#### All Permutations

p
```"""
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.

:param sequence: The input sequence for which permutations are generated.
:param current_sequence: The current permutation being built.
:param index: The current index in the sequence.
:param index_used: list to track which elements are used in permutation.

Example 1:
>>> sequence = [1, 2, 3]
>>> current_sequence = []
>>> index_used = [False, False, False]
>>> create_state_space_tree(sequence, current_sequence, 0, index_used)
[1, 2, 3]
[1, 3, 2]
[2, 1, 3]
[2, 3, 1]
[3, 1, 2]
[3, 2, 1]

Example 2:
>>> sequence = ["A", "B", "C"]
>>> current_sequence = []
>>> index_used = [False, False, False]
>>> create_state_space_tree(sequence, current_sequence, 0, index_used)
['A', 'B', 'C']
['A', 'C', 'B']
['B', 'A', 'C']
['B', 'C', 'A']
['C', 'A', 'B']
['C', 'B', 'A']

Example 3:
>>> sequence = [1]
>>> current_sequence = []
>>> index_used = [False]
>>> create_state_space_tree(sequence, current_sequence, 0, index_used)
[1]
"""

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)
```