why do you not simple do:
from itertools import permutations
perms = [''.join[p] for p in permutations[['s','t','a','c','k']]]
print perms
print len[perms]
print len[set[perms]]
you get no duplicate as you can see :
['stack', 'stakc', 'stcak', 'stcka', 'stkac', 'stkca', 'satck', 'satkc',
'sactk', 'sackt', 'saktc', 'sakct', 'sctak', 'sctka', 'scatk', 'scakt', 'sckta',
'sckat', 'sktac', 'sktca', 'skatc', 'skact', 'skcta', 'skcat', 'tsack',
'tsakc', 'tscak', 'tscka', 'tskac', 'tskca', 'tasck', 'taskc', 'tacsk', 'tacks',
'taksc', 'takcs', 'tcsak', 'tcska', 'tcask', 'tcaks', 'tcksa', 'tckas', 'tksac',
'tksca', 'tkasc', 'tkacs', 'tkcsa', 'tkcas', 'astck', 'astkc', 'asctk', 'asckt',
'asktc', 'askct', 'atsck', 'atskc', 'atcsk', 'atcks', 'atksc', 'atkcs', 'acstk',
'acskt', 'actsk', 'actks', 'ackst', 'ackts', 'akstc', 'aksct', 'aktsc', 'aktcs',
'akcst', 'akcts', 'cstak', 'cstka', 'csatk', 'csakt', 'cskta', 'cskat', 'ctsak',
'ctska', 'ctask', 'ctaks', 'ctksa', 'ctkas', 'castk', 'caskt', 'catsk', 'catks',
'cakst', 'cakts', 'cksta', 'cksat', 'cktsa', 'cktas', 'ckast', 'ckats', 'kstac',
'kstca', 'ksatc', 'ksact', 'kscta', 'kscat', 'ktsac', 'ktsca', 'ktasc', 'ktacs',
'ktcsa', 'ktcas', 'kastc', 'kasct', 'katsc', 'katcs', 'kacst', 'kacts', 'kcsta',
'kcsat', 'kctsa', 'kctas', 'kcast', 'kcats']
120
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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. Examples:
Input : str = 'ABC'
Output : ABC
ACB
BAC
BCA
CAB
CBAWe 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!]
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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']