Open wko opened 1 year ago
One interesting approach:
rotate :: Forest -> Forest
that rotates a matrix by 90 degrees lookAt :: Forest -> VisibilityMask
that finds the first local maximum looking from left to right through each row of the matrix, where a VisibilityMask
is a matrix of boolean values.lookAt
function from all directions using rotate
Iterative approach would be to traverse the matrix from four directions. For each position (i,j)
use a suc :: (Int, Int) -> (Int, Int)
function to store the current direction. Works best with mutable datastructures.
--- Day 8: Treetop Tree House ---
The expedition comes across a peculiar patch of tall trees all planted carefully in a grid. The Elves explain that a previous expedition planted these trees as a reforestation effort. Now, they're curious if this would be a good location for a tree house.
First, determine whether there is enough tree cover here to keep a tree house hidden. To do this, you need to count the number of trees that are visible from outside the grid when looking directly along a row or column.
The Elves have already launched a quadcopter to generate a map with the height of each tree (your puzzle input). For example:
30373 25512 65332 33549 35390
Each tree is represented as a single digit whose value is its height, where 0 is the shortest and 9 is the tallest.
A tree is visible if all of the other trees between it and an edge of the grid are shorter than it. Only consider trees in the same row or column; that is, only look up, down, left, or right from any given tree.
All of the trees around the edge of the grid are visible - since they are already on the edge, there are no trees to block the view. In this example, that only leaves the interior nine trees to consider:
With 16 trees visible on the edge and another 5 visible in the interior, a total of 21 trees are visible in this arrangement.
Consider your map; how many trees are visible from outside the grid?
InitialFeeling: This problem is more of an algorithmic challenge. The parsing should be trivial.