freeCodeCamp/curriculum/challenges/english/02-javascript-algorithms-and-data-structures/basic-javascript/use-recursion-to-create-a-range-of-numbers.english.md

id	title	challengeType	forumTopicId
5cc0bd7a49b71cb96132e54c	Use Recursion to Create a Range of Numbers	1	301180

Description

In a previous challenge, 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.

As mentioned in the previous challenge, there will be a base case. 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 recursive call which executes the original function with different arguments. If the function is written correctly, eventually the base case will be reached.

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:

function count(n) {
  if (n === 1) {
    return [1];
  } else {
    var numbers = count(n - 1); 
    numbers.push(n);
    return numbers;
  }
}

At first this is 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, count(n - 1) has already been evaluated and returned [1, 2, ..., n - 1].

Instructions

We have defined a function named rangeOfNumbers with two parameters. The function should return an array of integers which begins with a number represented by the startNum parameter and ends with a number represented by the endNum parameter. The starting number will always be less than or equal to the ending number. Your function must use recursion by calling itself and not use loops of any kind. It should also work for cases where both startNum and endNum are the same.

Tests

tests:
  - text: Your function should return an array.
    testString: assert(Array.isArray(rangeOfNumbers(5, 10)));
  - text: Your code should not use any loop syntax (<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: <code>rangeOfNumbers</code> should use recursion (call itself) to solve this challenge.
    testString: assert(removeJSComments(rangeOfNumbers.toString()).match(/rangeOfNumbers\s*\(.+\)/));
  - text: <code>rangeOfNumbers(1, 5)</code> should return <code>[1, 2, 3, 4, 5]</code>.
    testString: assert.deepStrictEqual(rangeOfNumbers(1, 5), [1, 2, 3, 4, 5]);
  - text: <code>rangeOfNumbers(6, 9)</code> should return <code>[6, 7, 8, 9]</code>.
    testString: assert.deepStrictEqual(rangeOfNumbers(6, 9), [6, 7, 8, 9]);
  - text: <code>rangeOfNumbers(4, 4)</code> should should return <code>[4]</code>.
    testString: assert.deepStrictEqual(rangeOfNumbers(4, 4), [4]);

Challenge Seed

function rangeOfNumbers(startNum, endNum) {
  return [];
};

After Test

const removeJSComments = str => str.replace(/\/\*[\s\S]*?\*\/|\/\/.*$/gm, '');

Solution

function rangeOfNumbers(startNum, endNum) {
  if (endNum - startNum === 0) {
    return [startNum];
  } else {
    var numbers = rangeOfNumbers(startNum, endNum - 1);
    numbers.push(endNum);
    return numbers;
  }
}