Added notebook solving the following: Check if a string is a permutation of another.

README.md

@@ -7,6 +7,7 @@ Data structures and algorithms practice problems.
 
 * [Check if a string contains unique characters](http://nbviewer.ipython.org/github/donnemartin/practice/blob/master/arrays-strings/unique_chars.ipynb)
 * [Reverse characters in a string](http://nbviewer.ipython.org/github/donnemartin/practice/blob/master/arrays-strings/reverse_string.ipynb)
+* [Check if a string is a permutation of another](http://nbviewer.ipython.org/github/donnemartin/practice/blob/master/arrays-strings/permutation.ipynb)
 
 ## License
 

arrays-strings/permutation.ipynb

@@ -0,0 +1,171 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Problem: Determine if a string is a permutation of another string\n",
+    "\n",
+    "* [Clarifying Questions](#Clarifying-Questions)\n",
+    "* [Test Cases](#Test-Cases)\n",
+    "* [Algorithm: Compare Sorted Strings](#Algorithm:-Compare-Sorted-Strings)\n",
+    "* [Code: Compare Sorted Strings](#Code:-Compare-Sorted-Strings)\n",
+    "* [Pythonic-Code](#Pythonic-Code)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Clarifying Questions\n",
+    "\n",
+    "* Is the string ASCII (extended)?  Or Unicode?\n",
+    "    * ASCII extended, which is 256 characters\n",
+    "* Is whitespace important?\n",
+    "    * Yes\n",
+    "* Is this case sensitive?  'Nib', 'bin' is not a match?\n",
+    "    * Yes"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Test Cases\n",
+    "\n",
+    "* One or more empty strings -> False\n",
+    "* 'Nib', 'bin' -> False\n",
+    "* 'act', 'cat' -> True\n",
+    "* 'a ct', 'ca t' -> True"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Algorithm: Compare Sorted Strings\n",
+    "\n",
+    "Anagrams contain the same strings but in different orders.  This approach could be slow for large strings due to sorting.\n",
+    "\n",
+    "* Sort both strings\n",
+    "* If both sorted strings are equal\n",
+    "    * return True\n",
+    "* Else\n",
+    "    * return False\n",
+    "\n",
+    "Complexity:\n",
+    "* Time: O(n log n) from the sort, in general\n",
+    "* Space: Additional O(l + m) is created by the sorting algorithm, where l is the length of one string and m is the length of the other"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Code: Compare Sorted Strings"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [],
+   "source": [
+    "def permutations(str1, str2):\n",
+    "    return sorted(str1) == sorted(str2)\n",
+    "\n",
+    "print(permutations('', 'foo'))\n",
+    "print(permutations('Nib', 'bin'))\n",
+    "print(permutations('act', 'cat'))\n",
+    "print(permutations('a ct', 'ca t'))"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Algorithm: Hash Map Lookup\n",
+    "\n",
+    "We'll keep a hash map (dict) to keep track of characters we encounter.  \n",
+    "\n",
+    "Steps:\n",
+    "* Scan each character\n",
+    "* For each character in each string:\n",
+    "    * If the character does not exist in a hash map, add the character to a hash map\n",
+    "    * Else, increment the character's count\n",
+    "* If the hash maps for each string are equal\n",
+    "    * Return True\n",
+    "* Else\n",
+    "    * Return False\n",
+    "\n",
+    "Notes:\n",
+    "* Since the characters are in ASCII, we could potentially use an array of size 128 (or 256 for extended ASCII)\n",
+    "* 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",
+    "* You can short circuit if the lengths of each string are not equal, len() in Python is generally O(1)\n",
+    "\n",
+    "Complexity:\n",
+    "* Time: O(n)\n",
+    "* Space: Additional O(m), where m is the number of unique characters in the hash map"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## Code"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [],
+   "source": [
+    "from collections import defaultdict\n",
+    "\n",
+    "def unique_counts(string):\n",
+    "    dict_chars = defaultdict(int)\n",
+    "    for char in string:\n",
+    "        dict_chars[char] += 1\n",
+    "    return dict_chars\n",
+    "\n",
+    "def permutations(str1, str2):\n",
+    "    if len(str1) != len(str2):\n",
+    "        return False\n",
+    "    unique_counts1 = unique_counts(str1)\n",
+    "    unique_counts2 = unique_counts(str2)\n",
+    "    return unique_counts1 == unique_counts2\n",
+    "\n",
+    "print(permutations('', 'foo'))\n",
+    "print(permutations('Nib', 'bin'))\n",
+    "print(permutations('act', 'cat'))\n",
+    "print(permutations('a ct', 'ca t'))"
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python 2",
+   "language": "python",
+   "name": "python2"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 2
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython2",
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+ "nbformat_minor": 0
+}