interactive-coding-challenges/arrays_strings/compress/compress_solution.ipynb

207 lines
5.2 KiB
Python

{
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{
"cell_type": "markdown",
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"source": [
"This notebook was prepared by [Donne Martin](http://donnemartin.com). Source and license info is on [GitHub](https://github.com/donnemartin/interactive-coding-challenges)."
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"# Solution Notebook"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Problem: Compress a string such that 'AAABCCDDDD' becomes 'A3BC2D4'. Only compress the string if it saves space.\n",
"\n",
"* [Constraints](#Constraints)\n",
"* [Test Cases](#Test-Cases)\n",
"* [Algorithm](#Algorithm)\n",
"* [Code](#Code)\n",
"* [Unit Test](#Unit-Test)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Constraints\n",
"\n",
"* Can we assume the string is ASCII?\n",
" * Yes\n",
" * Note: Unicode strings could require special handling depending on your language\n",
"* Is this case sensitive?\n",
" * Yes\n",
"* Can we use additional data structures? \n",
" * Yes\n",
"* Can we assume this fits in memory?\n",
" * Yes"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Test Cases\n",
"\n",
"* None -> None\n",
"* '' -> ''\n",
"* 'AABBCC' -> 'AABBCC'\n",
"* 'AAABCCDDDD' -> 'A3BC2D4'"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Algorithm\n",
"\n",
"* For each char in string\n",
" * If char is the same as last_char, increment count\n",
" * Else\n",
" * Append last_char and count to compressed_string\n",
" * last_char = char\n",
" * count = 1\n",
"* Append last_char and count to compressed_string\n",
"* If the compressed string size is < string size\n",
" * Return compressed string\n",
"* Else\n",
" * Return string\n",
"\n",
"Complexity:\n",
"* Time: O(n)\n",
"* Space: O(n)\n",
"\n",
"Complexity Note:\n",
"* Although strings are immutable in Python, appending to strings is optimized in CPython so that it now runs in O(n) and extends the string in-place. Refer to this [Stack Overflow post](http://stackoverflow.com/a/4435752)."
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Code"
]
},
{
"cell_type": "code",
"execution_count": 1,
"metadata": {},
"outputs": [],
"source": [
"class CompressString(object):\n",
"\n",
" def compress(self, string):\n",
" if string is None or not string:\n",
" return string\n",
" result = ''\n",
" prev_char = string[0]\n",
" count = 0\n",
" for char in string:\n",
" if char == prev_char:\n",
" count += 1\n",
" else:\n",
" result += self._calc_partial_result(prev_char, count)\n",
" prev_char = char\n",
" count = 1\n",
" result += self._calc_partial_result(prev_char, count)\n",
" return result if len(result) < len(string) else string\n",
"\n",
" def _calc_partial_result(self, prev_char, count):\n",
" return prev_char + (str(count) if count > 1 else '')"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Unit Test"
]
},
{
"cell_type": "code",
"execution_count": 2,
"metadata": {},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Overwriting test_compress.py\n"
]
}
],
"source": [
"%%writefile test_compress.py\n",
"import unittest\n",
"\n",
"\n",
"class TestCompress(unittest.TestCase):\n",
"\n",
" def test_compress(self, func):\n",
" self.assertEqual(func(None), None)\n",
" self.assertEqual(func(''), '')\n",
" self.assertEqual(func('AABBCC'), 'AABBCC')\n",
" self.assertEqual(func('AAABCCDDDDE'), 'A3BC2D4E')\n",
" self.assertEqual(func('BAAACCDDDD'), 'BA3C2D4')\n",
" self.assertEqual(func('AAABAACCDDDD'), 'A3BA2C2D4')\n",
" print('Success: test_compress')\n",
"\n",
"\n",
"def main():\n",
" test = TestCompress()\n",
" compress_string = CompressString()\n",
" test.test_compress(compress_string.compress)\n",
"\n",
"\n",
"if __name__ == '__main__':\n",
" main()"
]
},
{
"cell_type": "code",
"execution_count": 3,
"metadata": {},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Success: test_compress\n"
]
}
],
"source": [
"%run -i test_compress.py"
]
}
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