#!/usr/bin/env perl

use strict;
use warnings;
use v5.010;
use List::Util qw/ max /;

# Parse the input

my %games = ();

for my $line (<>) {
    $line =~ /Game (\d+): (.+)/;
    my $game_id = $1;
    my $game_str = $2;

    my @segments = split '; ', $game_str;
    my @game = ();
    for my $segment (@segments) {
        my @counts = split ', ', $segment;

        my %colors = (red => 0, blue => 0, green => 0);
        for my $count (@counts) {
            $count =~ /(\d+) (\w+)/;
            $colors{$2} = $1;
        }

        push @game, { %colors };
    }

    $games{$game_id} = [ @game ];
}

# Part 1

my $part1 = 0;

game: for my $game_id (keys %games) {
    for my $segment (@{$games{$game_id}}) {
        next game if $segment->{red} > 12 || $segment->{green} > 13 || $segment->{blue} > 14;
    }

    $part1 += $game_id;
}

say "Part 1: $part1";

# Part 2

my $part2 = 0;

for my $game (values %games) {
    my ($red, $green, $blue) = (0, 0, 0);

    for my $segment (@$game) {
        $red = max $segment->{red}, $red;
        $green = max $segment->{green}, $green;
        $blue = max $segment->{blue}, $blue;
    }

    $part2 += $red * $green * $blue;
}

say "Part 2: $part2";

  int? check3(String line, int index) {
    if (line.length < index + 3) {
      return null;
    }

    String sub = line.substring(index, index + 3);
    return sub == "one"
        ? 1
        : sub == "two"
            ? 2
            : sub == "six"
                ? 6
                : null;
  }

Anyway, a) is a pretty straightforward job-interview style question, except AoC doesn't care about efficiency. Still, we all have our own versions of pride, so I did it with a set (Though whether or not dart's native Set is tree or hash based is not known to me right now).

int matches(String line) {
  ({List wn, List n}) card = getNumbers(line);
  Set wn = Set.from(card.wn);

  return card.n.fold(0, (prev, e) => prev + (wn.contains(e) ? 1 : 0));
}

void day4a(List lines) {
  print(lines.fold(0, (acc, line) {
    int m = matches(line);
    return acc + (m != 0 ? (1 << (m - 1)) : 0);
  }));
}

So the general approach is to formulate it like this:

T_n(total cards won by card n) = M_n(total matches for card n) + CW_n(total cards won by the copies that card n wins).

and

CW_n =

• 0 if M_n = 0
• sum of T_i, where i = n + 1 ... n + M_n

Caching T_n is the DP trick to making this performant (though once again, it does not need to be)

Anyway, the above approach is the top-down version of the DP; the bottom-up version is what I actually started with in my head. I gave up on that approach because I felt like the logic was too hard for me to figure out.

void day4b(List lines) {
  List totalMatches = lines.map((e) => matches(e)).toList();

  // total copies won, including copies won from copies.
  List cachedWins = List.filled(totalMatches.length, -1);
  int totalWins(int i) {
    // return cached result if it exists
    if (cachedWins[i] > 0) {
      return cachedWins[i];
    }

    int count = totalMatches[i];
    // count the copies won from the subsequent copied cards
    for (int j = 1; j <= totalMatches[i]; j++) {
      count += totalWins(i + j);
    }

    // cache the result
    cachedWins[i] = count;
    return count;
  }

  int totalCards = totalMatches.length; // count all the originals
  // count all the wins
  for (int i = 0; i < totalMatches.length; i++) {
    totalCards += totalWins(i);
  }
  print(totalCards);
}

The approach this time was:

1. iterate through the characters to find a * symbol
2. Search the characters around it for a digit.
3. Get the value of the number associated with that digit by searching backwards until you find the start of a number
4. Use union-find to track whether or not you've seen this number before (because you can't assume that the same value is the same number)

A simpler approach would consider that you only have two numbers on the same line for the same gear if the character in the gear column is a non-digit; otherwise, if a number is adjacent to a gear, there is only one on that row. Union-find is completely overkill, but I like using it even when I don't need to.

Anyway, upon reflecting on this, while the general approach is fine, I didn't think too hard about the implementation and just ended up with globs of overly memoized spaghetti. I probably should check if Dart has a python-like tuple object or similar. Whatever. Behold!

void day3s() {
  List lines = [
    for (String? line = stdin.readLineSync();
        line != null;
        line = stdin.readLineSync())
      line
  ];

  List> digs = [for (int i = 0; i < lines.length; i++) Map()];
  int? isDigitMem(int r, int c) {
    return digs[r].putIfAbsent(c, () => isDigit(lines[r][c]));
  }

  // first entry is parentc, second is size
  List>> uf = List.generate(
      lines.length, (i) => List.generate(lines[0].length, (j) => [j, 1, -1]));

  int find(int r, int c) {
    if (uf[r][c][0] != c) {
      uf[r][c][0] = find(r, uf[r][c][0]);
      return uf[r][c][0];
    }
    return uf[r][c][0];
  }

  void union(int r, int cp, int c) {
    cp = find(r, cp);
    c = find(r, c);

    if (c == cp) {
      return;
    }

    if (uf[r][cp][1] >= uf[r][c][1]) {
      uf[r][c][0] = cp;
      uf[r][cp][1] += uf[r][c][1];
    } else {
      uf[r][cp][0] = c;
      uf[r][c][1] += uf[r][cp][1];
    }
  }

  int stoi(int row, int col) {
    int acc = 0;
    for (int i = col; i < lines[0].length; i++) {
      int? d = isDigitMem(row, i);
      if (d != null) {
        acc = (acc * 10) + d;
        union(row, col, i);
      } else {
        break;
      }
    }
    return acc;
  }

  int? stoiSearch(int row, int col) {
    assert(row >= 0 && col >= 0 && row < lines.length && col < lines[0].length);
    if (isDigitMem(row, col) == null) {
      return null;
    }
    int i = col - 1;
    while (i >= 0) {
      if (isDigitMem(row, i) == null) {
        return stoi(row, i + 1);
      }
      i--;
    }
    return stoi(row, 0);
  }

  List> s2i = [for (int i = 0; i < lines.length; i++) Map()];
  int? stoiSearchMem(int row, int col) {
    return s2i[row].putIfAbsent(col, () => stoiSearch(row, col));
  }

  int count = 0;
  for (int i = 0; i < lines.length; i++) {
    for (int j = 0; j < lines[0].length; j++) {
      if (lines[i][j] != "*") {
        continue;
      }

      List gearVals = List.empty(growable: true);
      for (int x = -1; x <= 1; x++) {
        if (i + x < 0 || i + x > lines.length) {
          continue;
        }

        Set parents = {};
        for (int y = -1; y <= 1; y++) {
          if (j + y < 0 || j + y > lines[0].length) {
            continue;
          }

          int? cur = stoiSearchMem(i + x, j + y);

          int parent = find(i + x, j + y);
          if (parents.contains(parent)) {
            continue;
          }

          parents.add(parent);

          if (cur != null) {
            gearVals.add(cur);
          }
        }
      }

      if (gearVals.length == 2) {
        count += gearVals[0] * gearVals[1];
      }
    }
  }

  print(count);
}

#!/usr/bin/env jq -n -R -f

# Dish to grid
[ inputs / "" ]

# Tilt UP
| transpose                       # Transpose, for easier RE use
| map(                            #
  ("#" + add) | [                 # For each column,   replace '^' with '#'
    scan("#[O.]*") | [            # From '#' get empty spaces and 'O' rocks
      "#", scan("O"), scan("\\.") # Let gravity do it's work.
    ]                             #
  ] | add[1:]                     # Add groups back together
 )                                #
| transpose                       # Transpose back

# For each row, count  'O'  rocks
| map(add | [scan("O")] | length)

# Add total load on "N" beam
| [0] + reverse | to_entries
| map( .key * .value ) | add

#!/usr/bin/env jq -n -R -f

# Get seeds
input | [ match("\\d+"; "g").string | tonumber ] as $seeds |

# Collect maps
reduce inputs as $line ({};
  if $line == "" then
    .
  elif $line | test(":") then
    .k = ( $line / " " | .[0] )
  else
    .[.k] += [[ $line | match("\\d+"; "g").string | tonumber ]]
  end
)

# For each map, apply transformation to all seeds.
# seed -＞ ... -＞ location
| reduce ( to_entries[] | select(.key != "k") .value) as $map ({s:$seeds};
  .s[] |= (
    # Only attempt transform if element is in one of the ranges
    [ . as $e | $map[] | select(. as  [$d,$s,$l] | $e ＞= $s and $e ＜ $s + $l) ] as $range |
    if ($range | length ) ＞ 0 then
      $range[0] as [$d,$s] |
      . - $s + $d
    else
      .
    end
  )
)

# Get lowest location
| .s | min

Some comments:

• A nice use of input first to get the seeds, then inputs to get remaining lines.

#!/usr/bin/env jq -n -R -f

# Utility function
def group_of($n):
  ( length / $n ) as $l |
  . as $arr |
  range($l) | $arr[.*$n:.*$n+$n]
;

# Get all seed ranges
input | [ match("\\d+"; "g").string | tonumber ] | [group_of(2)] as $seeds |

# Collect maps
reduce inputs as $line ({};
  if $line == "" then
    .
  elif $line | test(":") then
    .k = ( $line / " " | .[0] )
  else
    .[.k] += [[ $line | match("\\d+"; "g").string | tonumber ]]
  end
)

# For each map, apply transformation to all seeds ranges.
# Producing new seed ranges if applicable
# seed -＞ ... -＞ location
| reduce (to_entries[] | select(.key != "k") .value) as $map ({s:$seeds};
  .s |= [
    # Only attempt transform if seed range and map range instersect
    .[] | [.[0], add, .[1] ] as [$ea, $eb, $el] | [
      $map[] | select(.[1:] | [.[0], add ] as [$sa,$sb] |
        ( $ea ＞= $sa and $ea ＜ $sb ) or
        ( $eb ＞= $sa and $eb ＜ $sb ) or
        ( $sa ＞= $ea and $sa ＜ $eb )
      )
    ] as $range |
    if $range | length ＞ 0 then
      $range[0] as [$d,$s,$l] |
      # ( only end ) inside map range
      if $ea ＜ $s and $eb ＜ $s + $l then
        [$ea, $s - $ea], [$d, $eb - $s ]
      # ( both start, end ) outside map range
      elif $ea ＜ $s then
        [$ea, $s - $ea], [$d, $l], [ $s + $l, $eb ]
      # ( only start ) inside map range
      elif $eb ＞ $s + $l then
        [$ea + $d - $s, $l - $ea + $s ], [$s + $l, $eb - $s - $l]
      # ( both start, end ) inside map range
      else
        [$ea + $d - $s , $el]
      end
    else
      .
    end
  ]
)

# Get lowest location
| [.s[][0]] | min

Some comments:

• Since iterating across all seeds naively would take forever, iterating over seed ranges instead.
• It's nice that JQ can neatly produce extra elements: [1,2,3] | [ .[] | if . == 2 then . * 10 + 1 , . * 10 + 2 else . end ] -> [1, 21, 22, 3]
• There is probably a more compact way of expressing all the conditions and produced outputs.

b. Pretty much just changed the input parsing and ran the same algorithm. Thought that it might be too slow, but I put in some stopwatch ticks and found out that it should take about 5 minutes to run my a) algorithm to get the answer.

The actual time elapsed was 5m50s.

I miiiiight try make a faster version but my head is too fucked up to care right now, hence me letting the slow algorithm run.

I read the problem as follows: for N galaxies, find N/2 pairings such that the sum of distances is minimal.

At this point, I was like, wow, the difficulty has ramped up. A DP? That will probably run out of memory with most approaches, requiring a BitSet. I dove in, eager to implement a janky data structure to solve this humdinger.

I wrote the whole damn thing. The bitset, the DP, everything. I ran the code, and WOAH, that number was much smaller than the sample answer. I reread the prompt and realised I had it all wrong.

It wasn't all for naught, though. A lot of the convenience stuff I'd written was fine. Also, I implemented a sparse parser, which helped for b.

b: I was hoping they were asking for what I had accidentally implemented for a. My hopes were squandered.

Anyway, this was pretty trivial with a sparse representation of the galaxies.