In this problem, we want to determine all possible permutations of the given sequence. We use backtracking to solve this problem. Time complexity: On!  n, where n denotes the length of the given sequence. Creates a state space tree to iterate through each branch using DFS. We know that each state has exactly lensequence  index children. It terminates when it reaches the end of the given sequence. remove the comment to take an input from the user printEnter the elements sequence  listmapint, input.split
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)
