This notebook was prepared by Donne Martin. Source and license info is on GitHub.

Solution Notebook¶

Problem: Find all permutations of an input string.¶

Constraints¶

Can the input have duplicates?
- Yes
Can the output have duplicates?
- No
Is the output a list of strings?
- Yes
Do we have to output the results in sorted order?
- No
Can we assume the inputs are valid?
- No
Can we assume this fits memory?
- Yes

Test Cases¶

* None -> None
* '' -> ''
* 'AABC' -> ['AABC', 'AACB', 'ABAC', 'ABCA',
             'ACAB', 'ACBA', 'BAAC', 'BACA',
             'BCAA', 'CAAB', 'CABA', 'CBAA']

Algorithm¶

Build a dictionary of {chars: counts} where counts is the number of times each char is found in the input
Loop through each item in the dictionary
- If the counts is 0, continue
- Decrement the current char's count in the dictionary
- Add the current char to the current results
- If the current result is the same length as the input, add it to the results
- Else, recurse
- Backtrack by:
  - Removing the just added current char from the current results
  - Incrementing the current char's count in the dictionary

Complexity:

Time: O(n!)
Space: O(n!) since we are storing the results in an array, or O(n) if we are just printing each result

Code¶

In [1]:

from collections import OrderedDict


class Permutations(object):

    def _build_counts_map(self, string):
        counts_map = OrderedDict()
        for char in string:
            if char in counts_map:
                counts_map[char] += 1
            else:
                counts_map[char] = 1
        return counts_map

    def find_permutations(self, string):
        if string is None or string == '':
            return string
        counts_map = self._build_counts_map(string)
        curr_results = []
        results = []
        self._find_permutations(counts_map, curr_results, results, len(string))
        return results

    def _find_permutations(self, counts_map, curr_result,
                           results, input_length):
        for char in counts_map:
            if counts_map[char] == 0:
                continue
            curr_result.append(char)
            counts_map[char] -= 1
            if len(curr_result) == input_length:
                results.append(''.join(curr_result))
            else:
                self._find_permutations(counts_map, curr_result,
                                        results, input_length)
            counts_map[char] += 1
            curr_result.pop()

Unit Test¶

In [2]:

%%writefile test_permutations.py
import unittest


class TestPermutations(unittest.TestCase):

    def test_permutations(self):
        permutations = Permutations()
        self.assertEqual(permutations.find_permutations(None), None)
        self.assertEqual(permutations.find_permutations(''), '')
        string = 'AABC'
        expected = [
            'AABC', 'AACB', 'ABAC', 'ABCA',
            'ACAB', 'ACBA', 'BAAC', 'BACA',
            'BCAA', 'CAAB', 'CABA', 'CBAA'
        ]
        self.assertEqual(permutations.find_permutations(string), expected)
        print('Success: test_permutations')


def main():
    test = TestPermutations()
    test.test_permutations()


if __name__ == '__main__':
    main()

Overwriting test_permutations.py

In [3]:

%run -i test_permutations.py

Success: test_permutations