Finding all possible permutations of a given string in python

Solution 1:

The itertools module has a useful method called permutations(). The documentation says:

itertools.permutations(iterable[, r])

Return successive r length permutations of elements in the iterable.

If r is not specified or is None, then r defaults to the length of the iterable and all possible full-length permutations are generated.

Permutations are emitted in lexicographic sort order. So, if the input iterable is sorted, the permutation tuples will be produced in sorted order.

You'll have to join your permuted letters as strings though.

>>> from itertools import permutations
>>> perms = [''.join(p) for p in permutations('stack')]
>>> perms

['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']

If you find yourself troubled by duplicates, try fitting your data into a structure with no duplicates like a set:

>>> perms = [''.join(p) for p in permutations('stacks')]
>>> len(perms)
720
>>> len(set(perms))
360

Thanks to @pst for pointing out that this is not what we'd traditionally think of as a type cast, but more of a call to the set() constructor.

Solution 2:

You can get all N! permutations without much code

def permutations(string, step = 0):

    # if we've gotten to the end, print the permutation
    if step == len(string):
        print "".join(string)

    # everything to the right of step has not been swapped yet
    for i in range(step, len(string)):

        # copy the string (store as array)
        string_copy = [character for character in string]

        # swap the current index with the step
        string_copy[step], string_copy[i] = string_copy[i], string_copy[step]

        # recurse on the portion of the string that has not been swapped yet (now it's index will begin with step + 1)
        permutations(string_copy, step + 1)

Solution 3:

Here is another way of doing the permutation of string with minimal code based on bactracking. We basically create a loop and then we keep swapping two characters at a time, Inside the loop we'll have the recursion. Notice,we only print when indexers reaches the length of our string. Example: ABC i for our starting point and our recursion param j for our loop

here is a visual help how it works from left to right top to bottom (is the order of permutation)

the code :

def permute(data, i, length): 
    if i==length: 
        print(''.join(data) )
    else: 
        for j in range(i,length): 
            #swap
            data[i], data[j] = data[j], data[i] 
            permute(data, i+1, length) 
            data[i], data[j] = data[j], data[i]  
  

string = "ABC"
n = len(string) 
data = list(string) 
permute(data, 0, n)

Solution 4:

Stack Overflow users have already posted some strong solutions but I wanted to show yet another solution. This one I find to be more intuitive

The idea is that for a given string: we can recurse by the algorithm (pseudo-code):

permutations = char + permutations(string - char) for char in string

I hope it helps someone!

def permutations(string):
    """
    Create all permutations of a string with non-repeating characters
    """
    permutation_list = []
    if len(string) == 1:
        return [string]
    else:
        for char in string:
            [permutation_list.append(char + a) for a in permutations(string.replace(char, "", 1))]
    return permutation_list