interactive-coding-challenges/arrays-strings/permutation.ipynb

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    "<small><i>This notebook was prepared by [Donne Martin](http://donnemartin.com). Source and license info is on [GitHub](https://bit.ly/code-notes).</i></small>"
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    "## 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",
    "* [Algorithm: Hashmap Lookup](#Algorithm:-Hash-Map-Lookup)\n",
    "* [Code: Hashmap Lookup](#Code:-Hash-Map-Lookup)"
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    "## 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"
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    "## Test Cases\n",
    "\n",
    "* One or more empty strings -> False\n",
    "* 'Nib', 'bin' -> False\n",
    "* 'act', 'cat' -> True\n",
    "* 'a ct', 'ca t' -> True"
   ]
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    "## 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"
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   "source": [
    "## Code: Compare Sorted Strings"
   ]
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    "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'))"
   ]
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    "## 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"
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   "source": [
    "## Code: Hash Map Lookup"
   ]
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  {
   "cell_type": "code",
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   "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'))"
   ]
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