{ "cells": [ { "cell_type": "markdown", "metadata": {}, "source": [ "This notebook was prepared by [mrb00l34n](http://github.com/mrb00l34n). 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: Counting Ways of Making Change.\n", "\n", "* [Hints](#Hints)\n", "* [Algorithm](#Algorithm)\n", "* [Code](#Code)\n", "* [Unit Test](#Unit-Test)" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Hints\n", "\n", "* Can you think of a way to build up to a solution?\n", "* If there are 2 ways of making 3, and you are now given a coin of value v, how many ways can you make 3 + v?\n", "* Can you think of a way to divide the problem into smaller subproblems?" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Algorithm\n", "\n", "One possible solution using dynamic programming:\n", "* Create an array, s.t arr[i] = # of ways to make change for i\n", "* Initialize arr[0] = 1, arr[i>0] = 0\n", "* For each coin, and for each index from coin to n, increment arr[i] by arr[i - coin]\n", "\n", "How does this work?\n", "* As we iterate through each coin, we are adding the ways of making arr[i - coin] to arr[i]\n", "* If we have 2 ways of making 4, and are now iterating on a coin of value 3, there should be 2 ways of making 7.\n", "* We are essentially adding the coin we are iterating on to the # of ways of making arr[i].\n", "\n", "Complexity:\n", "* Time: O(mn); let the number of coins be m. We iterate from arr[coin] -> arr[n], or ~ n operations on each coin, hence n*m. \n", "* Space: O(n)" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Code" ] }, { "cell_type": "code", "execution_count": 1, "metadata": {}, "outputs": [], "source": [ "def change_ways(n, coins):\n", " arr = [1] + [0] * n\n", " for coin in coins:\n", " for i in range(coin, n + 1):\n", " arr[i] += arr[i - coin]\n", " return 0 if n == 0 else arr[n]" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Unit Test\n", "\n" ] }, { "cell_type": "code", "execution_count": 2, "metadata": {}, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Overwriting test_coin_change_ways.py\n" ] } ], "source": [ "%%writefile test_coin_change_ways.py\n", "import unittest\n", "\n", "\n", "class Challenge(unittest.TestCase):\n", "\n", " def test_coin_change_ways(self,solution):\n", " self.assertEqual(solution(0, [1, 2]), 0)\n", " self.assertEqual(solution(100, [1, 2, 3]), 884)\n", " self.assertEqual(solution(1000, range(1, 101)), \n", " 15658181104580771094597751280645)\n", " print('Success: test_coin_change_ways')\n", "\n", "\n", "def main():\n", " test = Challenge()\n", " test.test_coin_change_ways(change_ways)\n", "\n", "\n", "if __name__ == '__main__':\n", " main()" ] }, { "cell_type": "code", "execution_count": 3, "metadata": {}, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Success: test_coin_change_ways\n" ] } ], "source": [ "%run -i test_coin_change_ways.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 }