{
 "cells": [
  {
   "cell_type": "markdown",
   "metadata": {},
   "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",
    "<pre>\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",
    "</pre>"
   ]
  },
  {
   "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 acount 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",
    "<pre>\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",
    "</pre>"
   ]
  },
  {
   "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"
   ]
  }
 ],
 "metadata": {
  "kernelspec": {
   "display_name": "Python 3",
   "language": "python",
   "name": "python3"
  },
  "language_info": {
   "codemirror_mode": {
    "name": "ipython",
    "version": 3
   },
   "file_extension": ".py",
   "mimetype": "text/x-python",
   "name": "python",
   "nbconvert_exporter": "python",
   "pygments_lexer": "ipython3",
   "version": "3.4.3"
  }
 },
 "nbformat": 4,
 "nbformat_minor": 0
}
