You should mention that the answer should be given as a column vector, rather than row vector.
One at this time must examine the assessment code to give the right answer, while I think it should be self-contained in the question itself.
Thanks Richard. I have modified the problem description to mention this.
function good = find_good_hotels(hotels,ratings,cutoff)
good = hotels(cutoff<=ratings);
end
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.
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!
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