very easy kata!
Hint: Just use by simple math formula to count the odd numbers.

C++

The resulting array shouldn't necessarily match anyone in the tests, a function just checks if the array is now wave sorted.

The vector content should stay the same

It can't be both together. If the function only checks if the array is wave sorted, it shouldn't check if it stays the same.

It's even worse: i've added

if(is_wave_sorted(v))
    return;

at the beginning of my function, and still it returns "The vector content should stay the same". So, my function only runs if the check function fails - and still the vector should stay the same?

This problem can be reduced to the set cover problem, which is NP-hard, looking at size of tests I can assume that reference solution uses some king of greedy approach, which do not guaranty optimal solution.
Example A = [3, 8, 8, 6, 7], B = [2, 2, 3, 5, 6], k = 6. My solution is 3. First thief can take from jars 1 and 4 second from 2, third from 3 and 5

fixed in this fork

Merged

The easiest kyu

done there

approved

js fork

• made description language-agnostic (resolving this issue)
• used lodash for randomness
• provided inputs on failure

js fork 🤖🤖🤖

• updated to node 18
• used mocha/chai assertions + lodash for randomness
• provided inputs on failure

Can the password be empty (eg: no 'T' at the end of directions) ? Can grid be 1x1 and/or can directions be empty (thus password would also be empty) ?

I'd like to translate the kata and I think these cases could be interesting but I don't know if I should implement them or not. I also think it could be added to the description.

Merged the fork with the fix.

Merged.

fixed

Yes, it is. But perhaps this task is not for a beginner of the 8th kyu. This is the simplest level, where people learn how to write variables or if conditions, rather than how to choose the best algorithm