Generate all permutations of a sequence, using backtracking (#962)

* Fix typo

* Add all_permutations algorithm to backtracking directory
This commit is contained in:
Erfan Alimohammadi 2019-07-06 17:49:36 +04:30 committed by Andrii Siriak
parent cc4cf3ece7
commit 839160f83a

View File

@ -0,0 +1,45 @@
'''
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!),
where n denotes the length of the given sequence.
'''
def generate_all_permutations(sequence):
create_state_space_tree(sequence, [], 0, [0 for i in range(len(sequence))])
def create_state_space_tree(sequence, current_sequence, index, index_used):
'''
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 = [3, 1, 2, 4]
generate_all_permutations(sequence)
sequence = ["A", "B", "C"]
generate_all_permutations(sequence)