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* Pyupgrade to Python 3.9
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43 lines
1.2 KiB
Python
43 lines
1.2 KiB
Python
"""
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In this problem, we want to determine all possible subsequences
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of the given sequence. We use backtracking to solve this problem.
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Time complexity: O(2^n),
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where n denotes the length of the given sequence.
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"""
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from __future__ import annotations
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from typing import Any
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def generate_all_subsequences(sequence: list[Any]) -> None:
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create_state_space_tree(sequence, [], 0)
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def create_state_space_tree(
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sequence: list[Any], current_subsequence: list[Any], index: int
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) -> None:
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"""
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Creates a state space tree to iterate through each branch using DFS.
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We know that each state has exactly two children.
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It terminates when it reaches the end of the given sequence.
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"""
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if index == len(sequence):
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print(current_subsequence)
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return
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create_state_space_tree(sequence, current_subsequence, index + 1)
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current_subsequence.append(sequence[index])
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create_state_space_tree(sequence, current_subsequence, index + 1)
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current_subsequence.pop()
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if __name__ == "__main__":
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seq: list[Any] = [3, 1, 2, 4]
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generate_all_subsequences(seq)
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seq.clear()
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seq.extend(["A", "B", "C"])
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generate_all_subsequences(seq)
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