, you simply count the number of appearances
>>> of every
>>> | element in your set. It's just a single for-loop.
>>>
>> I am not on the Centos list either not do I understand the sorting
>> algorithm you describe above. Could you expand on the explanation above.
>>
>>
> FWIW, a judicious google search answered all of my questions on this
> sort of thing. :-)
>
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