db369fbed1
* Removes comments in js challanges by default * fix local-scope-and-functions test regex * fix all languages * revert language changes * removed unnecessary removeJSComments from challenges * fix challanges in other languages * removed removeJSComments from all challanges
93 lines
3.2 KiB
Markdown
93 lines
3.2 KiB
Markdown
---
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id: 5cd9a70215d3c4e65518328f
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title: Use Recursion to Create a Countdown
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challengeType: 1
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forumTopicId: 305925
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dashedName: use-recursion-to-create-a-countdown
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---
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# --description--
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In a [previous challenge](/learn/javascript-algorithms-and-data-structures/basic-javascript/replace-loops-using-recursion), you learned how to use recursion to replace a `for` loop. Now, let's look at a more complex function that returns an array of consecutive integers starting with `1` through the number passed to the function.
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As mentioned in the previous challenge, there will be a <dfn>base case</dfn>. The base case tells the recursive function when it no longer needs to call itself. It is a simple case where the return value is already known. There will also be a <dfn>recursive call</dfn> which executes the original function with different arguments. If the function is written correctly, eventually the base case will be reached.
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For example, say you want to write a recursive function that returns an array containing the numbers `1` through `n`. This function will need to accept an argument, `n`, representing the final number. Then it will need to call itself with progressively smaller values of `n` until it reaches `1`. You could write the function as follows:
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```javascript
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function countup(n) {
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if (n < 1) {
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return [];
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} else {
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const countArray = countup(n - 1);
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countArray.push(n);
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return countArray;
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}
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}
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console.log(countup(5));
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```
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The value `[1, 2, 3, 4, 5]` will be displayed in the console.
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At first, this seems counterintuitive since the value of `n` *decreases*, but the values in the final array are *increasing*. This happens because the push happens last, after the recursive call has returned. At the point where `n` is pushed into the array, `countup(n - 1)` has already been evaluated and returned `[1, 2, ..., n - 1]`.
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# --instructions--
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We have defined a function called `countdown` with one parameter (`n`). The function should use recursion to return an array containing the integers `n` through `1` based on the `n` parameter. If the function is called with a number less than 1, the function should return an empty array. For example, calling this function with `n = 5` should return the array `[5, 4, 3, 2, 1]`. Your function must use recursion by calling itself and must not use loops of any kind.
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# --hints--
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`countdown(-1)` should return an empty array.
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```js
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assert.isEmpty(countdown(-1));
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```
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`countdown(10)` should return `[10, 9, 8, 7, 6, 5, 4, 3, 2, 1]`
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```js
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assert.deepStrictEqual(countdown(10), [10, 9, 8, 7, 6, 5, 4, 3, 2, 1]);
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```
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`countdown(5)` should return `[5, 4, 3, 2, 1]`
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```js
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assert.deepStrictEqual(countdown(5), [5, 4, 3, 2, 1]);
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```
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Your code should not rely on any kind of loops (`for`, `while` or higher order functions such as `forEach`, `map`, `filter`, and `reduce`).
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```js
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assert(
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!code.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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countdown.toString().match(/countdown\s*\(.+\)/)
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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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// Only change code below this line
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function countdown(n){
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return;
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}
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// Only change code above this line
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```
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# --solutions--
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```js
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function countdown(n){
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return n < 1 ? [] : [n].concat(countdown(n - 1));
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}
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```
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