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}.
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.
Python3
# Python function to print permutations of a given listdef 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 functiondata = 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.
Python3
from itertools import permutationsl = 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)]
This article is contributed by Arpit Agarwal. If you like Lazyroar and would like to contribute, you can also write an article and mail your article to review-team@neveropen.co.za. See your article appearing on the Lazyroar main page and help other Geeks.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above
