{ "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: Find the difference of two integers without using the + or - sign.\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", "* Can we assume the inputs are valid?\n", " * No, check for None\n", "* Can we assume this fits memory?\n", " * Yes" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Test Cases\n", "\n", "
\n", "* None input -> TypeError\n", "* 7, 5 -> 2\n", "* -5, -7 -> 2\n", "* -5, 7 -> -12\n", "* 5, -7 -> 12\n", "" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Algorithm\n", "\n", "We'll look at the following example, subtracting a and b:\n", "\n", "
\n", "a 0110 = 6 \n", "b 0101 = 5\n", "\n", "\n", "First, subtract a and b, without worrying about the borrow (0-0=0, 0-1=1, 1-1=0):\n", "\n", "result = a ^ b = 0011\n", "\n", "Next, calculate the borrow (0-1=1). We'll need to left shift one to prepare for the next iteration when we move to the next most significant bit:\n", "\n", "~a = 1001\n", " b = 0101\n", "~a & b = 0001\n", "\n", "borrow = (~a&b) << 1 = 0010\n", "\n", "If the borrow is not zero, we'll need to subtract the borrow from the result. Recursively call the function, passing in result and borrow.\n", "\n", "Complexity:\n", "* Time: O(b), where b is the number of bits\n", "* Space: O(b), where b is the number of bits" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Code" ] }, { "cell_type": "code", "execution_count": 1, "metadata": {}, "outputs": [], "source": [ "class Solution(object):\n", "\n", " def sub_two(self, a, b):\n", " if a is None or b is None:\n", " raise TypeError('a or b cannot be None')\n", " result = a ^ b;\n", " borrow = (~a & b) << 1\n", " if borrow != 0:\n", " return self.sub_two(result, borrow)\n", " return result;" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Unit Test" ] }, { "cell_type": "code", "execution_count": 2, "metadata": {}, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Overwriting test_sub_two.py\n" ] } ], "source": [ "%%writefile test_sub_two.py\n", "import unittest\n", "\n", "\n", "class TestSubTwo(unittest.TestCase):\n", "\n", " def test_sub_two(self):\n", " solution = Solution()\n", " self.assertRaises(TypeError, solution.sub_two, None)\n", " self.assertEqual(solution.sub_two(7, 5), 2)\n", " self.assertEqual(solution.sub_two(-5, -7), 2)\n", " self.assertEqual(solution.sub_two(-5, 7), -12)\n", " self.assertEqual(solution.sub_two(5, -7), 12)\n", " print('Success: test_sub_two')\n", "\n", "\n", "def main():\n", " test = TestSubTwo()\n", " test.test_sub_two()\n", "\n", "\n", "if __name__ == '__main__':\n", " main()" ] }, { "cell_type": "code", "execution_count": 3, "metadata": { "scrolled": true }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Success: test_sub_two\n" ] } ], "source": [ "%run -i test_sub_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 }