Question 227958 in Javascript Question posted in
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Javascript recursion function that takes an argument n and recursively reduces it and stores the value of each in an array until it reaches 1

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.

function countdown(n){
    if (n < 1){
        return [];
    }else{
        const countArray = countdown(n - 1);
        countArray.push(n);
        return countArray;
    }
 }

I am getting the following errors:
countdown(10) should return [10, 9, 8, 7, 6, 5, 4, 3, 2, 1] countdown(5) should return [5, 4, 3, 2, 1] Global variables should not be used to cache the array.

Tags:

3

Answers


  1. 0

    You could use an optional argument to pass the result array into the recursive calls:

    function countdown(n, arr = []){
  return n < 1 ? arr : arr.push(n) && countdown(n - 1, arr);
}

console.log(...countdown(10));

    The spread is slooooow:

    enter image description here

    <script benchmark data-count="1000">

const count=1000;

// @benchmark spread/iterator
{
  function countdown(n) {
      const arr = [n];
      if( n > 1) arr.push( ...countdown(n-1));
      return arr;
  }
  countdown(count);
}

// @benchmark recursion
{
  function countdown(n, arr = []){
    return n < 1 ? arr : arr.push(n) && countdown(n - 1, arr);
  }
  countdown(count);
}

</script>
<script src="https://cdn.jsdelivr.net/gh/silentmantra/benchmark/loader.js"></script>
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  2. 0

    This works using the spread operator.

    function countdown(n) {
    const arr = [n];
    if( n > 1) arr.push( ...countdown(n-1));
    return arr;
}
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  3. 0

    There are many possible ways of going about this… here is just one:

    Here: I’ve used the following:

    function countdown(n) {
  if (n < 1) {
    return []
  } else {
    return [n, ...countdown(n - 1)]
  }
}
console.log(countdown(5));

    The spread operator unpacks elements of iterable objects such as arrays, sets, and maps into a list.
    So the line

      return [n, ...countdown(n - 1)]

    If we step through an example like countdown(2) it would look like this:

      // n = 2
  return [2, ...countdown(1)]
  // countdown(1) evaluates to [1]
  // meaning it looks like:
  // return [2, ...[1]]
  // which is: [2, 1]

      // n = 1
  return [1, ...countdown(0)]
  // countdown(0) evaluates to []
  // meaning it looks like:
  // return [1, ...[]]
  // which is: [1] (because `[]` is empty)
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