bbea699f06
Solves an InterviewCake.com problem that returns all of the permutations of a string input. The problem states that it's acceptable to assume that your input string will not have repeated characters, which is why using a Set is acceptable. I like this solution because it builds a permutations tree and then assembles all of the permutations by doing a DFT over that tree.
56 lines
1.3 KiB
Python
56 lines
1.3 KiB
Python
import unittest
|
|
from itertools import permutations
|
|
|
|
|
|
class Node(object):
|
|
def __init__(self, x):
|
|
self.value = x
|
|
self.children = []
|
|
|
|
|
|
def make_tree(c, xs):
|
|
root = Node(c)
|
|
for x in xs:
|
|
root.children.append(make_tree(x, xs - {x}))
|
|
return root
|
|
|
|
|
|
def get_permutations(xs):
|
|
xs = set(xs)
|
|
root = make_tree("", xs)
|
|
q, perms = [], set()
|
|
q.append(("", root))
|
|
while q:
|
|
c, node = q.pop()
|
|
if not node.children:
|
|
perms.add(c)
|
|
else:
|
|
for child in node.children:
|
|
q.append((c + child.value, child))
|
|
return perms
|
|
|
|
|
|
# Tests
|
|
class Test(unittest.TestCase):
|
|
def test_empty_string(self):
|
|
actual = get_permutations('')
|
|
expected = set([''])
|
|
self.assertEqual(actual, expected)
|
|
|
|
def test_one_character_string(self):
|
|
actual = get_permutations('a')
|
|
expected = set(['a'])
|
|
self.assertEqual(actual, expected)
|
|
|
|
def test_two_character_string(self):
|
|
actual = get_permutations('ab')
|
|
expected = set(['ab', 'ba'])
|
|
self.assertEqual(actual, expected)
|
|
|
|
def test_three_character_string(self):
|
|
actual = get_permutations('abc')
|
|
expected = set(['abc', 'acb', 'bac', 'bca', 'cab', 'cba'])
|
|
self.assertEqual(actual, expected)
|
|
|
|
|
|
unittest.main(verbosity=2)
|