2020-10-10 06:19:36 +00:00
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from typing import Any, List
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2019-10-05 05:14:13 +00:00
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"""
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2019-11-16 07:05:00 +00:00
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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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2019-07-06 13:13:50 +00:00
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2019-11-16 07:05:00 +00:00
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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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2019-10-05 05:14:13 +00:00
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"""
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2019-07-06 13:13:50 +00:00
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2020-10-10 06:19:36 +00:00
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def generate_all_subsequences(sequence: List[Any]) -> None:
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2019-10-05 05:14:13 +00:00
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create_state_space_tree(sequence, [], 0)
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2019-07-06 13:13:50 +00:00
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2020-10-10 06:19:36 +00:00
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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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2019-10-05 05:14:13 +00:00
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"""
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2020-09-10 08:31:26 +00:00
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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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2019-07-06 13:13:50 +00:00
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2019-10-05 05:14:13 +00:00
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if index == len(sequence):
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print(current_subsequence)
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return
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2019-07-06 13:13:50 +00:00
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2019-10-05 05:14:13 +00:00
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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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2019-07-06 13:13:50 +00:00
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2019-10-05 05:14:13 +00:00
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"""
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2019-11-16 07:05:00 +00:00
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remove the comment to take an input from the user
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2019-07-06 13:13:50 +00:00
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print("Enter the elements")
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sequence = list(map(int, input().split()))
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2019-10-05 05:14:13 +00:00
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"""
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2019-07-06 13:13:50 +00:00
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sequence = [3, 1, 2, 4]
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generate_all_subsequences(sequence)
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sequence = ["A", "B", "C"]
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generate_all_subsequences(sequence)
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