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

# Solution Notebook¶

## Constraints¶

• Should the resulting subsets be unique?
• Yes, treat 'ab' and 'bc' as the same
• Is the empty set included as a subset?
• Yes
• Are the inputs unique?
• No
• Can we assume the inputs are valid?
• No
• Can we assume this fits memory?
• Yes

## Test Cases¶

* None -> None
* '' -> ['']
* 'a' -> ['a', '']
* 'ab' -> ['a', 'ab', 'b', '']
* 'abc' -> ['a', 'ab', 'abc', 'ac',
'b', 'bc', 'c', '']
* 'aabc' -> ['a', 'aa', 'aab', 'aabc',
'aac', 'ab', 'abc', 'ac',
'b', 'bc', 'c', '']


## 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
• Keep track of the current index (first item will have current index 0)
• If the char's count is 0, continue
• Decrement the current char's count in the dictionary
• Add the current char to the current results
• Add the current result to the results
• Recurse, passing in the current index as the new starting point index
• When we recurse, we'll check if current index < starting point index, and if so, continue
• This avoids duplicate results such as 'ab' and 'bc'
• Backtrack by:
• Removing the just added current char from the current results
• Incrementing the current char's count in the dictionary

Complexity:

• Time: O(2^n)
• Space: O(2^n) if we are saving each result, or O(n) if we are just printing each result

We are doubling the number of operations every time we add an element to the results: O(2^n).

Note, you could also use the following method to solve this problem:

number binary  subset
0      000      {}
1      001      {c}
2      010      {b}
3      011      {b,c}
4      100      {a}
5      101      {a,c}
6      110      {a,b}
7      111      {a,b,c}


## Code¶

In [1]:
from collections import OrderedDict

class Combinatoric(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_power_set(self, string):
if string is None:
return string
if string == '':
return ['']
counts_map = self._build_counts_map(string)
curr_results = []
results = []
self._find_power_set(counts_map, curr_results,
results, index=0)
results.append('')
return results

def _find_power_set(self, counts_map, curr_result,
results, index):
for curr_index, char in enumerate(counts_map):
if curr_index < index or counts_map[char] == 0:
continue
curr_result.append(char)
counts_map[char] -= 1
results.append(''.join(curr_result))
self._find_power_set(counts_map, curr_result,
results, curr_index)
counts_map[char] += 1
curr_result.pop()


## Unit Test¶

In [2]:
%%writefile test_power_set.py
import unittest

class TestPowerSet(unittest.TestCase):

def test_power_set(self):
input_set = ''
expected = ['']
self.run_test(input_set, expected)
input_set = 'a'
expected = ['a', '']
self.run_test(input_set, expected)
input_set = 'ab'
expected = ['a', 'ab', 'b', '']
self.run_test(input_set, expected)
input_set = 'abc'
expected = ['a', 'ab', 'abc', 'ac',
'b', 'bc', 'c', '']
self.run_test(input_set, expected)
input_set = 'aabc'
expected = ['a', 'aa', 'aab', 'aabc',
'aac', 'ab', 'abc', 'ac',
'b', 'bc', 'c', '']
self.run_test(input_set, expected)
print('Success: test_power_set')

def run_test(self, input_set, expected):
combinatoric = Combinatoric()
result = combinatoric.find_power_set(input_set)
self.assertEqual(result, expected)

def main():
test = TestPowerSet()
test.test_power_set()

if __name__ == '__main__':
main()

Overwriting test_power_set.py

In [3]:
%run -i test_power_set.py

Success: test_power_set