Day 3: Lobby

Megathread guidelines

  • Keep top level comments as only solutions, if you want to say something other than a solution put it in a new post. (replies to comments can be whatever)
  • You can send code in code blocks by using three backticks, the code, and then three backticks or use something such as https://topaz.github.io/paste/ if you prefer sending it through a URL

FAQ

  • mykl@lemmy.world
    link
    fedilink
    arrow-up
    6
    ·
    edit-2
    1 month ago

    Uiua

    You can run this in Uiua Pad

    "987654321111111\n811111111111119\n234234234234278\n818181911112111"
    ⊜∘⊸≠@\n
    Jolt  (⊙⊃⋅∘∘+1(⊡|˜↘⊙+₁)⍜↘⟜⨂⊸/↥)))(+1¯)
    ∩/+⊃≡(Jolt2)(Jolt12)
    
    Explanation

    You’re always looking for the highest digit far enough from the end of the string that there’s space to pick the remaining number of digits. So the algorithm is very straightforward: just temporarily exclude the last n-1 characters off the end of the string and look for the first instance of the largest digit in that substring, store that char and behead the string at that point. Repeat n times.

      • mykl@lemmy.world
        link
        fedilink
        arrow-up
        2
        ·
        1 month ago

        Thanks, although it only really became clear after I’d written it. I am thinking about doing a more detailed explainer post based on one day’s answer to help people get a feel for it.