Added permutation challenge.

This commit is contained in:
Donne Martin
2015-07-03 19:56:38 -04:00
parent 0f34948626
commit fb7de503af
3 changed files with 248 additions and 22 deletions

View File

@@ -4,7 +4,14 @@
"cell_type": "markdown",
"metadata": {},
"source": [
"<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>"
"<small><i>This notebook was prepared by [Donne Martin](http://donnemartin.com). Source and license info is on [GitHub](https://github.com/donnemartin/coding-challenges).</i></small>"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"# Solution Notebook"
]
},
{
@@ -28,9 +35,9 @@
"source": [
"## Constraints\n",
"\n",
"*Problem statements are often intentionally ambiguous. Identifying constraints and stating assumptions can help to ensure you code the intended solution.*\n",
"*Problem statements are sometimes ambiguous. Identifying constraints and stating assumptions can help to ensure you code the intended solution.*\n",
"\n",
"* Can I assume the string is ASCII?\n",
"* Can we assume the string is ASCII?\n",
" * Yes\n",
" * Note: Unicode strings could require special handling depending on your language\n",
"* Is whitespace important?\n",
@@ -110,7 +117,7 @@
"Notes:\n",
"* Since the characters are in ASCII, we could potentially use an array of size 128 (or 256 for extended ASCII), where each array index is equivalent to an ASCII value\n",
"* Instead of using two hash maps, you could use one hash map and increment character values based on the first string and decrement based on the second string\n",
"* You can short circuit if the lengths of each string are not equal, len() in Python is generally O(1)\n",
"* You can short circuit if the lengths of each string are not equal, although len() in Python is generally O(1) unlike other languages like C where getting the length of a string is O(n)\n",
"\n",
"Complexity:\n",
"* Time: O(n)\n",
@@ -134,6 +141,7 @@
"source": [
"from collections import defaultdict\n",
"\n",
"\n",
"def unique_counts(string):\n",
" dict_chars = defaultdict(int)\n",
" for char in string:\n",
@@ -156,15 +164,51 @@
]
},
{
"cell_type": "markdown",
"metadata": {},
"cell_type": "code",
"execution_count": 3,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Overwriting test_permutation_solution.py\n"
]
}
],
"source": [
"*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.*"
"%%writefile test_permutation_solution.py\n",
"from nose.tools import assert_equal\n",
"\n",
"\n",
"class TestPermutation(object):\n",
" \n",
" def test_permutation(self, func):\n",
" assert_equal(func('', 'foo'), False)\n",
" assert_equal(func('Nib', 'bin'), False)\n",
" assert_equal(func('act', 'cat'), True)\n",
" assert_equal(func('a ct', 'ca t'), True)\n",
" print('Success: test_permutation')\n",
"\n",
"def main():\n",
" test = TestPermutation()\n",
" test.test_permutation(permutations)\n",
" try:\n",
" test.test_permutation(permutations_alt)\n",
" except NameError:\n",
" # Alternate solutions are only defined\n",
" # in the solutions file\n",
" pass\n",
" \n",
"if __name__ == '__main__':\n",
" main()"
]
},
{
"cell_type": "code",
"execution_count": 3,
"execution_count": 4,
"metadata": {
"collapsed": false
},
@@ -179,20 +223,7 @@
}
],
"source": [
"from nose.tools import assert_equal\n",
"\n",
"class Test(object):\n",
" def test_permutation(self, func):\n",
" assert_equal(func('', 'foo'), False)\n",
" assert_equal(func('Nib', 'bin'), False)\n",
" assert_equal(func('act', 'cat'), True)\n",
" assert_equal(func('a ct', 'ca t'), True)\n",
" print('Success: test_permutation')\n",
"\n",
"if __name__ == '__main__':\n",
" test = Test()\n",
" test.test_permutation(permutations)\n",
" test.test_permutation(permutations_alt)"
"run -i test_permutation_solution.py"
]
}
],