aoc2021

commit 928d2685b89788eb2f61823608fb015c0afdf13f
parent 77472dacbfefdef120a61b469fbe5849d7558f86
Author: Samir Parikh <noreply@samirparikh.com>
Date:   Fri, 17 Dec 2021 22:23:45 +0000

update day07-2.pl to provide proper solution for Day 07 Part 2

Diffstat:
M
day07-2.pl
 | 
18
+++++-------------

1 file changed, 5 insertions(+), 13 deletions(-)
diff --git a/day07-2.pl b/day07-2.pl
@@ -1,16 +1,9 @@
 #!/usr/bin/env perl
 
-# i am doing this wrong
-# I am assuming that the only positions they crabs can align on are ones
-# that they already occupy.  I got luck with Part 1 in that one of the crabs
-# already occupied the winning answer
-# for Part 2, it's possible that the winning position that consumes the least
-# amoutn of fuel is a position that no crab initially occupies
-# I need to find the range of positions (min/max) and use that as the ranges
-# for the x and y loops
 use strict;
 use warnings;
 use v5.22;
+use List::Util qw( min max );
 
 sub get_filehandle {
     if (@ARGV !=1) {
@@ -33,20 +26,19 @@ my $filehandle  = get_filehandle();
 my @numbers     = get_numbers($filehandle);
 my $min_fuel_cost;
 my $horiz_pos;
+my $low = min(@numbers);
+my $high = max(@numbers);
 
-foreach my $x (0 .. $#numbers) {
+foreach my $x ($low .. $high) {
     my $fuel_cost = 0;
     foreach my $y (0 .. $#numbers) {
-        my $n = abs($numbers[$x] - $numbers[$y]);
+        my $n = abs($x - $numbers[$y]);
         $fuel_cost += $n * ($n + 1) / 2;
-        say "move from $numbers[$y] to $numbers[$x]: $n";
     }
-    say "------------";
     if (!defined($min_fuel_cost) || $fuel_cost < $min_fuel_cost) {
         $min_fuel_cost = $fuel_cost;
         $horiz_pos = $x;
     }
-    say "TOTAL cost to move to position $x is $fuel_cost";
 }
 
 say "winning position is $horiz_pos with cost of $min_fuel_cost";