Hey guys! @Adám if I have a train returning a vector, how can I multiply that whole vector by ⍳≢?

Actually I gave up on that approach and found a new one that requires me to concatenate the result of a train with an earlier result of the same train. Is that even possible without storing to a variable?

@J.Sallé (⊢ × later train result)earlier train result ?

@Adám oh nice, that works. I was using the parenthesis at the wrong spot I guess

@Adám so I got this as my final code (⊢,11|(+/⊢×(⍳≢)))3↓¯1↓<arg>. Any way to get rid of some of those parens?

@J.Sallé (⍳≢) → ⍳∘≢

@J.Sallé If the argument is Boolean, you can replace (+/⊢×(⍳≢)) with 1⊥⍸.

@Adám thanks! Unfortunately it's not a boolean argument, that would've been too good hahahahahah

@J.Sallé Yeah, I think we should change the definition of ⍸ from {(~~,⍵)/,⍳⍴⍵} to {(,⍵)/,⍳⍴⍵}.

@Adám I just noticed there's a constraint to the challenge that you need to output a string or a single integer, not a list of digits. How do I turn a vector of digits into a single string or number?

(⊢,11|(+/⊢×⍳∘≢))3↓¯1↓⍎¨'<arg_as_string>' is the current code

ah, I can do 10⊥(⊢,11|(+/⊢×⍳∘≢))3↓¯1↓⍎¨

Is there a better way?

Actually that doesn't work because it removes leading zeroes D:

@J.Sallé ∊⍕¨(or ∊0⍕¨ if in a train)?

@Adám ah, I didn't remember that one :p Thanks!

@J.Sallé you can avoid the parens around (+/⊢×⍳∘≢) if you use A+.×B instead of +/A×B