interactive-coding-challenges/recursion_dynamic/steps/steps_solution.ipynb

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2017-03-27 17:19:37 +08:00
{
"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: You are running up n steps. If you can take a single, double, or triple step, how many possible ways are there to run up to the nth step?\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",
"* If n == 0, what should the result be?\n",
" * Go with 1, but discuss different approaches\n",
"* Can we assume the inputs are valid?\n",
" * No\n",
"* Can we assume this fits memory?\n",
" * Yes"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Test Cases\n",
"\n",
"* None or negative input -> Exception\n",
"* n == 0 -> 1\n",
"* n == 1 -> 1\n",
"* n == 2 -> 2\n",
"* n == 3 -> 4\n",
"* n == 4 -> 7\n",
"* n == 10 -> 274"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Algorithm\n",
"\n",
"To get to step n, we will need to have gone:\n",
"\n",
"* One step from n-1\n",
"* Two steps from n-2\n",
"* Three steps from n-3\n",
"\n",
"If we go the one step route above, we'll be at n-1 before taking the last step. To get to step n-1, we will need to have gone:\n",
"\n",
"* One step from n-1-1\n",
"* Two steps from n-1-2\n",
"* Three steps from n-1-2\n",
"\n",
"Continue this process until we reach the start.\n",
"\n",
"Base case:\n",
"\n",
"* If n < 0: return 0\n",
"* If n == 0: return 1\n",
"\n",
"Note, if we had chosen n == 0 to return 0 instead, we would need to add additional base cases. Otherwise we'd be adding multiple 0's once we hit the base cases and not get any result > 0.\n",
"\n",
"Recursive case:\n",
"\n",
"We'll memoize the solution to improve performance.\n",
"\n",
"* Use the memo if we've already processed the current step.\n",
"* Update the memo by adding the recursive calls to step(n-1), step(n-2), step(n-3)\n",
"\n",
"Complexity:\n",
"* Time: O(n), if using memoization\n",
"* Space: O(n), where n is the recursion depth\n",
"\n",
"Note: The number of ways will quickly overflow the bounds of an integer."
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Code"
]
},
{
"cell_type": "code",
"execution_count": 1,
"metadata": {},
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"outputs": [],
"source": [
"class Steps(object):\n",
"\n",
" def count_ways(self, num_steps):\n",
" if num_steps is None or num_steps < 0:\n",
" raise TypeError('num_steps cannot be None or negative')\n",
" cache = {}\n",
" return self._count_ways(num_steps, cache)\n",
"\n",
" def _count_ways(self, num_steps, cache):\n",
" if num_steps < 0:\n",
" return 0\n",
" if num_steps == 0:\n",
" return 1\n",
" if num_steps in cache:\n",
" return cache[num_steps]\n",
" cache[num_steps] = (self._count_ways(num_steps-1, cache) +\n",
" self._count_ways(num_steps-2, cache) +\n",
" self._count_ways(num_steps-3, cache))\n",
" return cache[num_steps]"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Unit Test"
]
},
{
"cell_type": "code",
"execution_count": 2,
"metadata": {},
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"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Overwriting test_steps.py\n"
]
}
],
"source": [
"%%writefile test_steps.py\n",
"import unittest\n",
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"\n",
"\n",
"class TestSteps(unittest.TestCase):\n",
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"\n",
" def test_steps(self):\n",
" steps = Steps()\n",
" self.assertRaises(TypeError, steps.count_ways, None)\n",
" self.assertRaises(TypeError, steps.count_ways, -1)\n",
" self.assertEqual(steps.count_ways(0), 1)\n",
" self.assertEqual(steps.count_ways(1), 1)\n",
" self.assertEqual(steps.count_ways(2), 2)\n",
" self.assertEqual(steps.count_ways(3), 4)\n",
" self.assertEqual(steps.count_ways(4), 7)\n",
" self.assertEqual(steps.count_ways(10), 274)\n",
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" print('Success: test_steps')\n",
"\n",
"\n",
"def main():\n",
" test = TestSteps()\n",
" test.test_steps()\n",
"\n",
"\n",
"if __name__ == '__main__':\n",
" main()"
]
},
{
"cell_type": "code",
"execution_count": 3,
"metadata": {},
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"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Success: test_steps\n"
]
}
],
"source": [
"%run -i test_steps.py"
]
}
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