{ "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": 1, "metadata": {}, "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": 2, "metadata": {}, "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": 3, "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('AAABCCDDDD'), 'A3BCCD4')\n", " self.assertEqual(\n", " func('aaBCCEFFFFKKMMMMMMP taaammanlaarrrr seeeeeeeeek tooo'),\n", " 'aaBCCEF4KKM6P ta3mmanlaar4 se9k to3',\n", " )\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": 4, "metadata": {}, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Success: test_compress\n" ] } ], "source": [ "%run -i test_compress.py" ] } ], "metadata": { "kernelspec": { "display_name": "Python 3", "language": "python", "name": "python3" }, "language_info": { "codemirror_mode": { "name": "ipython", "version": 3 }, "file_extension": ".py", "mimetype": "text/x-python", "name": "python", "nbconvert_exporter": "python", "pygments_lexer": "ipython3", "version": "3.7.2" } }, "nbformat": 4, "nbformat_minor": 1 }