{ "cells": [ { "cell_type": "markdown", "metadata": {}, "source": [ "This notebook was prepared by [Donne Martin](https://github.com/donnemartin). 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: Determine if a number is a power of two.\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", "* Is the input number an int?\n", " * Yes\n", "* Can we assume the inputs are valid?\n", " * No\n", "* Is the output a boolean?\n", " * Yes\n", "* Can we assume this fits memory?\n", " * Yes" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Test Cases\n", "\n", "* None -> TypeError\n", "* 0 -> False\n", "* 1 -> True\n", "* 2 -> True\n", "* 15 -> False\n", "* 16 -> True" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Algorithm\n", "\n", "We can use bit manipulation to determine if a number is a power of two. \n", "\n", "For a number to be a power of two, there must only be one bit that is a `1`. \n", "\n", "We can use the following bit manipulation trick to determine this:\n", "\n", "`n & (n - 1)`\n", "\n", "Here's an example why:\n", "\n", "
```\n",
"0000 1000 = n\n",
"0000 0001 = 1\n",
"0000 0111 = n-1\n",
"\n",
"0000 1000 = n\n",
"0000 0111 = n-1\n",
"0000 0000 = n & n-1, result = 0\n",
"```
\n", "\n", "Complexity:\n", "* Time: O(1)\n", "* Space: O(1)" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Code" ] }, { "cell_type": "code", "execution_count": 1, "metadata": {}, "outputs": [], "source": [ "class Solution(object):\n", "\n", " def is_power_of_two(self, n):\n", " if n is None:\n", " raise TypeError('n cannot be None')\n", " if n <= 0:\n", " return False\n", " return (n & (n - 1)) == 0" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Unit Test" ] }, { "cell_type": "code", "execution_count": 2, "metadata": {}, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Overwriting test_is_power_of_two.py\n" ] } ], "source": [ "%%writefile test_is_power_of_two.py\n", "import unittest\n", "\n", "\n", "class TestSolution(unittest.TestCase):\n", "\n", " def test_is_power_of_two(self):\n", " solution = Solution()\n", " self.assertRaises(TypeError, solution.is_power_of_two, None)\n", " self.assertEqual(solution.is_power_of_two(0), False)\n", " self.assertEqual(solution.is_power_of_two(1), True)\n", " self.assertEqual(solution.is_power_of_two(2), True)\n", " self.assertEqual(solution.is_power_of_two(15), False)\n", " self.assertEqual(solution.is_power_of_two(16), True)\n", " print('Success: test_is_power_of_two')\n", "\n", "\n", "def main():\n", " test = TestSolution()\n", " test.test_is_power_of_two()\n", "\n", "\n", "if __name__ == '__main__':\n", " main()" ] }, { "cell_type": "code", "execution_count": 3, "metadata": {}, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Success: test_is_power_of_two\n" ] } ], "source": [ "%run -i test_is_power_of_two.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 }