Python Program to print all permutations of a given string
A permutation also called an “arrangement number” or “order,” is a rearrangement of the elements of an ordered list S into a one-to-one correspondence with S itself. A string of length n has n! permutation.
Source: Mathword(http://mathworld.wolfram.com/Permutation.html)
Below are the permutations of string ABC.
ABC ACB BAC BCA CBA CAB
Here is a solution that is used as a basis in backtracking.
Python3
# Python program to print all permutations # with duplicates allowed def toString( List ): return ''.join( List ) # Function to print permutations # of string # This function takes three parameters: # 1. String # 2. Starting index of the string # 3. Ending index of the string. def permute(a, l, r): if l = = r: print (toString(a)) else : for i in range (l, r + 1 ): a[l], a[i] = a[i], a[l] permute(a, l + 1 , r) # backtrack a[l], a[i] = a[i], a[l] # Driver code string = "ABC" n = len (string) a = list (string) permute(a, 0 , n - 1 ) # This code is contributed by Bhavya Jain |
Output:
ABC ACB BAC BCA CBA CAB
Algorithm Paradigm: Backtracking
Time Complexity: O(n*n!) Note that there are n! permutations and it requires O(n) time to print a permutation.
Auxiliary Space: O(r – l)
Note: The above solution prints duplicate permutations if there are repeating characters in the input string. Please see the below link for a solution that prints only distinct permutations even if there are duplicates in input.
Print all distinct permutations of a given string with duplicates.
Permutations of a given string using STL
Another approach:
Python3
# Python program to implement # the above approach def permute(s, answer): if ( len (s) = = 0 ): print (answer, end = " " ) return for i in range ( len (s)): ch = s[i] left_substr = s[ 0 :i] right_substr = s[i + 1 :] rest = left_substr + right_substr permute(rest, answer + ch) # Driver Code answer = "" s = input ( "Enter the string : " ) print ( "All possible strings are : " ) permute(s, answer) # This code is contributed by Harshit Srivastava |
Output:
Enter the string : abc All possible strings are : abc acb bac bca cab cba
Time Complexity: O(n*n!) The time complexity is the same as the above approach, i.e. there are n! permutations and it requires O(n) time to print a permutation.
Auxiliary Space: O(|s|)