feat(curriculum): Add basic recursion lesson (#36200)

Co-Authored-By: Randell Dawson <5313213+RandellDawson@users.noreply.github.com>
Co-Authored-By: Manish Giri <manish.giri.me@gmail.com>
pull/36304/head
Oliver Eyton-Williams 2019-06-19 20:20:44 +02:00 committed by mrugesh
parent 063ca6d82c
commit 2024e238fa
2 changed files with 113 additions and 1 deletions

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"5a2efd662fb457916e1fe604",
"Iterate with JavaScript Do...While Loops"
],
[
"5cfa3679138e7d9595b9d9d4",
"Replace Loops using Recursion"
],
[
"5688e62ea601b2482ff8422b",
"Profile Lookup"
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],
"helpRoom": "HelpJavaScript",
"fileName": "02-javascript-algorithms-and-data-structures/basic-javascript.json"
}
}

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---
id: 5cfa3679138e7d9595b9d9d4
title: Replace Loops using Recursion
challengeType: 1
videoUrl: 'https://www.freecodecamp.org/news/how-recursion-works-explained-with-flowcharts-and-a-video-de61f40cb7f9/'
---
## Description
<section id='description'>
Recursion is the concept that a function can be expressed in terms of itself. To help understand this, start by thinking about the following task: multiply the first <code>n</code> elements in an array to create the product of the elements. Using a <code>for</code> loop, you could do this:
```js
function multiply(arr, n) {
var product = arr[0];
for (var i = 1; i <= n; i++) {
product *= arr[i];
}
return product;
}
```
However, notice that <code>multiply(arr, n) == multiply(arr, n - 1) * arr[n]</code>. That means you can rewrite <code>multiply</code> in terms of itself and never need to use a loop.
```js
function multiply(arr, n) {
if (n <= 0) {
return arr[0];
} else {
return multiply(arr, n - 1) * arr[n];
}
}
```
The recursive version of <code>multiply</code> breaks down like this. In the <dfn>base case</dfn>, where <code>n <= 0</code>, it returns the result, <code>arr[0]</code>. For larger values of <code>n</code>, it calls itself, but with <code>n - 1</code>. That function call is evaluated in the same way, calling <code>multiply</code> again until <code>n = 0</code>. At this point, all the functions can return and the original <code>multiply</code> returns the answer.
<strong>Note:</strong> Recursive functions must have a base case when they return without calling the function again (in this example, when <code>n <= 0</code>), otherwise they can never finish executing.
</section>
## Instructions
<section id='instructions'>
Write a recursive function, <code>sum(arr, n)</code>, that creates the sum of the first <code>n</code> elements of an array <code>arr</code>.
</section>
## Tests
<section id='tests'>
``` yml
tests:
- text: <code>sum([1], 0)</code> should equal 1.
testString: assert.equal(sum([1], 0), 1);
- text: <code>sum([2, 3, 4], 1)</code> should equal 5.
testString: assert.equal(sum([2, 3, 4], 1), 5);
- text: Your code should not rely on any kind of loops (<code>for</code> or <code>while</code> or higher order functions such as <code>forEach</code>, <code>map</code>, <code>filter</code>, or <code>reduce</code>.).
testString: assert(!removeJSComments(code).match(/for|while|forEach|map|filter|reduce/g));
- text: You should use recursion to solve this problem.
testString: assert(removeJSComments(sum.toString()).match(/sum\(.*\).*\{.*sum\(.*\).*\}/s));
```
</section>
## Challenge Seed
<section id='challengeSeed'>
<div id='js-seed'>
```js
function sum(arr, n) {
// Only change code below this line
// Only change code above this line
}
```
</div>
### After Test
<div id='js-teardown'>
```js
const removeJSComments = str => str.replace(/\/\*[\s\S]*?\*\/|\/\/.*$/gm, '');
```
</div>
</section>
## Solution
<section id='solution'>
```js
function sum(arr, n) {
// Only change code below this line
if(n <= 0) {
return arr[0];
} else {
return sum(arr, n - 1) + arr[n];
}
// Only change code above this line
}
```
</section>