Chunkey monkey clarification

I am trying to solve it recursively and wanna set proper base cases.

There are no test cases for size ===0.

But i guess chunkArrayInGroups(arr, size) should return empty array for the case above? Is that correct?

Edit. For moderators.
I accidentally posted it in General subforum. Is there an opportunity to move thread in JS subforum?

I didn’t write recursive step yet.

But I think I will use base case for size===1:

function chunkArrayInGroups(arr, size) {
  //base case
  if (size===1) {
    return [arr]
  //to do: figure out recursive step
  //try to grasp difference between cases size=== 3 and size===4
  //for some testArray

basecasecheck = chunkArrayInGroups(["a", "b", "c", "d"], 1);

console.log(basecasecheck);//[ [ 'a', 'b', 'c', 'd' ] ]

I think in recursive step will be some kind of call like below:

chunkArrayInGroups(arr, size + 1)

For now I didn’t figured out what methods I wanna use, but I have a general idea

So I need to know how function should behave when size < 1 to add some more base cases

Your size should not change at all during this process…

Makes sense, I’ll work on recursive step and will provide more info after that

I guess I gave little more info than needed in the original post.

My question:

What is expected behaviour for chunkArrayInGroups(arr, size) when size < 1?

There are no such tests in the challenge step. Should I be bothered with this at all?

The challenge does not specify but I would probably expect an empty array since an array of size zero has to be empty.

That is up to you.

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