Spoiler Day 5 Solution: <https://gitlab.com/abowe...
# advent-of-code
k
.map{}.flatten()
is
.flatMap {}
simple smile
a
true 🙂
k
Oh you calculate all possible pairs and check for those. That's an interesting approach!
Oh you calculate all possible pairs and check for those, that's an interesting approach!
a
Yep, but the approach that Todd took with the Stack is much more efficient.
k
It uses a stack, now I get it. I had some trouble with the fold obfustication simple smile
a
He is using a MutableList as a Stack where you check each char against the top of the Stack
k
Right, I'm going to try to do that in an optimized way now.