mirror of
https://github.com/donnemartin/interactive-coding-challenges.git
synced 2024-03-22 13:11:13 +08:00
264 lines
7.8 KiB
Plaintext
264 lines
7.8 KiB
Plaintext
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{
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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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"## Problem: Compress a String Such that 'AAABCCDDDD' Becomes 'A3B1C2D4'. Only Compress if it Saves Space.\n",
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"\n",
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"* [Clarifying Questions](#Clarifying-Questions)\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)"
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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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"## Clarifying Questions\n",
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"\n",
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"* Is the string ASCII (extended)? Or Unicode?\n",
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" * ASCII extended, which is 256 characters\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"
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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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"\n",
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"\n",
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"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",
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"\n",
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"* If string is empty return string\n",
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"* count = 0\n",
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"* size = 0\n",
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"* last_char = first char in string\n",
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"* For each char in string\n",
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" * If char == last_char\n",
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" count++\n",
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" * Else\n",
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" size += 2\n",
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" count++\n",
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" last_char = char\n",
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"* size += 2\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 == last_char\n",
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" 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(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: 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": null,
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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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" size = 0\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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" size += 2\n",
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" count = 1\n",
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" last_char = char\n",
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" size += 2\n",
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" if size >= len(string):\n",
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" return 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)\n",
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"\n",
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"string0 = None\n",
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"string1 = ''\n",
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"string2 = 'ABC'\n",
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"string3 = 'AAABCCDDDD'\n",
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"print(compress_string(string0))\n",
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"print(compress_string(string1))\n",
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"print(compress_string(string2))\n",
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"print(compress_string(string3))"
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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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"\n",
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"\n",
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"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",
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"\n",
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"* If bytearray is empty return bytearray\n",
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"* count = 0\n",
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"* size = 0\n",
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"* last_char_code = first char code in bytearray\n",
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"* For each char code in bytearray\n",
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" * If char code == last_char_code\n",
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" count++\n",
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" * Else\n",
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" size += 2\n",
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" count++\n",
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" last_char_code = char code\n",
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"* size += 2\n",
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"* If the compressed bytearray size is >= bytearray size, return string\n",
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"* Create compressed_bytearray\n",
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"* pos = 0\n",
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"* For each char code in bytearray\n",
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" * If char code == last_char_code\n",
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" count++\n",
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" * Else\n",
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" * compressed_bytearray[pos] = last_char_code\n",
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" * compressed_bytearray[pos + 1] = 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_bytearray[pos] = last_char_code\n",
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" * compressed_bytearray[pos + 1] = count\n",
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"* return compressed_bytearray\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": null,
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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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" size = 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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" size += 2\n",
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" count = 1\n",
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" last_char_code = char_code\n",
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" size += 2\n",
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" if size >= len(string):\n",
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" return 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\n",
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"\n",
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"string0 = None\n",
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"string1 = bytearray('')\n",
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"string2 = bytearray('ABC')\n",
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"string3 = bytearray('AAABCCDDDD')\n",
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"print(compress_string(string0))\n",
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"print(compress_string(string1))\n",
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"print(compress_string(string2))\n",
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"print(compress_string(string3))"
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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": "2.7.9"
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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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