未验证 提交 eaa7ef45 编写于 作者: C Cho Yin Yong 提交者: GitHub

kth order statistic divide and conquer algorithm (#3690)

* kth order statistics divide and conquer algorithm

* add explanation of algorithm.

* fix PEP8 line too long error

* update order to be compliant to isort

* add doctest

* make file black compliant
上级 61cb921d
"""
Find the kth smallest element in linear time using divide and conquer.
Recall we can do this trivially in O(nlogn) time. Sort the list and
access kth element in constant time.
This is a divide and conquer algorithm that can find a solution in O(n) time.
For more information of this algorithm:
https://web.stanford.edu/class/archive/cs/cs161/cs161.1138/lectures/08/Small08.pdf
"""
from random import choice
from typing import List
def random_pivot(lst):
"""
Choose a random pivot for the list.
We can use a more sophisticated algorithm here, such as the median-of-medians
algorithm.
"""
return choice(lst)
def kth_number(lst: List[int], k: int) -> int:
"""
Return the kth smallest number in lst.
>>> kth_number([2, 1, 3, 4, 5], 3)
3
>>> kth_number([2, 1, 3, 4, 5], 1)
1
>>> kth_number([2, 1, 3, 4, 5], 5)
5
>>> kth_number([3, 2, 5, 6, 7, 8], 2)
3
>>> kth_number([25, 21, 98, 100, 76, 22, 43, 60, 89, 87], 4)
43
"""
# pick a pivot and separate into list based on pivot.
pivot = random_pivot(lst)
# partition based on pivot
# linear time
small = [e for e in lst if e < pivot]
big = [e for e in lst if e > pivot]
# if we get lucky, pivot might be the element we want.
# we can easily see this:
# small (elements smaller than k)
# + pivot (kth element)
# + big (elements larger than k)
if len(small) == k - 1:
return pivot
# pivot is in elements bigger than k
elif len(small) < k - 1:
return kth_number(big, k - len(small) - 1)
# pivot is in elements smaller than k
else:
return kth_number(small, k)
if __name__ == "__main__":
import doctest
doctest.testmod()
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