Day 23 part 1

2024/Day_23/README.md

@@ -78,3 +78,26 @@ td,wh,yn
 ```
 
 Find all the sets of three inter-connected computers. **How many contain at least one computer with a name that starts with `t`?**
+
+---
+
+## --- Part Two ---
+
+There are still way too many results to go through them all. You'll have to find the LAN party another way and go there yourself.
+
+Since it doesn't seem like any employees are around, you figure they must all be at the LAN party. If that's true, the LAN party will be the **largest set of computers that are all connected to each other**. That is, for each computer at the LAN party, that computer will have a connection to every other computer at the LAN party.
+
+In the above example, the largest set of computers that are all connected to each other is made up of `co`, `de`, `ka`, and `ta`. Each computer in this set has a connection to every other computer in the set:
+
+```
+ka-co
+ta-co
+de-co
+ta-ka
+de-ta
+ka-de
+```
+
+The LAN party posters say that the **password** to get into the LAN party is the name of every computer at the LAN party, sorted alphabetically, then joined together with commas. (The people running the LAN party are clearly a bunch of nerds.) In this example, the password would be **`co,de,ka,ta`**.
+
+**What is the password to get into the LAN party?**

2024/Day_23/part_1.js

@@ -8,16 +8,30 @@ const __dirname = path.dirname(fileURLToPath(import.meta.url));
 // = Copyright (c) NullDev = //
 // ========================= //
 
+/* eslint-disable no-sequences */
+
 const INPUT = String(fs.readFileSync(path.join(__dirname, "input.txt"))).trim().split("\n");
 
 const pStart = performance.now();
 
-//
-// YOUR CODE HERE
-//
-const result = "...";
+const res = [
+    ...new Set(
+        (adj => Object.keys(adj).flatMap(n =>
+            [...adj[n]].flatMap((nx1, i, nxList) =>
+                nxList.slice(i + 1).filter(nx2 => adj[nx1].has(nx2))
+                    .map(nx2 => [n, nx1, nx2].sort().join(",")),
+            ))
+        )(INPUT.filter((x, i, arr) => !arr.some((y, j) =>
+            i > j && y.split("-").reverse().join("-") === x,
+        )).map(l => l.split("-")).reduce((acc, [a, b]) => (
+            ((acc[a] = (acc[a] || new Set()).add(b)) || 1) &&
+            (acc[b] = (acc[b] || new Set()).add(a)),
+            acc
+        ), {})),
+    ),
+].filter(e => e.split(",").some(n => n.startsWith("t"))).length;
 
 const pEnd = performance.now();
 
-console.log("<DESCRIPTION>: " + result);
+console.log("SETS WITH AT LEAST ONE T-NODE: " + res);
 console.log(pEnd - pStart);