Python/dynamic_programming/longest_increasing_subsequence.py
imp c1b15a86ba
[mypy] Fix type annotations for dynamic programming ()
* Fix mypy error for knapsack.py

* Fix mypy error for longest_increasing_subsequence

* Fix mypy error for fractional_knapsack_2.py
2021-09-03 11:49:23 +02:00

62 lines
1.8 KiB
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([9, 8, 7, 6, 5, 7])
[8]
>>> 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]
isFound = False
i = 1
longest_subseq: list[int] = []
while not isFound and i < array_length:
if array[i] < pivot:
isFound = True
temp_array = [element for element in array[i:] if element >= 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()