Use Recursion to Create a Countdown/ Help!

Hi guys! I almost finish the JS course, but I am stuck here

This is the code:

function countup(n) {
  if (n < 1) {
    return [];
  } else {
    const countArray = countup(n - 1); // HERE! HELP! I don't get it when it says: countup(n - 1) , okay, it uses the function, but it should be decreasing then, not increasing//
    countArray.push(n);
    return countArray;
  }
}
console.log(countup(5)); // [ 1, 2, 3, 4, 5 ]

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] . I do not get this part, could you please give me another example? I would appreciate it a lot!

Also this is the lesson. I have to submit a solution to the opposite type, when it is truly incrising, as I do no it get it, I am not able to move forward. Thanks in advance!

Let’s look at a series of function calls.

Your countup(0) returns []

Your countup(1) returns the result of countup(0) with 1 pushed onto it, or [1]

Your countup(2) returns the result of countup(1) with 2 pushed onto it, or [1, 2]

Your countup(3) returns the result of countup(2) with 3 pushed onto it, or [1, 2, 3]

Your countup(4) returns the result of countup(3) with 4 pushed onto it, or [1, 2, 3, 4]

Your countup(5) returns the result of countup(4) with 5 pushed onto it, or [1, 2, 3, 4, 5]

That’s really all there is to it. It is possible to overthink it in this case.

Edit: Woops, yeah, push is at the end. Time for more coffee.

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shouldn’t the numbers in the array be in increasing order if you are pushing to it?

if you push 2 to [1] doesn’t it become [1, 2]?

That’s exactly my question. The lesson affirms: 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] ., but I don’t get it

You are describing what happens exactly, so I’m confused about how you are confused?

    countArray.push(n);

isn’t evaluated until

const countArray = countup(n - 1); 

has returned.

1 Like

I do not understand why it increases. it should be something like Array: [5,4,3], instead of what I get [1,2,3]

Push added an element to the end of the array. Each function call gets completed in order.

Your countup(5) returns the result of countup(4) with 5 push ed onto it, or [1, 2, 3, 4, 5]. This means that countup(4) must be called and finish returning before countup(5) pushes 5 and returns the array.

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@JeremyLT I wholeheartedly appreciate your support. Please: Take a look, I am sure I can bypass this if I undertstand the following issue:

Let’s suppose this is my code:
function countdown(n) {
if (n < 1) {
return ;
} else {
const arr = countdown(n - 1);
arr.unshift(n);
return arr;
}
}
console.log(countdown(2));

OK. The function begins with 2.

function countdown(2) {
if (2 < 1) {
return ;
} else {
const arr = countdown(2 - 1); //Now it is remaning “1” one! Fine…
arr.unshift(2);
return arr;
}
}

It addes number “2” to a new array, which it is: [2].

Now:

function countdown(1) {
if (1 < 1) {
return ;
} else {
const arr = countdown(1 - 1); //Now it is remaning “0”, we are done, fin!
arr.unshift(1);
return arr;
}
}

I already have [2], and as it calls unshift function, which means it addes a new value at the beginning, so we now have: [1,2]. The console gives me: [2,1] :/, it’s quite the opposite! please

You have an incomplete call stack.

countdown(2) calls countdown(1).

countdown(1) calls countdown(0).

countdown(0) returns [].

countdown(1) then unshifts 1 onto [], returning [1].

countdown(2) then unshifts 2 onto [1], returning [2, 1].

1 Like

I got it! Many thanks Jeremy! Night night!

1 Like