The title says it all, this is a pretty standard interview question. Generate all permutations of a given string.

This may seem hard at first but it’s in fact pretty easy once we figure out the logic. Let’s say we’re given a string of length N, and we somehow generated some permutations of length N-1. How do we generate all permutations of length N? Demonstrating with a small example will help. Let the string be ‘LSE’, and we have length 2 permutations ‘SE’ and ‘ES’. How do we incorporate the letter L into these permutations? We just insert it into every possible location in both strings: beginning, middle, and the end. So for ‘SE’ the result is: ‘LSE’, ‘SLE’, ‘SEL’. And for the string ‘ES’ the results is: ‘LES’, ‘ELS’, ‘ESL’. We inserted the character L to every possible location in all the strings. This is it!. We will just use a recursive algorithm and we’re done. Recurse until the string is of length 1 by repeatedly removing the first character, this is the base case and the result is the string itself (the permutation of a string with length 1 is itself). Then apply the above algorithm, at each step insert the character you removed to every possible location in the recursion results and return. Here is the code:

def permutations(word): if len(word)<=1: return [word] #get all permutations of length N-1 perms=permutations(word[1:]) char=word[0] result=[] #iterate over all permutations of length N-1 for perm in perms: #insert the character into every possible location for i in range(len(perm)+1): result.append(perm[:i] + char + perm[i:]) return result

We remove the first character and recurse to get all permutations of length N-1, then we insert that first character into N-1 length strings and obtain all permutations of length N . The complexity is O(N!) because there are N! possible permutations of a string with length N, so it’s optimal. I wouldn’t recommend executing it for strings longer than 10-12 characters though, it’ll take a long time. Not because it’s inefficient, but inherently there are just too many permutations.

This is one of the most common interview questions, so very useful to know by heart.