Loading collection data...
Collections are a way for you to organize kata so that you can create your own training routines. Every collection you create is public and automatically sharable with other warriors. After you have added a few kata to a collection you and others can train on the kata contained within the collection.
Get started now by creating a new collection.
I get a SIGABRT with my code. Surely, my code is wrong but I do not allocate any memory so how is it possible?
Besides, I'm seeing
Expected: "*&ikh)ihg&j'kj%&hij'-&'*ih)h'"
which should not be possible ("Every String consists of only lowercase letters and spaces.")
Using the very big hammer, nice.
This won't handle empty lists. Tests have been updated to check for that.
no it's not, the reference solution was using a
+
instead of*
. fixed.Ruby version: For "" (empty string) the expected return is false although all gs are happy. Is that correct?
That's a common feeling, it's okay to feel that. Just remember: you aren't stupid, and you weren't. Sometimes a one's mind can play tricks and block a way to the solution, even the simplest soulutions. But now you've made it, you've learned something new. Cheers! :)
i feel so stupid right now
I did not remember which variable the number pi was stored in, I was too lazy to find out about it at the time of solving the problem
It's always good to have a π along ;-)
indeed
Thanks, that was fast.
The initial code problem has been fixed,
[Bool]
is now[Word]
. My apologies for this oversight.Yes, it is unclear why tube sizes are passed. This is not Haskell-specific though.
ETA: if you have a Haskell-specific issue, consider prodding the translation author under the translation
Suggestion
. I may or may not get notifications for a newIssue
, but I will get notifications for replies to mySuggestion
( because I authored that ).Haskell version
The code template says "balanced :: [Bool] -> Bool" but the required type is
"balanced :: [Word] -> Bool".
Also it is unclear why tube sizes are passed if they are all the same (or 0).
This also groups repeated digits in the integral part.
repeating_fractions(221, 1) => "(2)1.0"
The regexp should probably be /(\d)(\1+)(?!..)/.
Finally a solution in O(n).
Loading more items...