The solution that was good enough for ~50\%~ of points works under the assumption that a field's right and down neighbours are different. This way, we can start from the initial field and in each step compare the neighbours - we will always move the token to a field that is alphabetically smaller.

The problem arises when both neighbours are the same. This problem can be solved by keeping track of, in each step, a list of all optimal positions in which we could have ended up after a certain amount of steps. We start with the list containing the initial field ~(0, 0)~ and in each step update the list so that we first check the minimal value of the neighbours of all positions in the current list, and then create a new list that will contain all neighbouring positions with that value. Since we can reach a field in ~2~ ways, we must be careful not to add the same position twice to the list, because this way we would copy the number of appearances of the same position in each iteration.