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

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   "source": [
    "<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: Implement a function to reverse a string.\n",
    "\n",
    "* [Clarifying Questions](#Clarifying-Questions)\n",
    "* [Test Cases](#Test-Cases)\n",
    "* [Algorithm](#Algorithm)\n",
    "* [Code](#Code)\n",
    "* [C Algorithm](C-Algorithm)\n",
    "* [C Code]()\n",
    "* [Pythonic-Code](#Pythonic-Code)"
   ]
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   "source": [
    "## Clarifying Questions\n",
    "\n",
    "* Can I use the slice operator or the reversed function?\n",
    "    * No\n",
    "* Since Python string are immutable, can I use a list instead?\n",
    "    * Yes"
   ]
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   "metadata": {},
   "source": [
    "## Test Cases\n",
    "\n",
    "* NULL input\n",
    "* '' -> ''\n",
    "* 'foo bar' -> 'rab oof'"
   ]
  },
  {
   "cell_type": "code",
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   "source": [
    "from nose.tools import assert_equal\n",
    "\n",
    "class Test(object):\n",
    "    def test_reversed(self, func):\n",
    "        assert_equal(func(None), None)\n",
    "        assert_equal(func(''), '')\n",
    "        assert_equal(func('foo bar'), 'rab oof')\n",
    "\n",
    "def run_tests(func):\n",
    "    test = Test()\n",
    "    test.test_reversed(func)"
   ]
  },
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   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Algorithm\n",
    "\n",
    "* Convert the string to a list\n",
    "* Iterate len(string)/2 times, starting with i = 0:\n",
    "    * Swap i and len(string) - 1 - i\n",
    "    * Sncrement i\n",
    "* Convert the list to a string and return it\n",
    "\n",
    "Complexity:\n",
    "* Time: O(n)\n",
    "* Space: O(n), additional space converting to/from a list"
   ]
  },
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   "metadata": {},
   "source": [
    "## Code"
   ]
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   "source": [
    "def reverse_string(string):\n",
    "    if string is None:\n",
    "        return None\n",
    "    string_list = list(string)\n",
    "    string_length = len(string_list)\n",
    "    for i in xrange(string_length/2):\n",
    "        string_list[i], string_list[string_length - 1 - i] = \\\n",
    "            string_list[string_length - 1 - i], string_list[i]\n",
    "    return ''.join(string_list)\n",
    "\n",
    "run_tests(reverse_string)"
   ]
  },
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   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## C Algorithm\n",
    "\n",
    "This is a classic problem in C/C++\n",
    "\n",
    "We'll want to keep two pointers:\n",
    "* i is a pointer to the first char\n",
    "* j is a pointer to the last char\n",
    "\n",
    "To get a pointer to the last char, we need to loop through all characters, take note of null terminator.\n",
    "\n",
    "* while i < j\n",
    "    * swap i and j\n",
    "\n",
    "Complexity:\n",
    "* Time: O(n)\n",
    "* Space: In-place\n",
    "\n",
    "Note:\n",
    "* Instead of using i, you can use str instead, although this might not be as intuitive."
   ]
  },
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   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## C Code"
   ]
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  {
   "cell_type": "code",
   "execution_count": null,
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    "collapsed": false
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   "outputs": [],
   "source": [
    "# %load reverse_string.cpp\n",
    "#include <stdio.h>\n",
    "\n",
    "void Reverse(char* str) {\n",
    "    if (str) {\n",
    "        char* i = str;\t// first letter\n",
    "        char* j = str;\t// last letter\n",
    "        \n",
    "        // find the end of the string\n",
    "        while (*j) {\n",
    "            j++;\n",
    "        }\n",
    "        \n",
    "        // don't point to the null terminator\n",
    "        j--;\n",
    "        \n",
    "        char tmp;\n",
    "        \n",
    "        // swap chars to reverse the string\n",
    "        while (i < j) {\n",
    "            tmp = *i;\n",
    "            *i++ = *j;\n",
    "            *j-- = tmp;\n",
    "        }\n",
    "    }\n",
    "}\n",
    "\n",
    "int main() {\n",
    "    char test0[] = \"\";\n",
    "    char test1[] = \"foo\";\n",
    "    Reverse(NULL);\n",
    "    Reverse(test0);\n",
    "    Reverse(test1);\n",
    "    printf(\"%s \\n\", test0);\n",
    "    printf(\"%s \\n\", test1);\n",
    "    return 0;\n",
    "}"
   ]
  },
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   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Pythonic-Code\n",
    "\n",
    "This question has an artificial constraint that prevented the use of the slice operator and the reversed method.  For completeness, the solutions for these are provided below."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 7,
   "metadata": {
    "collapsed": false
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   "outputs": [],
   "source": [
    "def reverse_string_alt(string):\n",
    "    if string is None:\n",
    "        return None\n",
    "    return string[::-1]\n",
    "\n",
    "def reverse_string_alt2(string):\n",
    "    if string is None:\n",
    "        return None    \n",
    "    return ''.join(reversed(string))\n",
    "\n",
    "run_tests(reverse_string_alt)\n",
    "run_tests(reverse_string_alt2)"
   ]
  }
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