"""
In this problem, we want to determine all possible permutations
of the given sequence. We use backtracking to solve this problem.
Time complexity: O(n! * n),
where n denotes the length of the given sequence.
"""
from __future__ import annotations
def generate_all_permutations(sequence: list[int | str]) -> None:
create_state_space_tree(sequence, [], 0, [0 for i in range(len(sequence))])
def create_state_space_tree(
sequence: list[int | str],
current_sequence: list[int | str],
index: int,
index_used: list[int],
) -> None:
"""
Creates a state space tree to iterate through each branch using DFS.
We know that each state has exactly len(sequence) - index children.
It terminates when it reaches the end of the given sequence.
"""
if index == len(sequence):
print(current_sequence)
return
for i in range(len(sequence)):
if not index_used[i]:
current_sequence.append(sequence[i])
index_used[i] = True
create_state_space_tree(sequence, current_sequence, index + 1, index_used)
current_sequence.pop()
index_used[i] = False
"""
remove the comment to take an input from the user
print("Enter the elements")
sequence = list(map(int, input().split()))
"""
sequence: list[int | str] = [3, 1, 2, 4]
generate_all_permutations(sequence)
sequence_2: list[int | str] = ["A", "B", "C"]
generate_all_permutations(sequence_2)