Re-write day 2 solution to run in O(n*m) time.

2018/day02-2.go

@@ -1,3 +1,8 @@
+// This is my second solution, c.f. https://github.com/annabunches/adventofcode/commit/3ca1a1068b749bab4d0626af32549ce2385d57bb
+// for the original solution.
+//
+// This solution is far more elegant; it runs in O(n*m) time, instead of the original implementation's O(n^2 * m) time.
+// Thanks to @lizthegrey for (probably unintentionally) prompting me to rewrite this >_>
 package main
 
 import (
@@ -8,25 +13,22 @@ import (
 
 func main() {
 	ids := util.ReadInput()
+	subIDs := make(map[string]struct{})
 
-	for _, box1 := range ids {
-		for _, box2 := range ids {
-			diffIndex := -1 // diffIndex doubles as the letter to remove and a 'found' flag
-			for i := range box1 {
-				if box1[i] != box2[i] {
-					if diffIndex != -1 {
-						// If we reach here, we've detected more than one error, so we set
-						// diffIndex back to -1 here to avoid the conditional below this loop.
-						diffIndex = -1
-						break
-					}
-					diffIndex = i
-				}
-			}
-			if diffIndex != -1 {
-				fmt.Printf("%s%s\n", box1[0:diffIndex], box1[diffIndex+1:])
+	for _, id := range ids {
+		// iterate over each letter
+		for i := range id {
+			// create a subID with just letter id[i] omitted
+			subID := id[0:i] + id[i+1:]
+
+			// if it's already in our map, we've found the correct output
+			if _, found := subIDs[subID]; found {
+				fmt.Println(subID)
 				return
 			}
+
+			// otherwise, add the substring to our map
+			subIDs[subID] = struct{}{}
 		}
 	}
 }