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)]
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