mirror of
https://github.com/donnemartin/interactive-coding-challenges.git
synced 2024-03-22 13:11:13 +08:00
265 lines
7.4 KiB
Python
265 lines
7.4 KiB
Python
{
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"cells": [
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"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)."
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"# Solution Notebook"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"## Problem: Return all subsets of a set.\n",
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"\n",
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"* [Constraints](#Constraints)\n",
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"* [Test Cases](#Test-Cases)\n",
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"* [Algorithm](#Algorithm)\n",
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"* [Code](#Code)\n",
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"* [Unit Test](#Unit-Test)"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"## Constraints\n",
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"\n",
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"* Should the resulting subsets be unique?\n",
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" * Yes, treat 'ab' and 'bc' as the same\n",
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"* Is the empty set included as a subset?\n",
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" * Yes\n",
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"* Are the inputs unique?\n",
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" * No\n",
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"* Can we assume the inputs are valid?\n",
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" * No\n",
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"* Can we assume this fits memory?\n",
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" * Yes"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"## Test Cases\n",
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"\n",
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"<pre>\n",
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"* None -> None\n",
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"* '' -> ['']\n",
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"* 'a' -> ['a', '']\n",
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"* 'ab' -> ['a', 'ab', 'b', '']\n",
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"* 'abc' -> ['a', 'ab', 'abc', 'ac',\n",
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" 'b', 'bc', 'c', '']\n",
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"* 'aabc' -> ['a', 'aa', 'aab', 'aabc', \n",
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" 'aac', 'ab', 'abc', 'ac', \n",
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" 'b', 'bc', 'c', '']\n",
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"</pre>"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"## Algorithm\n",
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"\n",
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"* Build a dictionary of {chars: counts} where counts is the number of times each char is found in the input\n",
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"* Loop through each item in the dictionary\n",
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" * Keep track of the current index (first item will have current index 0)\n",
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" * If the char's count is 0, continue\n",
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" * Decrement the current char's count in the dictionary\n",
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" * Add the current char to the current results\n",
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" * Add the current result to the results\n",
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" * Recurse, passing in the current index as the new starting point index\n",
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" * When we recurse, we'll check if current index < starting point index, and if so, continue\n",
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" * This avoids duplicate results such as 'ab' and 'bc'\n",
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" * Backtrack by:\n",
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" * Removing the just added current char from the current results\n",
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" * Incrementing the current char's count in the dictionary\n",
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"\n",
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"Complexity:\n",
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"* Time: O(2^n)\n",
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"* Space: O(2^n) if we are saving each result, or O(n) if we are just printing each result\n",
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"\n",
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"We are doubling the number of operations every time we add an element to the results: O(2^n).\n",
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"\n",
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"Note, you could also use the following method to solve this problem:\n",
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"\n",
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"<pre>\n",
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"number binary subset\n",
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"0 000 {}\n",
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"1 001 {c}\n",
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"2 010 {b}\n",
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"3 011 {b,c}\n",
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"4 100 {a}\n",
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"5 101 {a,c}\n",
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"6 110 {a,b}\n",
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"7 111 {a,b,c}\n",
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"</pre>"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"## Code"
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]
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},
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{
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"cell_type": "code",
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"execution_count": 1,
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"metadata": {},
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"outputs": [],
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"source": [
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"from collections import OrderedDict\n",
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"\n",
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"\n",
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"class Combinatoric(object):\n",
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"\n",
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" def _build_counts_map(self, string):\n",
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" counts_map = OrderedDict()\n",
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" for char in string:\n",
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" if char in counts_map:\n",
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" counts_map[char] += 1\n",
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" else:\n",
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" counts_map[char] = 1\n",
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" return counts_map\n",
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"\n",
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" def find_power_set(self, string):\n",
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" if string is None:\n",
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" return string\n",
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" if string == '':\n",
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" return ['']\n",
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" counts_map = self._build_counts_map(string)\n",
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" curr_results = []\n",
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" results = []\n",
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" self._find_power_set(counts_map, curr_results,\n",
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" results, index=0)\n",
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" results.append('')\n",
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" return results\n",
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"\n",
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" def _find_power_set(self, counts_map, curr_result,\n",
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" results, index):\n",
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" for curr_index, char in enumerate(counts_map):\n",
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" if curr_index < index or counts_map[char] == 0:\n",
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" continue\n",
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" curr_result.append(char)\n",
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" counts_map[char] -= 1\n",
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" results.append(''.join(curr_result))\n",
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" self._find_power_set(counts_map, curr_result,\n",
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" results, curr_index)\n",
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" counts_map[char] += 1\n",
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" curr_result.pop()"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"## Unit Test"
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]
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},
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{
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"cell_type": "code",
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"execution_count": 2,
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"metadata": {},
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"outputs": [
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{
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"name": "stdout",
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"output_type": "stream",
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"text": [
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"Overwriting test_power_set.py\n"
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]
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}
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],
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"source": [
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"%%writefile test_power_set.py\n",
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"import unittest\n",
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"\n",
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"\n",
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"class TestPowerSet(unittest.TestCase):\n",
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"\n",
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" def test_power_set(self):\n",
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" input_set = ''\n",
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" expected = ['']\n",
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" self.run_test(input_set, expected)\n",
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" input_set = 'a'\n",
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" expected = ['a', '']\n",
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" self.run_test(input_set, expected)\n",
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" input_set = 'ab'\n",
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" expected = ['a', 'ab', 'b', '']\n",
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" self.run_test(input_set, expected)\n",
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" input_set = 'abc'\n",
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" expected = ['a', 'ab', 'abc', 'ac',\n",
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" 'b', 'bc', 'c', '']\n",
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" self.run_test(input_set, expected)\n",
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" input_set = 'aabc'\n",
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" expected = ['a', 'aa', 'aab', 'aabc', \n",
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" 'aac', 'ab', 'abc', 'ac', \n",
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" 'b', 'bc', 'c', '']\n",
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" self.run_test(input_set, expected)\n",
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" print('Success: test_power_set')\n",
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"\n",
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" def run_test(self, input_set, expected):\n",
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" combinatoric = Combinatoric()\n",
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" result = combinatoric.find_power_set(input_set)\n",
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" self.assertEqual(result, expected)\n",
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"\n",
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"\n",
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"def main():\n",
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" test = TestPowerSet()\n",
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" test.test_power_set()\n",
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"\n",
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"\n",
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"if __name__ == '__main__':\n",
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" main()"
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]
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},
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{
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"cell_type": "code",
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"execution_count": 3,
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"metadata": {},
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"outputs": [
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{
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"name": "stdout",
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"output_type": "stream",
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"text": [
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"Success: test_power_set\n"
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]
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}
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],
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"source": [
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"%run -i test_power_set.py"
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]
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}
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],
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"metadata": {
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"kernelspec": {
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"display_name": "Python 3",
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"language": "python",
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"name": "python3"
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},
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"language_info": {
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"codemirror_mode": {
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"name": "ipython",
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"version": 3
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},
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"file_extension": ".py",
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"mimetype": "text/x-python",
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"name": "python",
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"nbconvert_exporter": "python",
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"pygments_lexer": "ipython3",
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"version": "3.7.2"
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}
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},
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"nbformat": 4,
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"nbformat_minor": 1
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}
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