{
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"source": [
"This notebook was prepared by [Donne Martin](https://github.com/donnemartin). Source and license info is on [GitHub](https://github.com/donnemartin/interactive-coding-challenges)."
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"# Solution Notebook"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Problem: Return all subsets of a set.\n",
"\n",
"* [Constraints](#Constraints)\n",
"* [Test Cases](#Test-Cases)\n",
"* [Algorithm](#Algorithm)\n",
"* [Code](#Code)\n",
"* [Unit Test](#Unit-Test)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Constraints\n",
"\n",
"* Should the resulting subsets be unique?\n",
" * Yes, treat 'ab' and 'bc' as the same\n",
"* Is the empty set included as a subset?\n",
" * Yes\n",
"* Are the inputs unique?\n",
" * No\n",
"* Can we assume the inputs are valid?\n",
" * No\n",
"* Can we assume this fits memory?\n",
" * Yes"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Test Cases\n",
"\n",
"\n",
"* None -> None\n",
"* '' -> ['']\n",
"* 'a' -> ['a', '']\n",
"* 'ab' -> ['a', 'ab', 'b', '']\n",
"* 'abc' -> ['a', 'ab', 'abc', 'ac',\n",
" 'b', 'bc', 'c', '']\n",
"* 'aabc' -> ['a', 'aa', 'aab', 'aabc', \n",
" 'aac', 'ab', 'abc', 'ac', \n",
" 'b', 'bc', 'c', '']\n",
""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Algorithm\n",
"\n",
"* Build a dictionary of {chars: counts} where counts is the number of times each char is found in the input\n",
"* Loop through each item in the dictionary\n",
" * Keep track of the current index (first item will have current index 0)\n",
" * If the char's count is 0, continue\n",
" * Decrement the current char's count in the dictionary\n",
" * Add the current char to the current results\n",
" * Add the current result to the results\n",
" * Recurse, passing in the current index as the new starting point index\n",
" * When we recurse, we'll check if current index < starting point index, and if so, continue\n",
" * This avoids duplicate results such as 'ab' and 'bc'\n",
" * Backtrack by:\n",
" * Removing the just added current char from the current results\n",
" * Incrementing the current char's count in the dictionary\n",
"\n",
"Complexity:\n",
"* Time: O(2^n)\n",
"* Space: O(2^n) if we are saving each result, or O(n) if we are just printing each result\n",
"\n",
"We are doubling the number of operations every time we add an element to the results: O(2^n).\n",
"\n",
"Note, you could also use the following method to solve this problem:\n",
"\n",
"\n",
"number binary subset\n",
"0 000 {}\n",
"1 001 {c}\n",
"2 010 {b}\n",
"3 011 {b,c}\n",
"4 100 {a}\n",
"5 101 {a,c}\n",
"6 110 {a,b}\n",
"7 111 {a,b,c}\n",
""
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Code"
]
},
{
"cell_type": "code",
"execution_count": 1,
"metadata": {
"collapsed": true
},
"outputs": [],
"source": [
"from collections import OrderedDict\n",
"\n",
"\n",
"class Combinatoric(object):\n",
"\n",
" def _build_counts_map(self, string):\n",
" counts_map = OrderedDict()\n",
" for char in string:\n",
" if char in counts_map:\n",
" counts_map[char] += 1\n",
" else:\n",
" counts_map[char] = 1\n",
" return counts_map\n",
"\n",
" def find_power_set(self, string):\n",
" if string is None:\n",
" return string\n",
" if string == '':\n",
" return ['']\n",
" counts_map = self._build_counts_map(string)\n",
" curr_results = []\n",
" results = []\n",
" self._find_power_set(counts_map, curr_results,\n",
" results, index=0)\n",
" results.append('')\n",
" return results\n",
"\n",
" def _find_power_set(self, counts_map, curr_result,\n",
" results, index):\n",
" for curr_index, char in enumerate(counts_map):\n",
" if curr_index < index or counts_map[char] == 0:\n",
" continue\n",
" curr_result.append(char)\n",
" counts_map[char] -= 1\n",
" results.append(''.join(curr_result))\n",
" self._find_power_set(counts_map, curr_result,\n",
" results, curr_index)\n",
" counts_map[char] += 1\n",
" curr_result.pop()"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Unit Test"
]
},
{
"cell_type": "code",
"execution_count": 2,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Overwriting test_power_set.py\n"
]
}
],
"source": [
"%%writefile test_power_set.py\n",
"from nose.tools import assert_equal\n",
"\n",
"\n",
"class TestPowerSet(object):\n",
"\n",
" def test_power_set(self):\n",
" input_set = ''\n",
" expected = ['']\n",
" self.run_test(input_set, expected)\n",
" input_set = 'a'\n",
" expected = ['a', '']\n",
" self.run_test(input_set, expected)\n",
" input_set = 'ab'\n",
" expected = ['a', 'ab', 'b', '']\n",
" self.run_test(input_set, expected)\n",
" input_set = 'abc'\n",
" expected = ['a', 'ab', 'abc', 'ac',\n",
" 'b', 'bc', 'c', '']\n",
" self.run_test(input_set, expected)\n",
" input_set = 'aabc'\n",
" expected = ['a', 'aa', 'aab', 'aabc', \n",
" 'aac', 'ab', 'abc', 'ac', \n",
" 'b', 'bc', 'c', '']\n",
" self.run_test(input_set, expected)\n",
" print('Success: test_power_set')\n",
"\n",
" def run_test(self, input_set, expected):\n",
" combinatoric = Combinatoric()\n",
" result = combinatoric.find_power_set(input_set)\n",
" assert_equal(result, expected)\n",
"\n",
"\n",
"def main():\n",
" test = TestPowerSet()\n",
" test.test_power_set()\n",
"\n",
"\n",
"if __name__ == '__main__':\n",
" main()"
]
},
{
"cell_type": "code",
"execution_count": 3,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Success: test_power_set\n"
]
}
],
"source": [
"%run -i test_power_set.py"
]
}
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