.

7, with a pencil and a paper:

• 0 -> 1 -> 2 -> 3 -> 4 -> 5 -> 6 -> 0
• 0 -> 6 -> 5 -> 4 -> 3 -> 2 -> 1 -> 0

You are correct. This was a very naive brute force implementation which, while it passed the tests, was computationally super expensive.

Thanks for the feedback, rewriting this was a nice exercise :)

Yep, 'i' refers to the index of the row as the board is traversed with the 'every' function.
So it's mapping the ith element of each row and passing arrays representing individual columns into checkDigits().

check my solution. What think about that.

You treat this case as empty label ""

yes

Hi,
I cannot see your solution, so I cannot comment on that. Without more specific details or a solution already submitted, there's not so much I can tell you.

That said, if you're sure that the array of possible combinations you are creating is right (that is, the values pass the tests), then you are in the right track already; I'd suggest you stop using the array, and instead, just use a counter, a variable where you keep the count of the solutions you find. You don't need to keep the combinations themselves, just the number of them.

However, as mentioned at the beginning, as I don't see your solution and don't know what approach you are taking or how your array is generated, so I don't know how useful my suggestion will be.