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o
->{"o": ["only"]}
-> 1 letter prefix, 1 word -> 1 = 1 -> OKa
->{"a": ["are","answers]}
-> 1 letter prefix, 2 words -> 1 < 2 -> we will add more letters to make subdivisionsar
->{"ar": ["are"]}
-> 2 letters prefix, 1 word -> 2 > 1 -> Not OKan
->{"an": ["answers"]}
-> 2 letters prefix, 1 word -> 2 > 1 -> Not OKs
->{"s": ["some","solutions]}
-> 1 letter prefix, 2 words -> 1 < 2 -> we will add more letters to make subdivisionsso
->{"so": ["some","solutions]}
-> 2 letter prefix, 2 words -> 2 = 2 -> OK["only", "solutions", "some"]
only -> beginning = o (1 letter, 1 word) OK \\\ Why 'o'? Its only one word here wtf??
are -> beginning = ar (2 letters, 1 word) not OK \\\ Why 'ar'? Its only one word here wtf??
answers -> beginning = an (2 letters, 1 word) not OK \\\ Why 'an'? Its only one word here wtf??
Whats the logic here? Why dont author explain this in the description..
ONLY if their group size (or array length) is equal to the number of shared letters in each beginning
This comment is hidden because it contains spoiler information about the solution
I'm getting the same with my JS solution. I'm guessing python random tests (or the reference solution) have some flaws (maybe that's why it's in beta). Luckily JS has no random tests :D
Yes, the description is explicit about that.
Fixed 0- 0-
In this kata you can't.
Tests should take care of appropriate feedback on failure.
And I recommend you to use the spoiler flag, for this kind of things... 'x)
cheers (note: I already marked your message)
There is no such test, the first sample test is this:
Click reset, maybe you messed it up yourself.