Discussion on: AoC Day 15: Beverage Bandits

[Tiago Romero Garcia]

Also, to break the ties in the reading order, it's all in getAdjacents function, where I look for the following neighbor squares and return in their reading order.

Considering we're getting adjacents for position X,Y, N=max(X) and M=max(Y)

1. if X > 0, adjacent on X-1,Y
2. if Y > 0, adjacent on X,Y-1
3. if Y < M-1, adjacent on X,Y+1
4. if X < N-1, adjacent on X+1,Y

In other words,

1. Neighbor on the line above
2. Neighbor on the same line, to the left
3. Neighbor on the same line, to the right
4. Neighbor on the line below

[Javid Asgarov]

Oh, awesome. I got it. I want to come back after I've done others and revisit this problem with breadth-first-search. Maybe it'll be faster.