euyemura,
combination(k) returns an Enumerable, not an Array of all the combinations. The Enumerable is lazy and will only find out the next combination when you call for the next value. When calling an Enumerable method like reduce on the return value of combination(k) you can break out of the loop and return early in you want, hence not wasting cycles. In this kata, I would break out early if my current sum matches t exactly. The solution above does not do this so will always have to sum all combinations so there is a bit of room for performance improvement, but not much.

Sorry, late response. You can pass a block into ls.combination(k) instead of to the .map function, which would allow you to exit early. However, the target distance is a limit, not a minimum. Since each combination you try might be higher than your last max and lower than the limit, we must visit every single combination to be sure of our answer.

Good thinking though =D

Hello! yours is top voted so i wanted to ask you and everyone else, is there a better way to do this than to use .combination(k)? If i'm not not mistaken this method creates every single combination, as opposed to doing something like making every combination in order until it creates an array with a sum that is over the target distance and then stopping. I ended up using combination as well, but isn't the time efficiency of it pretty bad? Your solution is super elegant, just want to understand if there's more efficient ways.

you're kidding me.
edit: this is beautiful btw.

Thanks

good stuff!