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* Pyupgrade to Python 3.9
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66 lines
1.7 KiB
Python
66 lines
1.7 KiB
Python
"""
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Find the kth smallest element in linear time using divide and conquer.
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Recall we can do this trivially in O(nlogn) time. Sort the list and
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access kth element in constant time.
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This is a divide and conquer algorithm that can find a solution in O(n) time.
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For more information of this algorithm:
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https://web.stanford.edu/class/archive/cs/cs161/cs161.1138/lectures/08/Small08.pdf
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"""
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from __future__ import annotations
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from random import choice
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def random_pivot(lst):
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"""
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Choose a random pivot for the list.
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We can use a more sophisticated algorithm here, such as the median-of-medians
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algorithm.
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"""
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return choice(lst)
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def kth_number(lst: list[int], k: int) -> int:
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"""
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Return the kth smallest number in lst.
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>>> kth_number([2, 1, 3, 4, 5], 3)
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3
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>>> kth_number([2, 1, 3, 4, 5], 1)
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1
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>>> kth_number([2, 1, 3, 4, 5], 5)
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5
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>>> kth_number([3, 2, 5, 6, 7, 8], 2)
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3
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>>> kth_number([25, 21, 98, 100, 76, 22, 43, 60, 89, 87], 4)
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43
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"""
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# pick a pivot and separate into list based on pivot.
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pivot = random_pivot(lst)
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# partition based on pivot
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# linear time
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small = [e for e in lst if e < pivot]
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big = [e for e in lst if e > pivot]
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# if we get lucky, pivot might be the element we want.
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# we can easily see this:
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# small (elements smaller than k)
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# + pivot (kth element)
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# + big (elements larger than k)
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if len(small) == k - 1:
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return pivot
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# pivot is in elements bigger than k
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elif len(small) < k - 1:
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return kth_number(big, k - len(small) - 1)
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# pivot is in elements smaller than k
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else:
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return kth_number(small, k)
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if __name__ == "__main__":
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import doctest
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doctest.testmod()
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