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Using Dijkstra and starting from `z` the solution was 2x faster than BFS in my case. By starting from `z` it constraint is that you can go to a neighbor only if the height difference is 1, that is faster than starting from `a`. For the priority I used that the amount of steps from the start to the current node inspected, meaning the nodes with shortest paths are inspected first. I counted the number of cycles (each time an element is popped out the queue) For part 1 I get ~5000 cycles with Dijkstra and ~8000 with BFS. Cannot see if there is something wrong with my BFS implementation Solution in Go
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