OP solved it, closing. But with a terrible time complexity algorithm

It actually works because the ascii offset should be 96 but by adjusting it to 95 you are adding one additional point for each letter; thus tracking the length implicitly via ascii value.

how would you solve this recursively w/o R?

love your solution. plain and simple for a complicated problem.

really love this kata, thank you g964. i'm however a bit puzzled at the moment over how the sets should be grouped.... i'm tored. about it. on one hand. the 4, 4, 4, 4, 0 make sense. the 6, 6, 6, 6, 4 doesn't since 7, 7, 7, 7, 0 would be a better fit (at least to me). what about 1, 1, 1, 2, 1.... can the 4th number be bigger as long as the 5th number is smaller?

i know you have probably answer this a million times already... i'm just not clear about the instructions. if you can pls explain the significance of the math behind the 11/5 => 2.2 much appreciated. cheers!!!!

You're right, the length is ignored. I don't remember how I missed it, I solved this task a year ago. Apparently, the solution worked back then :)

You spotted correctly that he's ignoring length in his scoring. Which is incorrect, and the random tests mostly catch this. Which makes it amazing this is voted "Best Practice".

Not that I should be throwing stones here .. :P I've made similar mistakes aplenty.

thanks for explaining casting,
the brackets with the 3 dots in it [...]

i got it now!

love that you figure out a way to build each level with the correct elements on it

Which exactly part is not clear?
Let's say we have a name "John". Then

_ = 'John'
_.toLowerCase() // => 'john'
[..._.toLowerCase()] // basically casts the string to a list => ['j', 'o', 'h', 'n']
// Then `reduce` converts the letters to numbers and sums them.

i really like this chunk

if (st === '' || n > we.length) {
    return st === '' ? "No participants" : "Not enough participants";
  }

and the rest i'm still admiring. thanks for your beautiful code!!!

nice to see a solution from the author. thanks