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interactive-coding-challenges/arrays_strings/compress_alt/better_compress_solution.ipynb

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{
"cells": [
{
"cell_type": "markdown",
"metadata": {},
"source": [
"This notebook was prepared by [hashhar](https://github.com/hashhar), second solution added by [janhak] (https://github.com/janhak). 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 'A3BCCD4'. 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' -> 'A3BCCD4'"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Algorithm\n",
"\n",
"Since Python strings are immutable, we'll use a list of characters to build the compressed string representation. We'll then convert the list to a string.\n",
"\n",
"* Calculate the size of the compressed string\n",
" * Note the constraint about compressing only if it saves space\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",
" * If the count is more than 2\n",
" * Append last_char to compressed_string\n",
" * append count to compressed_string\n",
" * count = 1\n",
" * last_char = char\n",
" * If count is 1\n",
" * Append last_char to compressed_string\n",
" * count = 1\n",
" * last_char = char\n",
" * If count is 2\n",
" * Append last_char to compressed_string\n",
" * Append last_char to compressed_string once more\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(n)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Code"
]
},
{
"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",
"\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 original string\n",
" if size >= len(string):\n",
" return string\n",
"\n",
" # Create compressed_string\n",
" # New objective:\n",
" # Single characters are to be left as is\n",
" # Double characters are to be left as are\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",
" # Do the old compression tricks only if count exceeds two\n",
" if count > 2:\n",
" compressed_string.append(last_char)\n",
" compressed_string.append(str(count))\n",
" count = 1\n",
" last_char = char\n",
" # If count is either 1 or 2\n",
" else:\n",
" # If count is 1, leave the char as is\n",
" if count == 1:\n",
" compressed_string.append(last_char)\n",
" count = 1\n",
" last_char = char\n",
" # If count is 2, append the character twice\n",
" else:\n",
" compressed_string.append(last_char)\n",
" compressed_string.append(last_char)\n",
" count = 1\n",
" last_char = char\n",
" compressed_string.append(last_char)\n",
" compressed_string.append(str(count))\n",
"\n",
" # Convert the characters in the list to a string\n",
" return \"\".join(compressed_string)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Algorithm: Split to blocks and compress\n",
"\n",
"Let us split the string first into blocks of identical characters and then compress it block by block.\n",
"\n",
"* Split the string to blocks\n",
" * For each character in string\n",
" * Add this character to block\n",
" * If the next character is different\n",
" * Return block\n",
" * Erase the content of block\n",
"\n",
"\n",
"* Compress block\n",
" * If block consists of two or fewer characters\n",
" * Return block\n",
" * Else\n",
" * Append length of the block to the first character and return\n",
"\n",
"\n",
"* Compress string\n",
" * Split the string to blocks\n",
" * Compress blocks\n",
" * Join compressed blocks\n",
" * Return result if it is shorter than original string\n",
"\n",
"Complexity:\n",
"* Time: O(n)\n",
"* Space: O(n)"
]
},
{
"cell_type": "code",
"execution_count": null,
"metadata": {
"collapsed": true
},
"outputs": [],
"source": [
"def split_to_blocks(string):\n",
" block = ''\n",
" for char, next_char in zip(string, string[1:] + ' '):\n",
" block += char\n",
" if char is not next_char:\n",
" yield block\n",
" block = ''\n",
"\n",
"\n",
"def compress_block(block):\n",
" if len(block) <= 2:\n",
" return block\n",
" else:\n",
" return block[0] + str(len(block))\n",
"\n",
"\n",
"def compress_string(string):\n",
" if string is None or not string:\n",
" return string\n",
" compressed = (compress_block(block) for block in split_to_blocks(string))\n",
" result = ''.join(compressed)\n",
" return result if len(result) < len(string) else string"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Unit Test"
]
},
{
"cell_type": "code",
"execution_count": null,
"metadata": {
"collapsed": false
},
"outputs": [],
"source": [
"%%writefile test_compress.py\n",
"from nose.tools import assert_equal\n",
"\n",
"\n",
"class TestCompress(object):\n",
"\n",
" def test_compress(self, func):\n",
" assert_equal(func(None), None)\n",
" assert_equal(func(''), '')\n",
" assert_equal(func('AABBCC'), 'AABBCC')\n",
" assert_equal(func('AAABCCDDDD'), 'A3BCCD4')\n",
" assert_equal(func('aaBCCEFFFFKKMMMMMMP taaammanlaarrrr seeeeeeeeek tooo'), 'aaBCCEF4KKM6P ta3mmanlaar4 se9k to3')\n",
" print('Success: test_compress')\n",
"\n",
"\n",
"def main():\n",
" test = TestCompress()\n",
" test.test_compress(compress_string)\n",
"\n",
"\n",
"if __name__ == '__main__':\n",
" main()"
]
},
{
"cell_type": "code",
"execution_count": null,
"metadata": {
"collapsed": false
},
"outputs": [],
"source": [
"%run -i test_compress.py"
]
}
],
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