Use Recursion to Create a CountdownPassed (JS Algorithm)

Question

Question

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, newArr = []){
   
    if(n == 1){
    return newArr;
  }
    newArr.push(n);
    return countdown(n - 1)
  }


console.log(countdown(5));

My Question

Is there a way to fix this code so that it works?

I can provide an alternative solution, but I do not understand it:

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

In your recursive call, you only pass on n. For newArr the default parameter is taken, so you create a new array on every recursive step. — Jonas Wilms
– Jonas Wilms, Commented May 3, 2020 at 8:57
function countdown(n, newArr = []){ if(n == 1){ return newArr; } newArr.push(n); return countdown((n - 1), newArr) } console.log(countdown(5)); (tried this) — AndrewNeedsHelp
– AndrewNeedsHelp, Commented May 3, 2020 at 8:59

trincot · Accepted Answer · 2020-05-03 09:23:48Z

4

The problem is that you do not pass on the array to the recursive call, so each recursive execution creates a new, empty array. As a consequence, it does not return the array that had a value pushed to it, but the new, empty one that is coming back from the recursive calls.

Secondly, you never push value 1 to the array. So it would be better to stop the recursion at 0 instead of 1.

So taking those two fixes, you get this:

function countdown(n, newArr=[]) {
    if (n <= 0) {
        return newArr;
    }
    newArr.push(n);
    return countdown(n - 1, newArr)
}

console.log(countdown(5));

Your alternative solution is clean, because it does not need to pass an array as argument. It uses the returned array to add the next value to it (in front of it). It would have my preference.

To understand how it works, print out the intermediate values:

function countdown(n) {
  if (n < 1) {
    console.log("At the end of recursion. Creating and returning an empty array.");
    return [];
  } else {
    const arr = countdown(n - 1);
    console.log("Got the following array back from the recursive call:");
    console.log(JSON.stringify(arr));
    arr.unshift(n);
    console.log("Prefixing it with " + n + " and returning the result:");
    console.log(JSON.stringify(arr));
    return arr;
  }
}

var result = countdown(5);

answered May 3, 2020 at 9:23

trincot

357k38 gold badges282 silver badges338 bronze badges

Sign up to request clarification or add additional context in comments.

2 Comments

AndrewNeedsHelp Over a year ago

But what I don't understand is how does returning an empty array at the end solve anything? Surely it renders the whole function moot, as the end result is just: [empty array]

trincot Over a year ago

If you talk about the alternative solution: When the empty array is returned, the caller shifts a new value into it, so then it is no longer empty. And that is the array that is returned to its caller...etc. And so the array gets populated as execution backtracks out of the recursion. The last one to return the array will thus return the fully populated array. Just check the output that is logged at the end of my answer. The populating happens when getting out of recursion, not when getting into it.

Nina Scholz · Accepted Answer · 2020-05-03 09:25:24Z

2

You need to hand over the result array for the recursive call. And you need to check if no value is left, ten return the result array.

function countdown(n, result = []) {
    if (n < 1) return result;
    result.push(n);
    return countdown(n - 1, result);
}

console.log(countdown(5));

As another approach, you could return an array and for the exit condition take the final value, otherwise take n and the spreaded result of the recursive call.

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

console.log(countdown(5));

edited May 3, 2020 at 9:25

answered May 3, 2020 at 8:57

Nina Scholz

388k26 gold badges367 silver badges417 bronze badges

3 Comments

Jonas Wilms Over a year ago

And hows is that different from the OPs second version?

Nina Scholz Over a year ago

@JonasWilms, op uses the same array with mutaing this array. my solution take the array of the recursive call and spreads it into a new array, as i wrote above.

Jonas Wilms Over a year ago

The question was "Is there a way to fix this code so that it works?" ....

Abanoub Fathy · Accepted Answer · 2021-11-08 23:21:19Z

2

yes, you can modify your solution like that

function countdown(n){
       
        if(n == 0){
        // stop the function at 0 so it will not be included in the array
        return [];
      }
      
        // concat the value of n as an array with the value less than it
        return [n].concat(countdown(n - 1))
      }
    
    
    console.log(countdown(5));

the problem in your solution is that your array initialized as an empty array every time so the final answer will be an empty array

edited Nov 8, 2021 at 23:21

answered Nov 8, 2021 at 14:14

Abanoub Fathy

611 silver badge5 bronze badges

Comments

Khachik · Accepted Answer · 2021-05-16 15:38:53Z

0

At this point we will create the countdown function which call itself and called recursion.

function countdown(n) {
    if (n < 1) {
        return [];
    } else {
        console.log(n, "before calling");
        const arr = countdown(n - 1);
        console.log(n, "after calling");
        return arr;
    }
}
console.log(countdown(5));

And now when we know that the "before calling" is place where n is decrease and the "after calling" is place where n is increase, based on that we can do this.

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

answered May 16, 2021 at 15:38

Khachik

714 bronze badges

Collectives™ on Stack Overflow

Use Recursion to Create a CountdownPassed (JS Algorithm)

4 Answers 4

2 Comments

3 Comments

Comments

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

4 Answers 4

2 Comments

3 Comments

Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Related