mirror of
https://github.com/donnemartin/interactive-coding-challenges.git
synced 2024-03-22 13:11:13 +08:00
266 lines
7.0 KiB
Python
266 lines
7.0 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](http://donnemartin.com). 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: Determine if a string is a permutation of another string.\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: Compare Sorted Strings](#Algorithm:-Compare-Sorted-Strings)\n",
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"* [Code: Compare Sorted Strings](#Code:-Compare-Sorted-Strings)\n",
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"* [Algorithm: Hashmap Lookup](#Algorithm:-Hash-Map-Lookup)\n",
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"* [Code: Hashmap Lookup](#Code:-Hash-Map-Lookup)\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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"* Can we assume the string is ASCII?\n",
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" * Yes\n",
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" * Note: Unicode strings could require special handling depending on your language\n",
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"* Is whitespace important?\n",
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" * Yes\n",
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"* Is this case sensitive? 'Nib', 'bin' is not a match?\n",
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" * Yes\n",
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"* Can we use additional data structures?\n",
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" * Yes\n",
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"* Can we assume this fits in 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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"* One or more None inputs -> False\n",
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"* One or more empty strings -> False\n",
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"* 'Nib', 'bin' -> False\n",
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"* 'act', 'cat' -> True\n",
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"* 'a ct', 'ca t' -> True"
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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: Compare Sorted Strings\n",
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"\n",
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"Permutations contain the same strings but in different orders. This approach could be slow for large strings due to sorting.\n",
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"\n",
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"* Sort both strings\n",
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"* If both sorted strings are equal\n",
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" * return True\n",
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"* Else\n",
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" * return False\n",
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"\n",
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"Complexity:\n",
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"* Time: O(n log n) from the sort, in general\n",
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"* Space: O(n)"
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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: Compare Sorted Strings"
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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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"collapsed": false
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},
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"outputs": [],
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"source": [
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"class Permutations(object):\n",
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"\n",
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" def is_permutation(self, str1, str2):\n",
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" if str1 is None or str2 is None:\n",
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" return False\n",
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" return sorted(str1) == sorted(str2)"
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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: Hash Map Lookup\n",
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"\n",
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"We'll keep a hash map (dict) to keep track of characters we encounter. \n",
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"\n",
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"Steps:\n",
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"* Scan each character\n",
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"* For each character in each string:\n",
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" * If the character does not exist in a hash map, add the character to a hash map\n",
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" * Else, increment the character's count\n",
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"* If the hash maps for each string are equal\n",
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" * Return True\n",
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"* Else\n",
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" * Return False\n",
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"\n",
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"Notes:\n",
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"* Since the characters are in ASCII, we could potentially use an array of size 128 (or 256 for extended ASCII), where each array index is equivalent to an ASCII value\n",
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"* Instead of using two hash maps, you could use one hash map and increment character values based on the first string and decrement based on the second string\n",
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"* You can short circuit if the lengths of each string are not equal, although len() in Python is generally O(1) unlike other languages like C where getting the length of a string is O(n)\n",
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"\n",
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"Complexity:\n",
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"* Time: O(n)\n",
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"* Space: O(n)"
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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: Hash Map Lookup"
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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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"collapsed": false
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},
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"outputs": [],
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"source": [
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"from collections import defaultdict\n",
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"\n",
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"\n",
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"class PermutationsAlt(object):\n",
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"\n",
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" def is_permutation(self, str1, str2):\n",
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" if str1 is None or str2 is None:\n",
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" return False\n",
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" if len(str1) != len(str2):\n",
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" return False\n",
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" unique_counts1 = defaultdict(int)\n",
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" unique_counts2 = defaultdict(int)\n",
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" for char in str1:\n",
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" unique_counts1[char] += 1\n",
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" for char in str2:\n",
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" unique_counts2[char] += 1\n",
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" return unique_counts1 == unique_counts2"
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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": 3,
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"metadata": {
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"collapsed": false
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},
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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_permutation_solution.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_permutation_solution.py\n",
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"from nose.tools import assert_equal\n",
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"\n",
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"\n",
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"class TestPermutation(object):\n",
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"\n",
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" def test_permutation(self, func):\n",
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" assert_equal(func(None, 'foo'), False)\n",
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" assert_equal(func('', 'foo'), False)\n",
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" assert_equal(func('Nib', 'bin'), False)\n",
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" assert_equal(func('act', 'cat'), True)\n",
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" assert_equal(func('a ct', 'ca t'), True)\n",
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" print('Success: test_permutation')\n",
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"\n",
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"\n",
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"def main():\n",
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" test = TestPermutation()\n",
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" permutations = Permutations()\n",
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" test.test_permutation(permutations.is_permutation)\n",
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" try:\n",
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" permutations_alt = PermutationsAlt()\n",
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" test.test_permutation(permutations_alt.is_permutation)\n",
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" except NameError:\n",
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" # Alternate solutions are only defined\n",
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" # in the solutions file\n",
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" pass\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": 4,
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"metadata": {
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"collapsed": false
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},
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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_permutation\n",
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"Success: test_permutation\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_permutation_solution.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": "3.5.0"
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}
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},
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"nbformat": 4,
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"nbformat_minor": 0
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}
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