StarMirror/interactive-coding-challenges
1
0
Fork 0
mirror of https://github.com/donnemartin/interactive-coding-challenges.git synced 2024-03-22 13:11:13 +08:00
Code Issues Projects Releases Wiki Activity

Reworked notebook: Added more detail to clarifying questions and test cases. Reworked algorithm discussion, code, and unit test.

Browse Source
This commit is contained in:
Donne Martin 2015-06-24 18:20:20 -04:00
parent 369a073c85
commit afa2f60483
1 changed files with 83 additions and 73 deletions
Show Stats Download Patch File Download Diff File Expand all files Collapse all files

146
arrays-strings/compress.ipynb
View File

@ -18,7 +18,8 @@
"* [Algorithm: List](#Algorithm:-List)\n",
"* [Code: List](#Code:-List)\n",
"* [Algorithm: Byte Array](#Algorithm:-Byte-Array)\n",
"* [Code: Byte array](#Code:-Byte-Array)"
"* [Code: Byte array](#Code:-Byte-Array)\n",
"* [Unit Test](#Unit-Test)"
]
},
{
@ -27,8 +28,11 @@
"source": [
"## Clarifying Questions\n",
"\n",
"* Is the string ASCII (extended)? Or Unicode?\n",
" * ASCII extended, which is 256 characters\n",
"*Problem statements are sometimes intentionally ambiguous. Asking clarifying questions, identifying constraints, and stating assumptions help to ensure you code the intended solution.*\n",
"\n",
"* Can I assume the string is ASCII?\n",
" * Yes\n",
" * Note: Unicode strings could require special handling depending on your language\n",
"* Can you use additional data structures? \n",
" * Yes\n",
"* Is this case sensitive?\n",
@ -43,69 +47,37 @@
"source": [
"## Test Cases\n",
"\n",
"*Identifying and running through general and edge cases are important. You generally will not be asked to write a unit test like what is shown below.*\n",
"\n",
"* NULL\n",
"* '' -> ''\n",
"* 'ABC' -> 'ABC'\n",
"* 'AAABCCDDDD' -> 'A3B1C2D4'"
]
},
{
"cell_type": "code",
"execution_count": null,
"metadata": {
"collapsed": true
},
"outputs": [],
"source": [
"from nose.tools import assert_equal\n",
"\n",
"class Test(object):\n",
" def test_compress(self, func):\n",
" assert_equal(func(None), None)\n",
" assert_equal(func(''), '')\n",
" assert_equal(func('ABC'), 'ABC')\n",
" assert_equal(func('AAABCCDDDD'), 'A3B1C2D4')\n",
"\n",
"def run_tests(func):\n",
" test = Test()\n",
" test.test_compress(func)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Algorithm: List\n",
"\n",
"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",
"\n",
"![alt text](https://raw.githubusercontent.com/donnemartin/algorithms-data-structures/master/images/compress_string.jpg)\n",
"\n",
"Since Python strings are immutable, we'll use a list of characters to exercise string 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",
"\n",
"* If string is empty return string\n",
"* count = 0\n",
"* size = 0\n",
"* last_char = first char in string\n",
"* For each char in string\n",
" * If char == last_char\n",
" count++\n",
" * Else\n",
" size += 2\n",
" count++\n",
" last_char = char\n",
"* size += 2\n",
"* Calculate the size of the compressed string\n",
"* If the compressed string size is >= string size, return string\n",
"* Create compressed_string\n",
"* For each char in string\n",
" * If char == last_char\n",
" count++\n",
" * For each char in string\n",
" * If char is the same as last_char, increment count\n",
" * Else\n",
" * Append last_char to compressed_string\n",
" * append count to compressed_string\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",
" * Append count to compressed_string\n",
" * Return compressed_string\n",
"\n",
"Complexity:\n",
"* Time: O(n)\n",
@ -121,7 +93,7 @@
},
{
"cell_type": "code",
"execution_count": null,
"execution_count": 1,
"metadata": {
"collapsed": false
},
@ -130,19 +102,22 @@
"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",
" count = 0\n",
" last_char = string[0]\n",
" for char in string:\n",
" if char == last_char:\n",
" count += 1\n",
" else:\n",
" if char != last_char:\n",
" size += 2\n",
" count = 1\n",
" last_char = char\n",
" size += 2\n",
" \n",
" # If the compressed string size is greater than \n",
" # or equal to string size, return string\n",
" if size >= len(string):\n",
" return string\n",
"\n",
" # Create compressed_string\n",
" compressed_string = list()\n",
" count = 0\n",
" last_char = string[0]\n",
@ -156,9 +131,7 @@
" last_char = char\n",
" compressed_string.append(last_char)\n",
" compressed_string.append(str(count))\n",
" return \"\".join(compressed_string)\n",
"\n",
"run_tests(compress_string)"
" return \"\".join(compressed_string)"
]
},
{
@ -167,11 +140,7 @@
"source": [
"## Algorithm: Byte Array\n",
"\n",
"![alt text](https://raw.githubusercontent.com/donnemartin/algorithms-data-structures/master/images/compress_string.jpg)\n",
"\n",
"Since Python strings are immutable, we'll use a bytearray to exercise array manipulation. As seen above, 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",
"\n",
"The algorithm is the same, 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",
"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",
"\n",
"Complexity:\n",
"* Time: O(n)\n",
@ -187,7 +156,7 @@
},
{
"cell_type": "code",
"execution_count": null,
"execution_count": 2,
"metadata": {
"collapsed": false
},
@ -196,19 +165,22 @@
"def compress_string_alt(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",
" count = 0\n",
" last_char_code = string[0]\n",
" for char_code in string:\n",
" if char_code == last_char_code:\n",
" count += 1\n",
" else:\n",
" if char_code != last_char_code:\n",
" size += 2\n",
" count = 1\n",
" last_char_code = char_code\n",
" size += 2\n",
" \n",
" # If the compressed string size is greater than \n",
" # or equal to string size, return string \n",
" if size >= len(string):\n",
" return string\n",
" \n",
" # Create compressed_string\n",
" compressed_string = bytearray(size)\n",
" pos = 0\n",
" count = 0\n",
@ -218,15 +190,53 @@
" count += 1\n",
" else:\n",
" compressed_string[pos] = last_char_code\n",
" compressed_string[pos + 1] = ord(str(count))\n",
" compressed_string[pos+1] = ord(str(count))\n",
" pos += 2\n",
" count = 1\n",
" last_char_code = char_code\n",
" compressed_string[pos] = last_char_code\n",
" compressed_string[pos + 1] = ord(str(count))\n",
" return compressed_string\n",
" compressed_string[pos+1] = ord(str(count))\n",
" return compressed_string"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Unit Test"
]
},
{
"cell_type": "code",
"execution_count": 3,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Success: test_compress\n",
"Success: test_compress\n"
]
}
],
"source": [
"from nose.tools import assert_equal\n",
"\n",
"run_tests(compress_string_alt)"
"class Test(object):\n",
" def test_compress(self, func):\n",
" assert_equal(func(None), None)\n",
" assert_equal(func(''), '')\n",
" assert_equal(func('ABC'), 'ABC')\n",
" assert_equal(func('AAABCCDDDD'), 'A3B1C2D4')\n",
" print('Success: test_compress')\n",
"\n",
"if __name__ == '__main__':\n",
" test = Test()\n",
" test.test_compress(compress_string)\n",
" test.test_compress(compress_string_alt)"
]
}
],