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

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Anna Rose 2018-12-02 02:27:10 -05:00
parent 3ca1a1068b
commit d57d3ca80b
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@ -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{}{}
}
}
}