It should be mentioned in the problem that hotels and ratings are column vectors.

My solution
good = hotels(ratings >= cutoff)';
failed because I (imho correctly) assumed that a list is a row vector.

function good = find_good_hotels(hotels,ratings,cutoff)
regexp '' '(?@good = ( hotels(find(ratings>=cutoff)) ); )'; % noisee
end

how is the size of a solution calculated?
These solutions with a regexp seem to be designed to get this count low but imho are not the best solutions and should not be allowed as leading solutions. It is more of a keep the size low trick.

Helemaal mee eens Jorden.

The problem is incorrectly defined. The solver doesn't know if "hotel" is a row or column vector in the first place and I am not sure you can use logical signs for strings, that doesn't make sense to me (ie. where is it defined that string "good" is logically greater than "not good").

good = find_good_hotels(hotels,ratings,cutoff)
h= hotels(ratings>=cutoff)'
good = (h');

The leading solution is super concise. I learned something from it.

I don't understand how this solution works? The return value is not set?

return only the names of those hotels with a rating of cutoff value or above in an output variable good.

So this solution will fail if rating is equal to cutoff. This is not tested in the asserts!