Python/dynamic_programming/longest_increasing_subsequence.py
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Fix dynamic_programming/longest_increasing_subsequence.py (#12517)
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Python

"""
Author : Mehdi ALAOUI
This is a pure Python implementation of Dynamic Programming solution to the longest
increasing subsequence of a given sequence.
The problem is:
Given an array, to find the longest and increasing sub-array in that given array and
return it.
Example:
``[10, 22, 9, 33, 21, 50, 41, 60, 80]`` as input will return
``[10, 22, 33, 41, 60, 80]`` as output
"""
from __future__ import annotations
def longest_subsequence(array: list[int]) -> list[int]: # This function is recursive
"""
Some examples
>>> longest_subsequence([10, 22, 9, 33, 21, 50, 41, 60, 80])
[10, 22, 33, 41, 60, 80]
>>> longest_subsequence([4, 8, 7, 5, 1, 12, 2, 3, 9])
[1, 2, 3, 9]
>>> longest_subsequence([28, 26, 12, 23, 35, 39])
[12, 23, 35, 39]
>>> longest_subsequence([9, 8, 7, 6, 5, 7])
[5, 7]
>>> longest_subsequence([1, 1, 1])
[1, 1, 1]
>>> longest_subsequence([])
[]
"""
array_length = len(array)
# If the array contains only one element, we return it (it's the stop condition of
# recursion)
if array_length <= 1:
return array
# Else
pivot = array[0]
is_found = False
i = 1
longest_subseq: list[int] = []
while not is_found and i < array_length:
if array[i] < pivot:
is_found = True
temp_array = array[i:]
temp_array = longest_subsequence(temp_array)
if len(temp_array) > len(longest_subseq):
longest_subseq = temp_array
else:
i += 1
temp_array = [element for element in array[1:] if element >= pivot]
temp_array = [pivot, *longest_subsequence(temp_array)]
if len(temp_array) > len(longest_subseq):
return temp_array
else:
return longest_subseq
if __name__ == "__main__":
import doctest
doctest.testmod()