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