This was awesome, nicely designed and really makes you think.

Wow, impressive design. This really instills a great foundational understanding of control flow, validating inputs in a logical and correct way, and edge case testing.

Why tthis exercise dont complie in typescript?

An author?

Thank you friend, your advice help me alot and save me lot of time on last part of training.

Thank you for this kata, but frankly speaking it is not an 8kyu

Take a = "a", b = "b" as an example

The conditional statement will compare it like this ("A" === "a" || "b" === "B"). Clearly, both of them are not equal, so any alphabetical letters (a-z and A-Z) will not enter this part of the control flow statement. Whilst, if you compare any non-alphabetical letters, (~ / ! / ?), changing of casing does not affect them, so the condition will be as such:

'~' === '~' || '!' === '!' // assuming a = '~' and b = '!'

which will always evaluate to true, hence returning -1

Your conditional logic is wrong. Your code seems like pseudocode, though a human can understand it, a computer or interpreter cannot. So, let's break it down:

if a or b is not same_case:

First condition is a (will evaluate to True, unless it is an empty string "")

Second condition is b is not same_case (same_case is a function, and will always evaluate to True since b will never be a function object)

So, your code will always output 0 regardless of inputs since the or operator is used in which either one condition being True will direct the control flow to that part of the statement (return 0)

The same logic applies to elif a or b is not letter: (letter variable is not defined whatsoever though)

Approved

Approved

CoffeeScript translation

Scala translation

This comment was very helpful!

return -1