simple Haskell solution:
verify :: String -> Bool verify [] = True verify ('+':x:'+':xs) | isLetter x = verify $ '+':xs verify (x:xs) | isLetter x = False | otherwise = verify xs
Whoa, I was comming with something that involved using an indexedMap but your solution is very clever. Thanks for sharing your awesome answer!
Also, isn't isLetter part of Data.Char? Or no need to import that to use the isLetter function?
isLetter
Data.Char
yes, that function is from Data.Char, I omitted the import together with the module header.
I took your solution and simplified it:
verify :: String -> Bool verify [] = True verify ('+':x:'+':xs) = verify $ '+':xs verify ('+':xs) = verify xs verify ('=':xs) = verify xs verify (_) = False
Also note that nothing in the question indicates the characters other than = and + can only be letters. Your version would accept 1 even though 1 is a character and it is not surrounded by +s.
=
+
1
This solution assumes that the only other character is =. No idea if you can assume this
The string will be composed of + and = symbols with several characters between them.
If I'm reading this correctly, it means that the characters are categorized into three categories:
So any character other than + and = must be surrounded by +s - which is exactly what my version checks.
Are you sure you want to hide this comment? It will become hidden in your post, but will still be visible via the comment's permalink.
Hide child comments as well
Confirm
For further actions, you may consider blocking this person and/or reporting abuse
We're a place where coders share, stay up-to-date and grow their careers.
simple Haskell solution:
Whoa, I was comming with something that involved using an indexedMap but your solution is very clever. Thanks for sharing your awesome answer!
Also, isn't
isLetterpart ofData.Char? Or no need to import that to use theisLetterfunction?yes, that function is from
Data.Char, I omitted the import together with the module header.I took your solution and simplified it:
Also note that nothing in the question indicates the characters other than
=and+can only be letters. Your version would accept1even though1is a character and it is not surrounded by+s.This solution assumes that the only other character is =. No idea if you can assume this
If I'm reading this correctly, it means that the characters are categorized into three categories:
+=So any character other than
+and=must be surrounded by+s - which is exactly what my version checks.