Python: string permutations without duplicates

The problem

Given a non-empty string, return a list of all permutations — with duplicates removed when the string has repeated characters.

Examples:

• 'a' → ['a']
• 'ab' → ['ab', 'ba']
• 'abc' → ['abc', 'acb', 'bac', 'bca', 'cab', 'cba'] (6 items)
• 'aabb' → ['aabb', 'abab', 'abba', 'baab', 'baba', 'bbaa'] (6 unique, not 24)

Order in the result does not matter.

Approach

Recursive "pick a character → permute the rest":

1. If the string has ≤ 1 character, return it in a list
2. For each index i: take char = s[i], remainder = s[:i] + s[i+1:]
3. Recursively permute the remainder and prepend char to each permutation
4. list(set(...)) removes duplicates (e.g. for 'aabb')

Solution

def permutations(s):
    if len(s) <= 1:
        return [s]
 
    result = []
 
    for i, char in enumerate(s):
        remaining = s[:i] + s[i+1:]
        for p in permutations(remaining):
            result.append(char + p)
 
    return list(set(result))

Examples

print(permutations('a'))     # ['a']
print(permutations('ab'))    # ['ab', 'ba']  (order may vary)
print(permutations('abc'))   # 6 permutations
print(permutations('aabb'))  # 6 unique permutations
print(len(permutations('abc')))   # 6
print(len(permutations('aabb')))  # 6

Summary

Recursion that picks a "head" and permutes the "tail" — set at the end removes duplicates without complex logic.