ee1e8abd87
* feat(tools): add seed/solution restore script * chore(curriculum): remove empty sections' markers * chore(curriculum): add seed + solution to Chinese * chore: remove old formatter * fix: update getChallenges parse translated challenges separately, without reference to the source * chore(curriculum): add dashedName to English * chore(curriculum): add dashedName to Chinese * refactor: remove unused challenge property 'name' * fix: relax dashedName requirement * fix: stray tag Remove stray `pre` tag from challenge file. Signed-off-by: nhcarrigan <nhcarrigan@gmail.com> Co-authored-by: nhcarrigan <nhcarrigan@gmail.com>
108 lines
2.7 KiB
Markdown
108 lines
2.7 KiB
Markdown
---
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id: 5cfa3679138e7d9595b9d9d4
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title: Replace Loops using Recursion
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challengeType: 1
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videoUrl: >-
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https://www.freecodecamp.org/news/how-recursion-works-explained-with-flowcharts-and-a-video-de61f40cb7f9/
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forumTopicId: 301175
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dashedName: replace-loops-using-recursion
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---
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# --description--
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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 `n` elements of an array to create the product of those elements. Using a `for` loop, you could do this:
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```js
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function multiply(arr, n) {
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var product = 1;
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for (var i = 0; i < n; i++) {
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product *= arr[i];
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}
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return product;
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}
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```
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However, notice that `multiply(arr, n) == multiply(arr, n - 1) * arr[n - 1]`. That means you can rewrite `multiply` in terms of itself and never need to use a loop.
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```js
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function multiply(arr, n) {
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if (n <= 0) {
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return 1;
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} else {
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return multiply(arr, n - 1) * arr[n - 1];
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}
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}
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```
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The recursive version of `multiply` breaks down like this. In the <dfn>base case</dfn>, where `n <= 0`, it returns 1. For larger values of `n`, it calls itself, but with `n - 1`. That function call is evaluated in the same way, calling `multiply` again until `n <= 0`. At this point, all the functions can return and the original `multiply` returns the answer.
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**Note:** Recursive functions must have a base case when they return without calling the function again (in this example, when `n <= 0`), otherwise they can never finish executing.
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# --instructions--
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Write a recursive function, `sum(arr, n)`, that returns the sum of the first `n` elements of an array `arr`.
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# --hints--
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`sum([1], 0)` should equal 0.
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```js
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assert.equal(sum([1], 0), 0);
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```
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`sum([2, 3, 4], 1)` should equal 2.
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```js
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assert.equal(sum([2, 3, 4], 1), 2);
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```
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`sum([2, 3, 4, 5], 3)` should equal 9.
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```js
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assert.equal(sum([2, 3, 4, 5], 3), 9);
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```
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Your code should not rely on any kind of loops (`for` or `while` or higher order functions such as `forEach`, `map`, `filter`, or `reduce`.).
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```js
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assert(
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!__helpers
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.removeJSComments(code)
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.match(/for|while|forEach|map|filter|reduce/g)
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);
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```
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You should use recursion to solve this problem.
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```js
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assert(
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__helpers.removeJSComments(sum.toString()).match(/sum\(.*\)/g).length > 1
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);
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```
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# --seed--
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## --seed-contents--
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```js
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function sum(arr, n) {
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// Only change code below this line
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// Only change code above this line
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}
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```
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# --solutions--
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```js
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function sum(arr, n) {
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// Only change code below this line
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if(n <= 0) {
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return 0;
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} else {
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return sum(arr, n - 1) + arr[n - 1];
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}
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// Only change code above this line
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}
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```
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