isnt this O(3n) -> O(n). instead of O(2n) -> O(n). what i want to say is that concating both of these lists will make the algorithm slower. Since you are traversing the list 3 times instead of twice. This is the code which is faster than the one above: return arr1.reduce((sum, cur) => sum + cur) + arr2.reduce((sum, cur) => sum + cur);

cool

real🤣

Clean🔥

I second the approach but suggest creating the regular expression from the object keys.

I really like this idea of filtering out the negative items first!

That's how I did it :D

wow your solution is better than mine

Impressive)

practice more

goood job

Neat! Very clean.

Unrelated: reading your solution made Big O notation make sense to me. :)

Every item in the incoming array is processed twice, first to remove everything that's equal to or greater than zero to a fresh array, and then to sum every element of that new list. Every item is essentially touched twice so it's O squared, but technically maybe it's kind of smaller than that in an ideal case because the reducer only does its work on the resulting, probably smaller, subset of filtered numbers.

Thanks, I know this is sort of unrelated, I just needed to go off for a sec.

for initialitation
valeur a will begin with 0