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Given a string, write a Python program to find out all possible permutations of a string. Let’s discuss a few methods to solve the
problem.
Method #1: Using Naive Method
Python3
ini_str = "abc"
print["Initial string", ini_str]
result = []
def permute[data, i, length]:
if i == length:
result.append[''.join[data] ]
else:
for j in range[i, length]:
data[i], data[j] = data[j], data[i]
permute[data, i +
1, length]
data[i], data[j] = data[j], data[i]
permute[list[ini_str], 0, len[ini_str]]
print["Resultant permutations", str[result]]
Output:
Initial string abc Resultant permutations ['abc', 'acb', 'bac', 'bca', 'cba', 'cab']
Method #2: Using itertools
Python3
from itertools import permutations
ini_str = "abc"
print["Initial string", ini_str]
permutation = [''.join[p] for p in permutations[ini_str]]
print["Resultant List", str[permutation]]
Output:
Initial string abc Resultant List ['abc', 'acb', 'bac', 'bca', 'cab', 'cba']
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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[//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
def toString[List]:
return ''.join[List]
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]
a[l], a[i] = a[i], a[l]
string = "ABC"
n = len[string]
a = list[string]
permute[a, 0, n-1]
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
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]
answer = ""
s = input["Enter the string : "]
print["All possible strings are : "]
permute[s, answer]
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|]