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    "## Problem: Compress a String Such that 'AAABCCDDDD' Becomes 'A3B1C2D4'.  Only Compress if it Saves Space.\n",
    "\n",
    "* [Clarifying Questions](#Clarifying-Questions)\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)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Clarifying Questions\n",
    "\n",
    "* Is the string ASCII (extended)?  Or Unicode?\n",
    "    * ASCII extended, which is 256 characters\n",
    "* Can you use additional data structures?  \n",
    "    * Yes\n",
    "* Is this case sensitive?\n",
    "    * Yes"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Test Cases\n",
    "\n",
    "* NULL\n",
    "* '' -> ''\n",
    "* 'ABC' -> 'ABC'\n",
    "* 'AAABCCDDDD' -> 'A3B1C2D4'"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Algorithm: List\n",
    "\n",
    "![alt text](https://raw.githubusercontent.com/donnemartin/algorithms-data-structures/master/images/compress_string.jpg)\n",
    "\n",
    "Since Python strings are immutable, we'll use a list to exercise array manipulation.  Note using a list vs a bytearray will will result in additional space to create the list and to convert the list to a string.\n",
    "\n",
    "* If string is empty return string\n",
    "* count = 0\n",
    "* size = 0\n",
    "* last_char = first char in string\n",
    "* For each char in string\n",
    "    * If char == last_char\n",
    "        count++\n",
    "    * Else\n",
    "        size += 2\n",
    "        count++\n",
    "        last_char = char\n",
    "* size += 2\n",
    "* If the compressed string size is >= string size, return string\n",
    "* Create compressed_string\n",
    "* For each char in string\n",
    "    * If char == last_char\n",
    "        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(m) where m is the size of the compressed bytearray"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Code: List"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
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   "outputs": [],
   "source": [
    "def compress_string(string):\n",
    "    if string is None or len(string) == 0:\n",
    "        return string\n",
    "    size = 0\n",
    "    count = 0\n",
    "    last_char = string[0]\n",
    "    for char in string:\n",
    "        if char == last_char:\n",
    "            count += 1\n",
    "        else:\n",
    "            size += 2\n",
    "            count = 1\n",
    "            last_char = char\n",
    "    size += 2\n",
    "    if size >= len(string):\n",
    "        return 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)\n",
    "\n",
    "string0 = None\n",
    "string1 = ''\n",
    "string2 = 'ABC'\n",
    "string3 = 'AAABCCDDDD'\n",
    "print(compress_string(string0))\n",
    "print(compress_string(string1))\n",
    "print(compress_string(string2))\n",
    "print(compress_string(string3))"
   ]
  },
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   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Algorithm: Byte Array\n",
    "\n",
    "![alt text](https://raw.githubusercontent.com/donnemartin/algorithms-data-structures/master/images/compress_string.jpg)\n",
    "\n",
    "Since Python strings are immutable, we'll use a bytearray to exercise array manipulation.  We could use a list of characters to create the compressed string then convert it to a string in the end, but this will result in additional space.\n",
    "\n",
    "* If bytearray is empty return bytearray\n",
    "* count = 0\n",
    "* size = 0\n",
    "* last_char_code = first char code in bytearray\n",
    "* For each char code in bytearray\n",
    "    * If char code == last_char_code\n",
    "        count++\n",
    "    * Else\n",
    "        size += 2\n",
    "        count++\n",
    "        last_char_code = char code\n",
    "* size += 2\n",
    "* If the compressed bytearray size is >= bytearray size, return string\n",
    "* Create compressed_bytearray\n",
    "* pos = 0\n",
    "* For each char code in bytearray\n",
    "    * If char code == last_char_code\n",
    "        count++\n",
    "    * Else\n",
    "        * compressed_bytearray[pos] = last_char_code\n",
    "        * compressed_bytearray[pos + 1] = count\n",
    "        * pos += 2\n",
    "        * count = 1\n",
    "        * last_char_code = char code\n",
    "    * compressed_bytearray[pos] = last_char_code\n",
    "    * compressed_bytearray[pos + 1] = count\n",
    "* return compressed_bytearray\n",
    "\n",
    "Complexity:\n",
    "* Time: O(n)\n",
    "* Space: O(m) where m is the size of the compressed bytearray"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Code: Byte Array"
   ]
  },
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   "cell_type": "code",
   "execution_count": null,
   "metadata": {
    "collapsed": false
   },
   "outputs": [],
   "source": [
    "def compress_string(string):\n",
    "    if string is None or len(string) == 0:\n",
    "        return string\n",
    "    size = 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",
    "            size += 2\n",
    "            count = 1\n",
    "            last_char_code = char_code\n",
    "    size += 2\n",
    "    if size >= len(string):\n",
    "        return 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\n",
    "\n",
    "string0 = None\n",
    "string1 = bytearray('')\n",
    "string2 = bytearray('ABC')\n",
    "string3 = bytearray('AAABCCDDDD')\n",
    "print(compress_string(string0))\n",
    "print(compress_string(string1))\n",
    "print(compress_string(string2))\n",
    "print(compress_string(string3))"
   ]
  }
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