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* Pyupgrade to Python 3.9 * updating DIRECTORY.md Co-authored-by: github-actions <${GITHUB_ACTOR}@users.noreply.github.com>
52 lines
1.5 KiB
Python
52 lines
1.5 KiB
Python
"""
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In this problem, we want to determine all possible permutations
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of the given sequence. We use backtracking to solve this problem.
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Time complexity: O(n! * 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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def generate_all_permutations(sequence: list[int | str]) -> None:
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create_state_space_tree(sequence, [], 0, [0 for i in range(len(sequence))])
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def create_state_space_tree(
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sequence: list[int | str],
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current_sequence: list[int | str],
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index: int,
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index_used: list[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 len(sequence) - index 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_sequence)
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return
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for i in range(len(sequence)):
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if not index_used[i]:
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current_sequence.append(sequence[i])
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index_used[i] = True
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create_state_space_tree(sequence, current_sequence, index + 1, index_used)
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current_sequence.pop()
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index_used[i] = False
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"""
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remove the comment to take an input from the user
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print("Enter the elements")
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sequence = list(map(int, input().split()))
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"""
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sequence: list[int | str] = [3, 1, 2, 4]
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generate_all_permutations(sequence)
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sequence_2: list[int | str] = ["A", "B", "C"]
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generate_all_permutations(sequence_2)
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