Python | Permutation of a given string using inbuilt function
Last Updated :
07 Apr, 2023
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. Examples:
Input : str = 'ABC'
Output : ABC
ACB
BAC
BCA
CAB
CBA
We have existing solution for this problem please refer Permutations of a given string using STL link. We can also solve this problem in python using inbuilt function permutations(iterable).
Python3
from itertools import permutations
def allPermutations(str):
permList = permutations(str)
for perm in list(permList):
print (''.join(perm))
if __name__ == "__main__":
str = 'ABC'
allPermutations(str)
|
Output:
ABC
ACB
BAC
BCA
CAB
CBA
Permutation and Combination in Python Permutations of a given string with repeating characters The idea is to use dictionary to avoid printing duplicates.
Python3
from itertools import permutations
import string
s = "GEEK"
a = string.ascii_letters
p = permutations(s)
d = []
for i in list(p):
if (i not in d):
d.append(i)
print(''.join(i))
|
Output:
GEEK
GEKE
GKEE
EGEK
EGKE
EEGK
EEKG
EKGE
EKEG
KGEE
KEGE
KEEG
Time Complexity: O(n!) where n is the size of the string.
Auxiliary Space: O(n!)
Approach: Using Recursion
This code generates all possible permutations of a given string using recursion.
Algorithm
1. Take input string from the user
2. Define a recursive function to generate all possible permutations of the given string
3. Base case: If the length of the input string is 1, return the string
4. Recursive case: For each character in the string, fix it and recursively find all the permutations of the remaining characters
5. Combine the fixed character with each of the permutations obtained from the recursive call and append it to the list of permutations
6. Return the list of permutations
Python3
def generate_permutations(string):
if len(string) == 1:
return [string]
permutations = []
for i in range(len(string)):
fixed_char = string[i]
remaining_chars = string[:i] + string[i+1:]
for perm in generate_permutations(remaining_chars):
permutations.append(fixed_char + perm)
return permutations
string ='GEEK'
permutations_list = generate_permutations(string)
z=set(permutations_list)
for perm in z:
print(perm)
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Output
GKEE
KEEG
KEGE
KGEE
EKGE
EGEK
EEGK
GEKE
EGKE
EKEG
GEEK
EEKG
Time complexity: O(n*n!), where n is the length of the input string.
Auxiliary Space: O(n*n!).
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