interactive-coding-challenges/arrays_strings/compress/compress_solution.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: Compress a string such that 'AAABCCDDDD' becomes 'A3B1C2D4'\n",
    "\n",
    "* [Constraints](#Constraints)\n",
    "* [Test Cases](#Test-Cases)\n",
    "* [Algorithm: List](#Algorithm:-List)\n",
    "* [Code: List](#Code:-List)\n",
    "* [Algorithm: Byte Array](#Algorithm:-Byte-Array)\n",
    "* [Code: Byte array](#Code:-Byte-Array)\n",
    "* [Unit Test](#Unit-Test)"
   ]
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   "source": [
    "## Constraints\n",
    "\n",
    "*Problem statements are often intentionally ambiguous.  Identifying constraints and stating assumptions can help to ensure you code the intended solution.*\n",
    "\n",
    "* Can I assume the string is ASCII?\n",
    "    * Yes\n",
    "    * Note: Unicode strings could require special handling depending on your language\n",
    "* Can you use additional data structures?  \n",
    "    * Yes\n",
    "* Is this case sensitive?\n",
    "    * Yes\n",
    "* Do you compress even if it doesn't save space?\n",
    "    * No"
   ]
  },
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   "cell_type": "markdown",
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   "source": [
    "## Test Cases\n",
    "\n",
    "* NULL\n",
    "* '' -> ''\n",
    "* 'ABC' -> 'ABC'\n",
    "* 'AAABCCDDDD' -> 'A3B1C2D4'"
   ]
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   "cell_type": "markdown",
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    "## Algorithm: List\n",
    "\n",
    "Since Python strings are immutable, we'll use a list of characters instead to exercise in-place string manipulation as you would get with a C string (which is null terminated, as seen in the diagram below).  Python does not use a null-terminator.\n",
    "\n",
    "![alt text](https://raw.githubusercontent.com/donnemartin/algorithms-data-structures/master/images/compress_string.jpg)\n",
    "\n",
    "* Calculate the size of the compressed string\n",
    "* If the compressed string size is >= string size, return string\n",
    "* Create compressed_string\n",
    "    * For each char in string\n",
    "        * If char is the same as last_char, increment count\n",
    "        * Else\n",
    "            * Append last_char to compressed_string\n",
    "            * append count to compressed_string\n",
    "            * count = 1\n",
    "            * last_char = char\n",
    "        * Append last_char to compressed_string\n",
    "        * Append count to compressed_string\n",
    "    * Return compressed_string\n",
    "\n",
    "Complexity:\n",
    "* Time: O(n)\n",
    "* Space: O(2m) where m is the size of the compressed list and the resulting string copied from the list"
   ]
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   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Code: List"
   ]
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   "source": [
    "def compress_string(string):\n",
    "    if string is None or len(string) == 0:\n",
    "        return string\n",
    "    \n",
    "    # Calculate the size of the compressed string\n",
    "    size = 0\n",
    "    last_char = string[0]\n",
    "    for char in string:\n",
    "        if char != last_char:\n",
    "            size += 2\n",
    "            last_char = char\n",
    "    size += 2\n",
    "    \n",
    "    # If the compressed string size is greater than \n",
    "    # or equal to string size, return string\n",
    "    if size >= len(string):\n",
    "        return string\n",
    "\n",
    "    # Create compressed_string\n",
    "    compressed_string = list()\n",
    "    count = 0\n",
    "    last_char = string[0]\n",
    "    for char in string:\n",
    "        if char == last_char:\n",
    "            count += 1\n",
    "        else:\n",
    "            compressed_string.append(last_char)\n",
    "            compressed_string.append(str(count))\n",
    "            count = 1\n",
    "            last_char = char\n",
    "    compressed_string.append(last_char)\n",
    "    compressed_string.append(str(count))\n",
    "    return \"\".join(compressed_string)"
   ]
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   "source": [
    "## Algorithm: Byte Array\n",
    "\n",
    "The byte array algorithm similar when using a list, except we will need to work with the bytearray's character codes instead of the characters as we did above when we implemented this solution with a list.\n",
    "\n",
    "Complexity:\n",
    "* Time: O(n)\n",
    "* Space: O(m) where m is the size of the compressed bytearray"
   ]
  },
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   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Code: Byte Array"
   ]
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   "source": [
    "def compress_string_alt(string):\n",
    "    if string is None or len(string) == 0:\n",
    "        return string\n",
    "    \n",
    "    # Calculate the size of the compressed string\n",
    "    size = 0\n",
    "    last_char_code = string[0]\n",
    "    for char_code in string:\n",
    "        if char_code != last_char_code:\n",
    "            size += 2\n",
    "            last_char_code = char_code\n",
    "    size += 2\n",
    "    \n",
    "    # If the compressed string size is greater than \n",
    "    # or equal to string size, return string    \n",
    "    if size >= len(string):\n",
    "        return string\n",
    "    \n",
    "    # Create compressed_string\n",
    "    compressed_string = bytearray(size)\n",
    "    pos = 0\n",
    "    count = 0\n",
    "    last_char_code = string[0]\n",
    "    for char_code in string:\n",
    "        if char_code == last_char_code:\n",
    "            count += 1\n",
    "        else:\n",
    "            compressed_string[pos] = last_char_code\n",
    "            compressed_string[pos+1] = ord(str(count))\n",
    "            pos += 2\n",
    "            count = 1\n",
    "            last_char_code = char_code\n",
    "    compressed_string[pos] = last_char_code\n",
    "    compressed_string[pos+1] = ord(str(count))\n",
    "    return compressed_string"
   ]
  },
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   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Unit Test"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "*It is important to identify and run through general and edge cases from the [Test Cases](#Test-Cases) section by hand.  You generally will not be asked to write a unit test like what is shown below.*"
   ]
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  {
   "cell_type": "code",
   "execution_count": 3,
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     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Success: test_compress\n",
      "Success: test_compress\n"
     ]
    }
   ],
   "source": [
    "from nose.tools import assert_equal\n",
    "\n",
    "class Test(object):\n",
    "    def test_compress(self, func):\n",
    "        assert_equal(func(None), None)\n",
    "        assert_equal(func(''), '')\n",
    "        assert_equal(func('ABC'), 'ABC')\n",
    "        assert_equal(func('AAABCCDDDD'), 'A3B1C2D4')\n",
    "        print('Success: test_compress')\n",
    "\n",
    "if __name__ == '__main__':\n",
    "    test = Test()\n",
    "    test.test_compress(compress_string)\n",
    "    test.test_compress(compress_string_alt)"
   ]
  }
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