The "else" does nothing in this code. Otherwise, great!

Cheaky ;)

Can't believe this is highest for best practice and clever.

You can do this:

• without iterating over both lists twice
• without finding mod 2 of every list item twice (only have to find it 3 times, total)
• without calculating the length of both arrays (if one of them is 1, clearly it's element 0 from that one, or the other if not)

Simply:

• Check if there are more odds or evens in the first 3 list items to determine the list orientation
• iterate over list until you've found the item that is opposite of the list orientation

O(n) instead of O(n*2)

Clever and best performance by far, but if your average developer has to stop and think too much about how it works, it's probably not a best practice. Yet... this is the highest rated best practice. Why?

There are submissions which fail the description. I've seen a couple that would fail on an array of [-2], just because if there's only 1 element, they return that element. The details clearly state for all negative numbers, return 0. Basically, you cannot return a negative and the tests do not explicitly include an array with only negative results, even though that's explicitly stated in the details.