mirror of
https://github.com/donnemartin/interactive-coding-challenges.git
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270 lines
7.9 KiB
Plaintext
270 lines
7.9 KiB
Plaintext
{
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"cells": [
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"<small><i>This notebook was prepared by [Donne Martin](http://donnemartin.com). Source and license info is on [GitHub](https://bit.ly/code-notes).</i></small>"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"## Problem: Compress a string such that 'AAABCCDDDD' becomes 'A3B1C2D4'\n",
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"\n",
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"* [Constraints and Assumptions](#Constraints-and-Assumptions)\n",
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"* [Test Cases](#Test-Cases)\n",
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"* [Algorithm: List](#Algorithm:-List)\n",
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"* [Code: List](#Code:-List)\n",
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"* [Algorithm: Byte Array](#Algorithm:-Byte-Array)\n",
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"* [Code: Byte array](#Code:-Byte-Array)\n",
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"* [Unit Test](#Unit-Test)"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"## Constraints\n",
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"\n",
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"*Problem statements are often intentionally ambiguous. Identifying constraints and stating assumptions can help to ensure you code the intended solution.*\n",
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"\n",
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"* Can I assume the string is ASCII?\n",
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" * Yes\n",
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" * Note: Unicode strings could require special handling depending on your language\n",
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"* Can you use additional data structures? \n",
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" * Yes\n",
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"* Is this case sensitive?\n",
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" * Yes\n",
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"* Do you compress even if it doesn't save space?\n",
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" * No"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"## Test Cases\n",
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"\n",
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"* NULL\n",
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"* '' -> ''\n",
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"* 'ABC' -> 'ABC'\n",
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"* 'AAABCCDDDD' -> 'A3B1C2D4'"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"## Algorithm: List\n",
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"\n",
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"Since Python strings are immutable, we'll use a list of characters instead to exercise in-place string manipulation as you would get with a C string (which is null terminated, as seen in the diagram below). Python does not use a null-terminator.\n",
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"\n",
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"\n",
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"\n",
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"* Calculate the size of the compressed string\n",
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"* If the compressed string size is >= string size, return string\n",
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"* Create compressed_string\n",
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" * For each char in string\n",
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" * If char is the same as last_char, increment count\n",
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" * Else\n",
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" * Append last_char to compressed_string\n",
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" * append count to compressed_string\n",
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" * count = 1\n",
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" * last_char = char\n",
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" * Append last_char to compressed_string\n",
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" * Append count to compressed_string\n",
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" * Return compressed_string\n",
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"\n",
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"Complexity:\n",
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"* Time: O(n)\n",
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"* Space: O(2m) where m is the size of the compressed list and the resulting string copied from the list"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"## Code: List"
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]
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},
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{
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"cell_type": "code",
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"execution_count": 1,
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"metadata": {
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"collapsed": false
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},
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"outputs": [],
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"source": [
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"def compress_string(string):\n",
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" if string is None or len(string) == 0:\n",
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" return string\n",
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" \n",
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" # Calculate the size of the compressed string\n",
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" size = 0\n",
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" last_char = string[0]\n",
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" for char in string:\n",
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" if char != last_char:\n",
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" size += 2\n",
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" last_char = char\n",
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" size += 2\n",
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" \n",
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" # If the compressed string size is greater than \n",
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" # or equal to string size, return string\n",
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" if size >= len(string):\n",
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" return string\n",
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"\n",
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" # Create compressed_string\n",
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" compressed_string = list()\n",
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" count = 0\n",
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" last_char = string[0]\n",
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" for char in string:\n",
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" if char == last_char:\n",
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" count += 1\n",
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" else:\n",
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" compressed_string.append(last_char)\n",
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" compressed_string.append(str(count))\n",
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" count = 1\n",
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" last_char = char\n",
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" compressed_string.append(last_char)\n",
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" compressed_string.append(str(count))\n",
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" return \"\".join(compressed_string)"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"## Algorithm: Byte Array\n",
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"\n",
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"The byte array algorithm similar when using a list, except we will need to work with the bytearray's character codes instead of the characters as we did above when we implemented this solution with a list.\n",
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"\n",
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"Complexity:\n",
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"* Time: O(n)\n",
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"* Space: O(m) where m is the size of the compressed bytearray"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"## Code: Byte Array"
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]
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},
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{
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"cell_type": "code",
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"execution_count": 2,
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"metadata": {
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"collapsed": false
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},
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"outputs": [],
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"source": [
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"def compress_string_alt(string):\n",
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" if string is None or len(string) == 0:\n",
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" return string\n",
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" \n",
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" # Calculate the size of the compressed string\n",
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" size = 0\n",
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" last_char_code = string[0]\n",
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" for char_code in string:\n",
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" if char_code != last_char_code:\n",
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" size += 2\n",
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" last_char_code = char_code\n",
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" size += 2\n",
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" \n",
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" # If the compressed string size is greater than \n",
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" # or equal to string size, return string \n",
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" if size >= len(string):\n",
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" return string\n",
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" \n",
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" # Create compressed_string\n",
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" compressed_string = bytearray(size)\n",
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" pos = 0\n",
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" count = 0\n",
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" last_char_code = string[0]\n",
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" for char_code in string:\n",
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" if char_code == last_char_code:\n",
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" count += 1\n",
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" else:\n",
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" compressed_string[pos] = last_char_code\n",
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" compressed_string[pos+1] = ord(str(count))\n",
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" pos += 2\n",
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" count = 1\n",
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" last_char_code = char_code\n",
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" compressed_string[pos] = last_char_code\n",
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" compressed_string[pos+1] = ord(str(count))\n",
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" return compressed_string"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"## Unit Test"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"*It is important to identify and run through general and edge cases from the [Test Cases](#Test-Cases) section by hand. You generally will not be asked to write a unit test like what is shown below.*"
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]
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},
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{
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"cell_type": "code",
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"execution_count": 3,
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"metadata": {
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"collapsed": false
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},
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"outputs": [
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{
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"name": "stdout",
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"output_type": "stream",
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"text": [
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"Success: test_compress\n",
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"Success: test_compress\n"
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]
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}
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],
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"source": [
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"from nose.tools import assert_equal\n",
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"\n",
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"class Test(object):\n",
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" def test_compress(self, func):\n",
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" assert_equal(func(None), None)\n",
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" assert_equal(func(''), '')\n",
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" assert_equal(func('ABC'), 'ABC')\n",
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" assert_equal(func('AAABCCDDDD'), 'A3B1C2D4')\n",
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" print('Success: test_compress')\n",
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"\n",
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"if __name__ == '__main__':\n",
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" test = Test()\n",
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" test.test_compress(compress_string)\n",
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" test.test_compress(compress_string_alt)"
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]
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}
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],
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"metadata": {
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"kernelspec": {
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"display_name": "Python 2",
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"language": "python",
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"name": "python2"
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},
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"language_info": {
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"codemirror_mode": {
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"name": "ipython",
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"version": 2
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},
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"file_extension": ".py",
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"mimetype": "text/x-python",
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"name": "python",
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"nbconvert_exporter": "python",
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"pygments_lexer": "ipython2",
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"version": "2.7.10"
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}
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},
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"nbformat": 4,
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"nbformat_minor": 0
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}
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