this post was submitted on 12 Dec 2024
16 points (90.0% liked)

Advent Of Code

980 readers
19 users here now

An unofficial home for the advent of code community on programming.dev!

Advent of Code is an annual Advent calendar of small programming puzzles for a variety of skill sets and skill levels that can be solved in any programming language you like.

AoC 2024

Solution Threads

M T W T F S S
1
2 3 4 5 6 7 8
9 10 11 12 13 14 15
16 17 18 19 20 21 22
23 24 25

Rules/Guidelines

Relevant Communities

Relevant Links

Credits

Icon base by Lorc under CC BY 3.0 with modifications to add a gradient

console.log('Hello World')

founded 1 year ago
MODERATORS
 

Day 12: Garden Groups

Megathread guidelines

  • Keep top level comments as only solutions, if you want to say something other than a solution put it in a new post. (replies to comments can be whatever)
  • You can send code in code blocks by using three backticks, the code, and then three backticks or use something such as https://topaz.github.io/paste/ if you prefer sending it through a URL

FAQ

you are viewing a single comment's thread
view the rest of the comments
[–] [email protected] 4 points 1 week ago (2 children)

C

No big trouble today, just a bit of careful debugging of my part 2 logic for which I was greatly helped by some Redditor’s testcase πŸ™

Happy to have gotten it all in the single flood fill function without any extra passes.

Code

#include "common.h"

#define GZ 144
static char g[GZ][GZ];
static char seen[GZ][GZ];

static void
count(char c, int x, int y, int *area, int *perim, int *sides)
{
	if (g[y][x] != c) { (*perim)++; return; }
	if (seen[y][x]) return;

	*area += 1;
	seen[y][x] = 1;

	/* count start of top/left edges, end of bottom/right edges */
	*sides += g[y-1][x]!=c && ((g[y-1][x-1]==c) || (g[y][x-1]!=c));
	*sides += g[y+1][x]!=c && ((g[y+1][x+1]==c) || (g[y][x+1]!=c));
	*sides += g[y][x-1]!=c && ((g[y-1][x-1]==c) || (g[y-1][x]!=c));
	*sides += g[y][x+1]!=c && ((g[y+1][x+1]==c) || (g[y+1][x]!=c));

	count(c, x, y-1, area, perim, sides);
	count(c, x, y+1, area, perim, sides);
	count(c, x-1, y, area, perim, sides);
	count(c, x+1, y, area, perim, sides);
}

int
main(int argc, char **argv)
{
	int p1=0,p2=0, x,y, area, perim, sides;

	if (argc > 1)
		DISCARD(freopen(argv[1], "r", stdin));

	for (y=1; fgets(g[y]+1, GZ-2, stdin); y++)
		assert(y+1 < GZ);

	for (y=1; y<GZ-1; y++)
	for (x=1; x<GZ-1; x++)
		if (isalpha(g[y][x]) && !seen[y][x]) {
			area  = perim = sides = 0;
			count(g[y][x], x, y, &area, &perim, &sides);
			p1 += area * perim;
			p2 += area * sides;
		}

	printf("12: %d %d\n", p1, p2);
}

https://github.com/sjmulder/aoc/blob/master/2024/c/day12.c

[–] [email protected] 3 points 1 week ago

Clean and concise. Admirable!

[–] [email protected] 2 points 1 week ago

Woah! That solution is a work of art!