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Advent of Code 2022 Day 8: https://adventofcode.com/2022/day/8
Overall this solution is a little bit ugly and a lot of brute force; there
are probably better ways to solve this. We start by parsing the input (we
do an initial pass n*n complexity) to convert the strings in the input
into a vector of integer vectors which saves us having to do the parse for
every comparison. Then we look at interior trees only. In part one we can
do a simple calculation based on the dimensions to account for them
without needing to loop through and account for them if x/y==0/len and in
part two we can ignore them because as edge trees they have one side with
a viewing distance of zero which when multiplied by any other viewing
distance will obviously result in a scenic score of zero too. For part two
we use the usual std::collections::BinaryHeap to keep track of the
tree with the highest scenic score value.
Functions§
- The solution for the day eight challenge.