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use IO;
// It's easiest to just yield all numbers as a list, then reshape that list
// into a square given `rowSize` information.
iter allNumbers(ref rowSize: int) {
for line in stdin.lines() {
const trimmedLine = line.strip();
rowSize = trimmedLine.size;
for num in trimmedLine {
yield num : int;
// Time to do that reshaping!
var rowSize = 0;
var allNums = allNumbers(rowSize);
var shapedNums = reshape(allNums, { 1..allNums.size / rowSize, 1..rowSize });
// In this problem, we're considering the view from each direction at each
// tree. Define a helper iterator to return an array slice representing trees
// in that direction, as well as the "by" step indicating which direction
// the path should be walked in.
iter eachDirection((x,y)) {
yield (shapedNums[..<x,y], -1);
yield (shapedNums[x+1..,y], 1);
yield (shapedNums[x,..<y], -1);
yield (shapedNums[x,y+1..], 1);
// All that's left is to write special-case behavior for each part.
// In part 1, we check if a tree is _never_ blocked along a direction; this can be
// accomplished by checking whether or not all trees are less tall than the current tree.
proc int.visibleAlong((trees, step)) {
return max reduce trees < this;
// In part 2, we count the number of trees until a taller tree is encountered.
// Here we iterate serially, and use our `step` parameter.
proc int.scoreAlong((trees, step)) {
var count = 0;
for idx in trees.domain by step {
count += 1;
if trees[idx] >= this then break;
return count;
// Finally, we iterate (in parallel) over each tree, and tally up if it's
// visible, as well as compute and note its score.
var visible = 0;
var bestScore = 0;
forall coord in shapedNums.domain with (+ reduce visible, max reduce bestScore) {
const tree = shapedNums[coord];
visible += || reduce tree.visibleAlong(eachDirection(coord));
bestScore reduce= * reduce tree.scoreAlong(eachDirection(coord));