Added notebook solving the following: Compress a string such that aabbb becomes a2b3.

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Donne Martin 2015-05-04 05:54:50 -04:00
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@ -9,6 +9,7 @@ Continually updated IPython Notebooks containing algorithms and data structures.
* [Reverse characters in a string](http://nbviewer.ipython.org/github/donnemartin/algorithms-data-structures/blob/master/arrays-strings/reverse_string.ipynb)
* [Check if a string is a permutation of another](http://nbviewer.ipython.org/github/donnemartin/algorithms-data-structures/blob/master/arrays-strings/permutation.ipynb)
* [Encode spaces in a string in-place](http://nbviewer.ipython.org/github/donnemartin/algorithms-data-structures/blob/master/arrays-strings/replace_char.ipynb)
* [Compress a string](http://nbviewer.ipython.org/github/donnemartin/algorithms-data-structures/blob/master/arrays-strings/compress.ipynb)
## License

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@ -0,0 +1,263 @@
{
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{
"cell_type": "markdown",
"metadata": {},
"source": [
"## 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,
"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 = 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))"
]
},
{
"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"
]
},
{
"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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