Nice!

Neeeeat!

Interesting way of thinking about it, but your solution makes sense. Clever!

Wasn't specified in the kata description but that's a good edge case to consider, I suppose.

Kudos for including multiple ways to do this. Second one is simplest but the first one is really clever!

Smart!

This is very clever! But two questions: 1) how does this take care of the scenario when you don't have subsequent negative numbers? (i.e. [3, -30, 1, -20 ...so on]), and 2) on the same note, with the c <= 0 conditonal, I see that you are trying to take care of the 0 scenario plus the negative number scenario. But.. when if the current is 0 and the next number is a positive one? Would that mess up the computation since you need to do p[0] += 1, not p[1] += c? Hope I'm making sense here.

Nice! This is easy to read and understand, so that's a plus. A few tips for optimization:

• Personally I would have leave out the 'continue' block since your two following conditionals take care of checking that inadvertedly.
• Instead of creating a newArr variable, you could have used array literal syntax and put [] and [positiveNumber, negativeNumber] in replacement of where you used the newArr variable. Saves unnecessary computation from push(), and reduces variables being stored in your namespace.
• Speaking of push(), you also didn't have to call it twice as just using newArr.push(positiveNumber, negativeNumber); would have just done the job as well.

That's mainly it but I do think this is a nice, straightforward solution. Good job!

Clever! Agree with the prev comments, but this was a smart solution nonetheless.