Generate all permutation of a set in Python

• Difficulty Level : Easy
• Last Updated : 08 Jun, 2021

Permutation is an arrangement of objects in a specific order. Order of arrangement of object is very important. The number of permutations on a set of n elements is given by  n!.  For example, there are 2! = 2*1 = 2 permutations of {1, 2}, namely {1, 2} and {2, 1}, and 3! = 3*2*1 = 6 permutations of {1, 2, 3}, namely {1, 2, 3}, {1, 3, 2}, {2, 1, 3}, {2, 3, 1}, {3, 1, 2} and {3, 2, 1}.

Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

Method 1 (Backtracking)
We can use the backtracking based recursive solution discussed here.
Method 2
The idea is to one by one extract all elements, place them at first position and recur for remaining list.

Python

# Python function to print permutations of a given list
def permutation(lst):

# If lst is empty then there are no permutations
if len(lst) == 0:
return []

# If there is only one element in lst then, only
# one permutation is possible
if len(lst) == 1:
return [lst]

# Find the permutations for lst if there are
# more than 1 characters

l = [] # empty list that will store current permutation

# Iterate the input(lst) and calculate the permutation
for i in range(len(lst)):
m = lst[i]

# Extract lst[i] or m from the list.  remLst is
# remaining list
remLst = lst[:i] + lst[i+1:]

# Generating all permutations where m is first
# element
for p in permutation(remLst):
l.append([m] + p)
return l

# Driver program to test above function
data = list('123')
for p in permutation(data):
print p

Output:

['1', '2', '3']
['1', '3', '2']
['2', '1', '3']
['2', '3', '1']
['3', '1', '2']
['3', '2', '1']

Method 3 (Direct Function)
We can do it by simply using the built-in permutation function in itertools library. It is the shortest technique to find the permutation.

Python

from itertools import permutations
l = list(permutations(range(1, 4)))
print l

Output:

[(1, 2, 3), (1, 3, 2), (2, 1, 3), (2, 3, 1), (3, 1, 2), (3, 2, 1)]

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