day 4 done

04/index.ts

@@ -0,0 +1,103 @@
+import fs from "node:fs";
+import { printTime, now } from "../util";
+
+const test = false;
+
+const data = fs.readFileSync(
+  test ? "./inputs/testinput" : "./inputs/input",
+  "utf8"
+);
+
+let timer = now.instant();
+console.log(
+  "part one:",
+  data
+    .split("\n")
+    .slice(0, -1)
+    .reduce((coll, curr, index, array) => {
+      // 2D string matching is a pain. Let's turn it into a 1D problem!
+      // create the down rows
+      curr.split("").forEach((s, i) => {
+        coll[i] = coll[i] ? coll[i] + s : s;
+      });
+      // create the down-left-diagonals
+      curr.split("").forEach((s, i) => {
+        coll[i + array.length + index] = coll[i + array.length + index]
+          ? coll[i + array.length + index] + s
+          : s;
+      });
+      // create the down-right-diagonals
+      curr.split("").forEach((s, i) => {
+        coll[array.length * 4 + index - i - 2] = coll[
+          array.length * 4 + index - i - 2
+        ]
+          ? coll[array.length * 4 + index - i - 2] + s
+          : s;
+      });
+      // include the original crosses
+      return [...coll, curr];
+    }, Array.from({ length: test ? 10 + (10 * 4 - 2) : 140 + 140 * 4 - 2 }) as string[])
+    .reduce((total, curr) => {
+      // and now just do some regexes over the set of 1D strings
+      let matchesForward = curr.match(/XMAS/g);
+      let matchesBackward = curr.match(/SAMX/g);
+      if (matchesForward) {
+        total += matchesForward.length;
+      }
+      if (matchesBackward) {
+        total += matchesBackward.length;
+      }
+      return total;
+    }, 0),
+  printTime(now.instant().since(timer))
+);
+
+timer = now.instant();
+console.log(
+  "part two:",
+  data
+    .split("\n")
+    .slice(0, -1)
+    .reduce((coll, curr, index, array) => {
+      if (index === 0 || index === array.length - 1) {
+        // we can entirely skip the first and last rows, since there's no way to get a cross started
+        return coll;
+      }
+      // look for A's
+      curr.split("").forEach((s, i, a) => {
+        if (i === 0 || i === a.length - 1) {
+          // same as above, edges are useless
+          return;
+        }
+        if (s === "A") {
+          // 1 2
+          //  A
+          // 3 4
+          coll = [
+            ...coll,
+            array[index - 1][i - 1] +
+              array[index - 1][i + 1] +
+              array[index + 1][i - 1] +
+              array[index + 1][i + 1],
+          ];
+        }
+      });
+      return coll;
+    }, [] as string[])
+    .reduce((coll, curr) => {
+      // we could have just done this check above,
+      // but it's a little more clear, although slower,
+      // to iterate again.
+      switch (curr) {
+        case "MSMS":
+        case "SSMM":
+        case "SMSM":
+        case "MMSS":
+          coll += 1;
+        default:
+          break;
+      }
+      return coll;
+    }, 0),
+  printTime(now.instant().since(timer))
+);