Solve 20b

20_javascript/20b.js

@@ -0,0 +1,30 @@
+#!/usr/bin/env node
+
+const fs = require('fs');
+const input = fs.readFileSync('/dev/stdin').toString();
+const lines = input.split('\n').filter(line => line.indexOf('-') > 0);
+const ranges = lines.map((line) => {
+  const parts = line.split('-');
+  return {
+    start: parseInt(parts[0], 10),
+    end: parseInt(parts[1], 10),
+  };
+});
+ranges.sort((a, b) => {
+  if (a.start < b.start) return -1;
+  if (a.start > b.start) return 1;
+  return 0;
+});
+const MAX_IP = 4294967295;
+ranges.push({ start: MAX_IP + 1, end: MAX_IP + 1 });
+
+let candidate = 0;
+let count = 0;
+for (let i = 0; i < ranges.length; i++) {
+  const range = ranges[i];
+  if (range.start > candidate) {
+    count += range.start - candidate;
+  }
+  candidate = Math.max(candidate, range.end + 1);
+}
+console.log(count);

20_javascript/README.md

@@ -6,3 +6,7 @@ well.
 Part one is easy: just sort the ranges by start value, run through increasing
 your candidate IP address to be beyond each of the ranges, and if you encounter
 one that starts after the candidate, you're done.
+
+Part two has a similar approach, just with a bit more careful bookkeeping. For
+convenience, I pushed a sentinel range onto the end which is one past the final
+possible IP address. JavaScript's 64-bit doubles let you do that.