Is there a way of getting the permutations of a string without using loops and list comprehension in python - Just recursion

Question

Is there a way of doing the above without using loops and list comprehension - Just recursion?

This is my solution with for loops:

def permutations(s):        
    if(len(s)==1):
        return [s]
    combs = []
    for i in range(len(s)):
        for perm in permutations(s[:i]+s[i+1:]):
            combs += [s[i]+perm]
    return combs

example:

input :"bca"

output: ["abc", "acb", "bca", "bac", "cab", "cba"]

Any loop can be converted to recursion. See stackoverflow.com/questions/11295346/… — Barmar
– Barmar, Commented Dec 8, 2021 at 17:29
Repost of Get all the permutations of a string without using loops in python - Just recursion. Please ask a question once. If the question is closed, please edit it based on the comments and closure reason rather than asking it again verbatim. The problem is that there's no attempt here. Try making the conversion yourself, then ask a concrete question when and if you get stuck. — ggorlen
– ggorlen, Commented Dec 8, 2021 at 18:49

flakes · Accepted Answer · 2021-12-08 17:44:46Z

1

Basically you move the end-conditions of your loop to an end-condition in the recursion. You can also pass the intermediary result down the chain to avoid for loops appending data to results.

def permutations(s, i=0, curr=""):
    # end what was the for loop
    if i == len(s):
        return []
    if len(s) == 1:
        return [curr + s]
    return [
        *permutations(s[:i] + s[i+1:], 0, curr + s[i]),
        *permutations(s, i+1, curr),
    ]

answered Dec 8, 2021 at 17:44

flakes

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vaizki · Accepted Answer · 2021-12-08 17:43:35Z

0

No need to reinvent the weel:

from itertools import permutations

combs = [ ''.join(p) for p in permutations(s) ]

answered Dec 8, 2021 at 17:43

vaizki

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