"""
 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)