CODEWITHSUNDEEP
pythonloopsrecursionpermutationpython-itertools
kanyeeast
kanyeeast

Reputation: 45

Permutations without itertools for two values (using recursion!)

Stackoverflow, I am once again asking for your help.

I'm aware there are other threads about this but I'll explain what makes my assignment different.

Basically my function would get a list of 0s and 1s, and return all the possible orders for the string. For example for "0111" we will get "0111", "1011", "1101", "1110".

Here's my code:

def permutations(string):
    if len(string) == 1:
        return [string]

    lst = []

    for j in range(len(string)):
        remaining_elements = ''.join([string[i] for i in range(len(string)) if i != j])
        mini_perm = permutations(remaining_elements)

        for perm in mini_perm:
            new_str = string[j] + perm
            if new_str not in lst:
                lst.append(new_str)

    return lst

The problem is when I run a string like "000000000011" it takes a very long time to process. There is supposed to be a more efficient way to do it because it's just two numbers. So I shouldn't be using the indexes?

Please help me if you can figure out a more efficient say to do this.

(I am allowed to use loops just have to use recursion as well!)

Upvotes: 0

Views: 175

Answers (3)

kanyeeast
kanyeeast

Reputation: 45

posting an answer someone gave me. Thanks for your responses!:

def permutations(zeroes, ones, lst, perm):
    if zeroes == 0 and ones == 0:
        lst.append(perm)
        return
    elif zeroes < 0 or ones < 0:
        return
    
    permutations(zeroes - 1, ones, lst, perm + '0')
    permutations(zeroes, ones - 1, lst, perm + '1')

Upvotes: 0

Ajax1234
Ajax1234

Reputation: 71451

You can also use collections.Counter with a recursive generator function:

from collections import Counter
def permute(d):
   counts = Counter(d)
   def get_permuations(c, s = []):
     if len(s) == sum(counts.values()):
        yield ''.join(s)
     else:
        for a, b in c.items():
           for i in range(1, b+1):
              yield from get_permuations({**c, a:b - i}, s+([a]*i))
   return list(set(get_permuations(counts)))

print(permute("0111"))
print(permute("000000000011"))

Output:

['0111', '1110', '1101', '1011']
['010000100000', '100000000001', '010000001000', '000000100001', '011000000000', '100000000010', '001001000000', '000000011000', '100000001000', '100000100000', '100001000000', '001000100000', '100010000000', '000000001100', '000100000100', '010010000000', '000000000011', '000000100010', '101000000000', '110000000000', '100000010000', '000100001000', '000001001000', '000000000101', '000000100100', '010000000001', '001000000100', '001000000010', '000110000000', '000011000000', '000001100000', '000000110000', '001000000001', '000010001000', '000100100000', '000001000001', '000010000001', '001100000000', '000100000001', '001000001000', '010000000100', '010000010000', '000000010001', '001000010000', '010001000000', '100000000100', '100100000000', '000000001001', '010100000000', '000010100000', '010000000010', '000000001010', '000010000100', '001010000000', '000000010010', '000001000010', '000100000010', '000101000000', '000000010100', '000100010000', '000000000110', '000001000100', '000010010000', '000000101000', '000001010000', '000010000010']

Upvotes: 1

flooo
flooo

Reputation: 26

Here is an example for creating permutations with recursion that is more efficient:

def permute(string):
    string = list(string)
    n = len(string)

    # Base conditions
    # If length is 0 or 1, there is only 1 permutation
    if n in [0, 1]:
        return [string]

    # If length is 2, then there are only two permutations
    # Example: [1,2] and [2,1]
    if n == 2:
        return [string, string[::-1]]

    res = []
    # For every number in array, choose 1 number and permute the remaining
    # by calling permute recursively
    for i in range(n):
        permutations = permute(string[:i] + string[i+1:])
        for p in permutations:
            res.append([''.join(str(n) for n in [string[i]] + p)])

    return res

This should also work for permute('000000000011') - hope it helps!

Upvotes: 1

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