No-no) Maybe I did not explain clearly enough: in that basic test, when 1-st work-chain ends at "9", "10" is just exactly the pass-step we need. So at the next sircle we start from [0]-index element again; if the last item in a basic chain was "9", to pass 1 step for keeping the order we would have to pass "2" and start the next chain from "4". And keep doing this till the arr has length =2, after what we would add these 2 last items in reverse mode.
Hope I did not confuse You)

In your solution of this test there is only one last item at the and so the way of adding it is only 1 too. But as I think, according 2 description there should always be one pass-step so when U start working with the rest of the arr again and again, U shold watch 4 this order: after first circle next 1 is ['2','[]',the] after which we have ['big','out','!'] and take from it ['out'] and ['!','big'] reversed at the end.
And what I definitely understand, this kata needs more detailed description.

U R missing next thing: when there are 2 last values of the arr, they should be added in reverse mode ( arr[1] than arr[0] ), according 2 description.

Hi there.

After successful basic tests I've got some problems with random tests in 'JavaScript' like these 2 below:

Expected: '['6', 'L']', instead got: '['X', 'n', 'o', '6', 'L']'
Expected: '['H']', instead got: '['R', 'o', 'H']'.

As I can see in comments it is a sort of a common problem. Can anybody explain me why is that happening ?

And question/ request to the author of this kata: can U fix the test or make the kata description more detailed to prevent problems like those 2 above?

Agree)
That's why I solved this kata taking into account the conditions U R talking about without waiting till the description will B changed.