fix(curriculum): Improved Use Recursion to Create a Countdown challenge (#37548)
* fix: improved recursive challenge * fix: rewording of two challenges * fix: add code tags Co-Authored-By: Tom <20648924+moT01@users.noreply.github.com>
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Oliver Eyton-Williams
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@ -8,15 +8,34 @@ forumTopicId: 305925
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## Description
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<section id='description'>
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Continuing from the previous challenge, we provide you another opportunity to create a recursive function to solve a problem.
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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 <code>1</code> 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 <code>1</code> through <code>n</code>. This function will need to accept an argument, <code>n</code>, representing the final number. Then it will need to call itself with progressively smaller values of <code>n</code> until it reaches <code>1</code>. 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)); // [ 1, 2, 3, 4, 5 ]
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```
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At first, this seems counterintuitive since the value of `n` <em>decreases</em>, but the values in the final array are <em>increasing</em>. This happens because the push happens last, after the recursive call has returned. At the point where `n` is pushed into the array, `count(n - 1)` has already been evaluated and returned `[1, 2, ..., n - 1]`.
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</section>
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## Instructions
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<section id='instructions'>
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We have defined a function called <code>countdown</code> with two parameters. The function should take an array in the <code>myArray</code> parameter and append the numbers n through 1 based on the <code>n</code> parameter.
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For example, calling this function with <code>n = 5</code> will pad the array with the numbers <code>[5, 4, 3, 2, 1]</code> inside of it.
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We have defined a function called <code>countdown</code> with one parameter (<code>n</code>). The function should use recursion to return an array containing the integers <code>n</code> through <code>1</code> based on the <code>n</code> parameter. If the function is called with a number less than 1, the function should return an empty array.
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For example, calling this function with <code>n = 5</code> should return the array <code>[5, 4, 3, 2, 1]</code>.
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Your function must use recursion by calling itself and must not use loops of any kind.
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</section>
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@ -26,13 +45,13 @@ Your function must use recursion by calling itself and must not use loops of any
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``` yml
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tests:
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- text: After calling <code>countdown(myArray, -1)</code>, myArray should be empty.
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testString: assert.isEmpty(padArray([], -1));
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- text: After calling <code>countdown(myArray, 10)</code>, myArray should contain <code>[10, 9, 8, 7, 6, 5, 4, 3, 2, 1]</code>
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testString: assert.deepStrictEqual(padArray([], 10), [10, 9, 8, 7, 6, 5, 4, 3, 2, 1]);
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- text: After calling <code>countdown(myArray, 5)</code>, myArray should contain <code>[5, 4, 3, 2, 1]</code>
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testString: assert.deepStrictEqual(padArray([], 5), [5, 4, 3, 2, 1]);
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- 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>.).
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- text: <code>countdown(-1)</code> should return an empty array.
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testString: assert.isEmpty(countdown(-1));
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- text: <code>countdown(10)</code> should return <code>[10, 9, 8, 7, 6, 5, 4, 3, 2, 1]</code>
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testString: assert.deepStrictEqual(countdown(10), [10, 9, 8, 7, 6, 5, 4, 3, 2, 1]);
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- text: <code>countdown(5)</code> should return <code>[5, 4, 3, 2, 1]</code>
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testString: assert.deepStrictEqual(countdown(5), [5, 4, 3, 2, 1]);
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- text: Your code should not rely on any kind of loops (<code>for</code>, <code>while</code> or higher order functions such as <code>forEach</code>, <code>map</code>, <code>filter</code>, and <code>reduce</code>).
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testString: assert(!removeJSComments(code).match(/for|while|forEach|map|filter|reduce/g));
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- text: You should use recursion to solve this problem.
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testString: assert(removeJSComments(countdown.toString()).match(/countdown\s*\(.+\)\;/));
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@ -49,10 +68,10 @@ tests:
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//Only change code below this line
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function countdown(myArray, n){
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function countdown(n){
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return;
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}
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console.log(countdown(5)); // [5, 4, 3, 2, 1]
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```
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</div>
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@ -62,11 +81,6 @@ function countdown(myArray, n){
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```js
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const removeJSComments = str => str.replace(/\/\*[\s\S]*?\*\/|\/\/.*$/gm, '');
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function padArray(arr, n){
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countdown(arr, n);
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return arr;
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}
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```
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</div>
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@ -78,14 +92,8 @@ function padArray(arr, n){
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```js
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//Only change code below this line
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function countdown(myArray, n){
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if(n <= 0){
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return;
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}
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else{
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myArray.push(n);
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countdown(myArray, n - 1);
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}
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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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@ -8,25 +8,7 @@ forumTopicId: 301180
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## Description
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<section id='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 <code>1</code> 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 <code>n</code> representing the final number. Then it will need to call itself with progressively smaller values of <code>n</code> until it reaches 1. You could write the function as follows:
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```js
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function count(n) {
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if (n === 1) {
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return [1];
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} else {
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var numbers = count(n - 1);
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numbers.push(n);
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return numbers;
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}
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}
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```
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At first this is counterintuitive since the value of `n` <em>decreases</em>, but the values in the final array are <em>increasing</em>. This happens because the push happens last, after the recursive call has returned. At the point where `n` is pushed into the array, `count(n - 1)` has already been evaluated and returned `[1, 2, ..., n - 1]`.
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Continuing from the previous challenge, we provide you another opportunity to create a recursive function to solve a problem.
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</section>
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