 # AoC Day 2: Inventory Management System

OK, the first day was awesome, I'm super excited about all of the solutions people posted. And I'm learning a lot! We've got a solid 20 people on the DEV leaderboard, which means there are still spots for 180 more -- use code 224198-25048a19 to join!

On to Day 2!

Day 2 of Advent of Code, and I'm pretty sure that Google is tired of me asking it questions every 15 seconds about "How Do I Do X in Rust."

Today's challenge involves an inventory system. Boxes have IDs that are a jumble of letters, and we've got a warehouse full of boxes to check. The first part asks us to analyze the frequency of letters in each ID. The second part gets into Hamming Distances, which are a familiar sight after mentoring on Exercism.

I got both parts working, and even part 2 ran pretty fast, but I'm not algorithmically happy with the double-loop (O(n^2)) runtime. Did anybody come up with anything tricky to do it more efficiently?

I'm going to post my solution in the comments like the rest of the cool kids.

How'd everybody else do?

Posted on by: ### Ryan Palo

Ryan is an engineer in the Sacramento Area with a focus in Python, Ruby, and Rust. Bash/Python Exercism mentor. Coding, physics, calculus, music, woodworking. Message me on DEV!

### Discussion Python solutions!

part 1

from collections import Counter

with open('input.txt', 'r') as f:
twice = 0
thrice = 0
for line in f:
counts = Counter(line).values()
if 2 in counts:
twice += 1
if 3 in counts:
thrice += 1
print(twice * thrice)


part 2 -- this one feels clunky to me.

with open('input.txt', 'r') as f:
boxes = [box.strip() for box in f]

def check_differences(box1, box2):
difference_idx = -1
for idx, letter in enumerate(box1):
if letter != box2[idx]:
if difference_idx < 0:
difference_idx = idx
else:
return False
return box1[0:difference_idx] + box1[difference_idx+1:]

def find_one_difference(boxes):
for idx, box in enumerate(boxes):
for box2 in boxes[idx+1:]:
diff = check_differences(box, box2)
if diff:
return diff

print(find_one_difference(boxes))


10 points for the variable name “thrice!”

My part one ended up looking super similar!

#!/usr/bin/env python

from collections import Counter

if __name__ == '__main__':
box_ids_with_two_duplicate_letters = 0
box_ids_with_three_duplicate_letters = 0
with open('1-input.txt') as box_file:
for box_id in box_file:
fingerprint = Counter(box_id)
if 2 in fingerprint.values():
box_ids_with_two_duplicate_letters += 1
if 3 in fingerprint.values():
box_ids_with_three_duplicate_letters += 1

print(box_ids_with_two_duplicate_letters * box_ids_with_three_duplicate_letters)


I ended up using zip to help with the comparison if the strings in part 2:

#!/usr/bin/env python

from collections import Counter

def find_similar_box_ids(all_box_ids):
for box_id_1 in all_box_ids:
for box_id_2 in all_box_ids:
# This is kind of a naive Levenshtein distance!
letter_pairs = zip(box_id_1, box_id_2)
duplicates = list(filter(lambda pair: pair == pair, letter_pairs))
if len(duplicates) == len(box_id_1) - 1:
return ''.join(pair for pair in duplicates)

if __name__ == '__main__':
with open('2-input.txt') as box_file:
print(find_similar_box_ids(all_box_ids))


Nice! Definitely think that's the easiest way to do number 1. Zip also makes sense for the second, though not using it allowed me to do the early return!

True!

As a lispy thinker, my brain wants to tell you to make those for loops into comprehensions, but my work brain is telling me that my coworkers would have me drawn and quartered to make your "find-one-difference" a one-liner!

Kudos on how neat this looks. I find python and ruby to be difficult to make look "clean" when I write it.

## Part 1

You can probably see my Python shining through as I implemented a custom Counter struct to help me out.

use std::collections::HashMap;

// Part 1

/// A histogram of the characters in a string.
struct Counter {
letters: HashMap<char, usize>,
}

impl Counter {
pub fn new(word: &str) -> Self {
let mut letters = HashMap::new();
for letter in word.chars() {
let current_reference = letters.entry(letter).or_insert(0);
*current_reference += 1;
}
Self { letters }
}

pub fn count_value(&self, number: usize) -> usize {
self.letters.values().filter(|count| **count == number).count()
}
}

/// Calculates a checksum for id strings.
///
/// The checksum is the number of id's with at least one set of exactly
/// two of a letter times the number of id's with at least one set of
/// exactly three of a letter.  If it has more than one
pub fn checksum(text: &str) -> usize {
let mut twos = 0;
let mut threes = 0;
text.lines()
.map(|id| Counter::new(id))
.for_each(|counter| {
if counter.count_value(2) != 0 {
twos += 1;
}
if counter.count_value(3) != 0 {
threes += 1;
}
});
twos * threes
}


## Part 2

Double for loop boooooooo! But it works and it's fast enough for now. I'm pretty happy with the mileage I got out of Iterators for this part.

// Part 2

/// Finds the letters that are shared between the two prototype fabric
/// box ids.
///
/// These ids are the only two that differ from each other by exactly
/// one letter.
pub fn prototype_ids_common_letters(text: &str) -> String {
let ids: Vec<&str> = text.lines().collect();
for (i, s1) in ids.iter().enumerate() {
for s2 in ids.iter().skip(i) {
if hamming_distance(s1, s2) == 1 {
return common_letters(s1, s2);
}
}
}
String::new()
}

/// Calculates the "Hamming Distance" between two strings
///
/// Hamming distance is the number of characters who are different
/// between the two strings when the corresponding indices are compared
/// in each string
fn hamming_distance(s1: &str, s2: &str) -> usize {
s1.chars().zip(s2.chars())
.filter(|(c1, c2)| c1 != c2)
.count()
}

/// Returns the letters that are the same (and in the same place)
/// between the two strings
fn common_letters(s1: &str, s2: &str) -> String {
s1.chars().zip(s2.chars())
.filter(|(c1, c2)| c1 == c2)
.map(|(c1, _c2)| c1)
.collect()
}


This is very well documented and clear, easy-to-read code. This also makes me want to jump into Rust again (I've only hobbied around with it here and there).

Thanks! That really encouraging!

I love how you've used enumerate and skip together in your nested for loop. I struggled to find a clean solution like this.

Thanks! Yeah, I originally had a very manual nested for-loop set up, but after I got the tests passing, I decided to make an effort to do everything I could with iterators instead :) I've decided that the iterator module in Rust is where most of the magic that I'm missing from Python and Ruby lives.

This was still bothering me, but I found the Itertools crate and the tuple_combinations method. Check out my updated solution in the thread. Itertools makes iterators even more powerful.

PHP

Ended up learning about a bunch of useful array functions like array_values, array_count_values and array_diff_assoc for this one!

Part 1:

$list = file_get_contents($argv);
$boxes = explode("\n", trim($list));
$twos = 0;$threes = 0;
foreach ($boxes as$box) {
$letters = str_split($box);
$values = array_values(array_count_values($letters));
if (in_array(2, $values)) {$twos++;
}
if (in_array(3, $values)) {$threes++;
}
}
echo $twos*$threes;
die(1);


Part 2:

<?php
$list = file_get_contents($argv);
$boxes = explode("\n", rtrim($list));
foreach ($boxes as$i=>$box) { if ($i != count($boxes)-1) { for ($j = $i+1;$j < count($boxes);$j++) {
$one = str_split($box);
$two = str_split($boxes[$j]); if (count(array_diff_assoc($one, $two)) == 1) { echo join("", array_intersect_assoc($one, $two)); die(1); } } } }  I’m trying to use a broader range of languages than I do usually, so I figured I’d try not to use one I’d already used before through the days. I use bash all the time, so I thought I’d get it out of the way early. This was not one of my better decisions, but worked fine! Part 1 uses regular expressions; I could have used an associative array like some other people in the thread, but for some reason I went here first. The sort function wasn’t necessary, but helped with debugging. #!/bin/bash twos=0 threes=0 function sort_str () { # Bubble Sort ^_^ local sl="$1"
changed=1
while [[ "$changed" -eq 1 ]]; do changed=0 for i in$(seq 1 $((${#sl} - 1 ))); do
if [[ ${sl:$(( i - 1 )):1} > ${sl:$i:1} ]]; then
echo "${sl:$(( i - 1 )):1} > ${sl:$i:1}" >>sort_progress
changed=1
if [[ $i -eq 1 ]]; then sl=${sl:1:1}${sl:0:1}${sl:2}
else
sl=${sl:0:$(( i - 1 ))}${sl:$i:1}${sl:$((i-1)):1}${sl:$i+1}
fi
fi
done
done
echo $sl } re_2_str="" re_3_str="" for a in a b c d e f g h i j k l m n o p q r s t u v w x y z; do re_2_str+="^[^$a]*$a[^$a]*$a[^$a]*$|" re_3_str+="^[^$a]*$a[^$a]*$a[^$a]*$a[^$a]*$|" done re_2_str="(${re_2_str%|})"
re_3_str="(${re_3_str%|})" while read line do echo -n "$line: $(sort_str$line): "
if [[ "$line" =~$re_2_str ]]; then
twos=$(( twos + 1 )) echo -n "2 " fi if [[ "$line" =~ $re_3_str ]]; then threes=$(( threes + 1 ))
echo -n "3 "
fi
echo
done <2.1.input

echo "Twos: $twos" echo "Threes:$threes"
echo "x: $(( twos * threes ))"  Part 2 uses the same double for loop lamented elsewhere, but it gets the job done. #!/bin/bash while read line do while read comp_line do same_string="" ndiffs=0 for i in$(seq 0 $((${#line} - 1 )) )
do
if [[ "${line:$i:1}" == "${comp_line:$i:1}" ]]; then
same_string+=${line:$i:1}
else
ndiffs=$(( ndiffs+1 )) fi done if [[ "$ndiffs" -eq 1 ]]; then
echo "$line:$comp_line: $same_string" exit fi done <2.2.input done <2.2.input  Neither of these is what I’d call “fast”. Woah, nice! It's always really cool to see bigger things done with Bash :) P.S. Depending on your Bash version, you can save yourself some typing with {a..z}. Oh yes, good call, I missed that compaction. Clojure (inefficient part 2) Part 1: (->> (utils/read-file (str utils/resources-path "day2.txt")) (map frequencies) (map vals) (map (juxt (fn [coll] (some #(= 2 %) coll)) (fn [coll] (some #(= 3 %) coll)))) (reduce (fn [acc [_2 _3]] (-> acc (update :2 #(+ (if _2 1 0) %)) (update :3 #(+ (if _3 1 0) %)))) {:2 0 :3 0}) ((fn [coll] (* (:2 coll) (:3 coll)))))  Part 2: (->> (utils/read-file (str utils/resources-path "day2.txt")) (map #(map-indexed (fn [idx itm] [idx itm]) %)) (map set) ((fn [coll] (combinatorics/combinations coll 2))) (map (fn [[f s]] {:diff (clj-set/difference f s) :orig [f s]})) (filter #(= (count (:diff %)) 1)) ((fn [[{:keys [diff orig]}]] (apply str (map second (sort-by first (clj-set/difference (first orig) diff)))))))  I like the threaded use of update here in part 1 - my method used a transient map and returned a persistent copy at the end: (ns aoc.aoc2) (defn reduce-twos-threes "check the given frequency map n for twos or threes matches, and update the memo map to indicate if the string has a match. Used for a reducer." [memo n] (let [t-memo (transient memo)] (if (some (fn [[k v]] (= v 2)) n) (assoc! t-memo :twos (inc (:twos t-memo)))) (if (some (fn [[k v]] (= v 3)) n) (assoc! t-memo :threes (inc (:threes t-memo)))) (persistent! t-memo))) (defn checksum [input] (let [sum-maps (map frequencies input) twos-threes (reduce reduce-twos-threes {:twos 0 :threes 0} sum-maps)] (* (:twos twos-threes) (:threes twos-threes))))  Nice one. Is definitely faster than mine. Javascript lazy solution I don't have much time to solve the challenges :( So I'm just trying to get the stars. part 1  let twoAppears = 0; let threeAppears = 0; for(ID of input) { const letters = new Set(ID.split('')); let twoAppearing = false; let threeAppearing = false; for(letter of letters) { const length = ID.match(new RegExp(letter, 'g')).length; switch (length) { case 2: if(!twoAppearing) { twoAppears++; twoAppearing = true; } break; case 3: if(!threeAppearing) { threeAppears++; threeAppearing = true; } break; } } } console.log(twoAppears * threeAppears);  Part 2  function getCommonLetters(string1, string2) { let commonLetters = []; for(let i=0; i<string1.length; i++) { if(string1[i] === string2[i] && string1[i] !== '\r') { commonLetters.push(string1[i]); } } return commonLetters; } let mostOcurrencies = [0, 0]; for(let i=0; i<input.length; i+=1) { for(let j=i+1; j<input.length; j+=1) { commonLetters = getCommonLetters(input[i], input[j]); if(commonLetters.length >= 1 && commonLetters.length > mostOcurrencies) { mostOcurrencies = commonLetters.length; mostOcurrencies = commonLetters; } } } console.log(mostOcurrencies.join().replace(/,/g,''))  Thanks for hosting the private leaderboard! Never been on a leaderboard before lol so that'll be fun. :) I am curious - how is everyone posting their code? Is there a code tag on here, like there is on Slack? Is everyone sharing screenshots? I haven't posted a whole lot on here yet, so I'm not sure of the best way to share code. I'm using JS this year, so here's my day 2 solutions: not the prettiest / most succinct, but they work! Part 1: let countDouble = 0; let countTriple = 0; for (let i = 0; i < input.length; i++) { let label = input[i].split(''); let letterCount = {}; label.reduce((letters, letter) => { if (letter in letterCount) { letterCount[letter]++; } else { letterCount[letter] = 1; } return letters; }, 0); let checkCounts = Object.values(letterCount); if (checkCounts.includes(2)) { countDouble++; } if (checkCounts.includes(3)) { countTriple++; } } let checksum = countDouble * countTriple; console.log(checksum);  Part 2: for (let i = 0; i < input.length; i++) { let root = input[i]; for (let j = i+1; j < input.length; j++) { let currName = input[j]; if (compareNames(root, currName)) { return; } } } function compareNames(first, second) { let differences = 0; let locations = []; for (let k = 0; k < first.length; k++) { if (first[k] === second[k]) { continue; } else { differences++; locations.push(k); } } if (differences === 1) { const letterArray = first.split(''); const removeLetter = letterArray.splice(locations, 1); const matchingLetters = letterArray.join(''); console.log(same letters:${matchingLetters});
return true;
} else {
return false;
}

differences = 0;
locations = [];
}


There's an enhanced form of markdown for code blocks: triple backticks for start and end, and if you immediately follow the opening backticks with the language you get syntax highlighting. Difficult to show raw markdown in markdown unfortunately.

Excellent, thank you! Much better than screenshots. :)

Enjoyed this one. My Kotlin solution:

package adventofcode2018.day2

import java.io.File

fun repeatCounts(s: String): Set<Int> =
s.groupBy { it }.values.map { it.size }.toSet()

fun difference(s1: String, s2: String): Int =
s1.zip(s2, { x, y -> if (x == y) 0 else 1 }).sum()

fun common(s1: String, s2: String): String =
s1.zip(s2, { x, y -> if (x == y) x.toString() else "" }).joinToString(separator="")

fun <T> pairs(xs: Collection<T>): List<Pair<T, T>> = when {
xs.isEmpty() -> listOf()
else -> {
val tail = xs.drop(1)
(tail.map { head to it }) + pairs(tail)
}
}

fun main(args: Array<String>) {
val file = if (args.isEmpty()) "input.txt" else args

// Part 1
val counts = input.map(::repeatCounts)
val numPairs = counts.filter { s -> s.contains(2) }.size
val numTriples = counts.filter { s -> s.contains(3) }.size
println("Part 1 checksum: ${numPairs * numTriples}") // Part 2 val differentByOnePairs = pairs(input).filter { (x, y) -> difference(x, y) == 1 } println(differentByOnePairs.map { (x, y) -> common(x, y) }) }  Were you also annoyed that Kotlin has .groupBy but not .frequencies? Have you thought about looking into Sequence? You could make your pairs function lazily? Using List means you're materializing the entirety of your double for-loop right away. The lack of frequencies didn't bother me - it's easy to add. And yes, I've been thinking for the rest of the day that I should use lazy sequences. In this case the execution time remains O(N²) but as you say the memory footprint becomes more like constant. Definitely a good practice when you can't find a better algorithm. A little late, to the party. I tried really hard to think of a solution to part 2 that only involved iterating the list once, but no luck. Here is my solution in Elixir. Part one:  def part_1() do %{"3" => total_3, "2" => total_2} = AOC.read_input("2018/day_2" |> Enum.reduce(%{"2" => 0, "3" => 0}, fn id, acc -> counts = String.split(id, "", trim: true) |> Enum.reduce(%{}, &Map.update(&2, &1, 1, fn val -> val + 1 end)) |> Enum.reduce(%{"2" => 0, "3" => 0}, fn {_, 2}, count -> Map.update!(count, "2", fn _ -> 1 end) {_, 3}, count -> Map.update!(count, "3", fn _ -> 1 end) _, count -> count end) %{acc | "2" => acc["2"] + counts["2"], "3" => acc["3"] + counts["3"]} end) total_2 * total_3 end  Part 2: def part_2() do AOC.read_input("2018/day_2") |> traverse_list() end def traverse_list([head | tail]) do traverse_list(head, tail, tail) end def traverse_list(_, [], [head | tail]) do traverse_list(head, tail, tail) end def traverse_list(compare, [current | tail] = remaining, rest_list) when length(remaining) > 0 do case compare_strings(compare, current) do :notfound -> traverse_list(compare, tail, rest_list) common_chars -> common_chars end end def compare_strings(s1, s2) do s1 = String.split(s1, "", trim: true) s2 = String.split(s2, "", trim: true) zipped = Enum.zip(s1, s2) case Enum.reduce(zipped, %{misses: 0, common: ""}, fn {s, s}, acc -> %{acc | common: acc.common <> s} _, acc -> %{acc | misses: acc.misses + 1} end) do %{misses: 1, common: common} -> common _ -> :notfound end end  So this was a pain. I also ended up with a double loop (O(n2)) and couldn't think of anything better. One thing I discovered during part two was that Crystal has Levenshtein distance as part of the standard library. It might have been a bit heavy going for what I needed, but it did the trick! require "levenshtein" class FabricBox getter id @letter_map : Hash(String, Int32) def initialize(@id : String) @letter_map = @id.split("").reduce(Hash(String, Int32).new(default_value: 0)) do |acc, letter| acc[letter] = acc[letter] + 1 acc end end def has_exactly?(letters : Int32) : Bool @letter_map.values.any? { |count| count == letters } end end class FabricBoxCollection getter boxes @boxes : Array(FabricBox) def initialize(ids : Array(String)) @boxes = ids.map { |id| FabricBox.new(id) } end def checksum @boxes.count { |box| box.has_exactly? 2 } * @boxes.count { |box| box.has_exactly? 3 } end def find_close_id @boxes.each do |box_i| @boxes.each do |box_j| if Levenshtein.distance(box_i.id, box_j.id) == 1 characters_i = box_i.id.split("") characters_j = box_j.id.split("") char_pairs = characters_i.zip(characters_j) return char_pairs.reduce("") do |acc, (char_i, char_j)| acc += char_i if char_i == char_j acc end end end end end end puts "--- Day 2: Inventory Management System ---" input = File.read_lines("./02/input.txt") collection = FabricBoxCollection.new(input) puts "Checksum: #{collection.checksum}" puts "Common letters: #{collection.find_close_id}"  Bring on day 3! High five for a beefy standard library! That’s awesome 😎 from collections import defaultdict from difflib import SequenceMatcher from operator import itemgetter def similar(a, b): return SequenceMatcher(None, a, b).ratio() def main(): part_one() part_two() def part_one(): with open('input.txt', 'r') as input_file: lines = [line for line in input_file] word_twice_count = 0 word_three_times_count = 0 for line in lines: word_count_dic = defaultdict(int) has_char_twice = False has_char_three_times = False for char in line: word_count_dic[char] += 1 for key in word_count_dic.keys(): if word_count_dic[key] == 2: has_char_twice = True elif word_count_dic[key] == 3: has_char_three_times = True if has_char_twice: word_twice_count += 1 if has_char_three_times: word_three_times_count += 1 checksum = word_twice_count * word_three_times_count print 'checksum is ' + str(checksum) def part_two(): with open('input.txt', 'r') as input_file: lines = [line for line in input_file] similarity_strings = [(line, comparing_line, similar(line, comparing_line)) for line in lines for comparing_line in lines if line is not comparing_line] most_similair_strings = max(similarity_strings, key=itemgetter(2)) result_word = ''.join([char_word_one for char_word_one, char_word_two in zip(most_similair_strings, most_similair_strings) if char_word_one == char_word_two]) print result_word if __name__ == '__main__': main()  My solution with Python, not sure about the second part, not the most fastest solution I think regarding of performance. Nice! Don’t forget about collections.Counter for part 1! I didn’t know about difflib though. Cool! Thanks for the hint! Will do that later! def part_one(): word_twice_count = 0 word_three_times_count = 0 with open('input.txt', 'r') as input_file: for line in input_file: line_counter = Counter(line) if 2 in line_counter.values(): word_twice_count += 1 if 3 in line_counter.values(): word_three_times_count += 1 checksum = word_twice_count * word_three_times_count print 'checksum is ' + str(checksum)  My edited solution for part one with collections Counter Part 1: C# + LINQ = one-liner return (input.Count(id => id.GroupBy(c => c).Any(group => group.Count() == 2)) * input.Count(id => id.GroupBy(c => c).Any(group => group.Count() == 3)))  Part 2 for (int i = 0; i < input.Length; i++) { var commonIds = input.Select(id => id.Remove(i, 1)).GroupBy(id => id).FirstOrDefault(group => group.Count() > 1); if (commonIds != null) { return commonIds.First(); } }  Part 1 from collections import Counter with open("input.txt") as f: ids = [Counter(l.strip()) for l in f] count2 = 0 count3 = 0 for c in ids: if 2 in c.values(): count2 += 1 if 3 in c.values(): count3 += 1 print(count2 * count3)  Part 2 from editdistance import eval as dist from itertools import product with open("input.txt") as f: ids = [l.strip() for l in f] for a, b in product(ids, ids): d = dist(a, b) if d == 1: print(a, b) break  And from the output, I just copied and pasted the necessary characters that matched. That was faster than comming up with a custom method to do so. Nice! Did you implement editdistance yourself, or is that an external library? It is external. I found it via a quick google search. The edit distance measures how many operations - insertion, deletion or substitution - it takes to get from one string to the other. Since all the strings in the puzzle input have the same length, insertion and deletion do not come into play and it works out perfectly. Ok that’s cool, just checking. Neat! Terrible C++ solution for part 1 ! #include <iostream> #include <fstream> #include <list> #include <map> std::list<std::string> read_file(std::string filename) { auto lines = std::list<std::string>(); std::ifstream file(filename); for (std::string line; std::getline(file, line);) { lines.push_back(line); } file.close(); return lines; } class WordCounter { public: static std::map<char, int> CalculateLine(const std::string line) { auto map = std::map<char, int>(); if (line.length() == 0) { return map; } for (int i = 0; i < line.length(); i++) { std::map<char, int>::iterator it = map.find(line[i]); if (it != map.end()) { it->second = it->second + 1; } else { map.insert(std::pair<char, int>(line[i], 1)); } } return map; } static const std::pair<int, int> Sum(std::map<char, int> map) { int twos = 0; int threes = 0; for (auto line = map.begin(); line != map.end(); line++) { if (line->second == 2) { twos = 1; } else if (line->second == 3) { threes = 1; } } return std::pair<int, int>(twos, threes); } }; int main() { auto input = read_file("./input.txt"); int two = 0; int three = 0; for (auto i = input.begin(); i != input.end(); i++) { auto letterMap = WordCounter::CalculateLine(*i); auto calculated = WordCounter::Sum(letterMap); two += calculated.first; three += calculated.second; } std::cout << two * three << std::endl; return 0; }  Terrible is better than never finished! And this looks pretty good to me, not knowing C++ if that makes you feel better 😄 I did my solutions at midnight last night, but I was surviving on very little sleep at the time, so the resulting code was below standard. I tried again this morning and felt better about it. For anyone reading this, I'm still using the simple lines function from Day 1 which reads a file into a sequence of strings. ### Part 1 This was my solution last night: (defn nums [s] (let [cs (->> s (group-by identity) vals (map count))] [(some #(= 2 %) cs) (some #(= 3 %) cs)])) (defn star [input-file] (let [tt (->> (lines input-file) (map nums))] (* (count (filter first tt)) (count (filter second tt)))))  This is embarrassing code. I totally forgot about the frequencies function, which is why I used group-by followed by count. But the 2 filter operations in the final calculation meant that the results of the map get processed twice. My morning attempt fixed these: (defn counts [[two-count three-count] s] (let [cs (-> s frequencies vals)] [(if (some #(= 2 %) cs) (inc two-count) two-count) (if (some #(= 3 %) cs) (inc three-count) three-count)])) (defn star [input-file] (let [[two-count three-count] (->> (lines input-file) (reduce counts [0 0]))] (* two-count three-count)))  This time I accumulated the 2/3 count values while processing the list, so it only goes through it once. ### Part 2 Since each element needs to be compared to every other element, I can't see any way around the O(n2 ) complexity. Of course, each element should only be compared to the ones after it, so as to avoid comparing each one twice (since comparing A/B has the same result as comparing B/A). When doing list processing, the only ways I know to refer to everything following an element are by using indexes (yuck) or with a loop. Unfortunately, I got fixated on the loop construct, and nested it: (defn close [a b] (let [sim (filter identity (map #(when (= %1 %2) %1) a b))] (when (= (count sim) (dec (count a))) sim))) (defn cmp-close [ll] (loop [[f & fr] ll] (when (seq fr) (or (loop [[s & sr] fr] (when s (or (close f s) (recur sr)))) (recur fr))))) (defn star2 [input-file] (let [ll (lines input-file)] (apply str (cmp-close ll))))  The other way you can tell that I wrote this on very little sleep was the use of ridiculously terse var names. On reflection this morning, I realized that the inner loop should have been a some operation. This does a predicate test and terminates processing early, which is what I was doing manually with the inner loop. Also, my close function has several issues. First is the name! I was thinking of "A is close to B", but when I saw it again this morning I realized that it's the same function name for closing a resource. Something else that bothers me is that it processes the entirety of each string before returning, when a false result can usuall terminate early. Finally, a minor issue is that the anonymous function #(when (= %1 %2) %1) would be more idiomatic to use a singleton set on %1 to compare to %2. (defn nearly= [left right] (let [same (filter identity (map #(#{%1} %2) left right))] (when (= (count same) (dec (count left))) (apply str same)))) (defn compare-lines [ll] (loop [[line & xlines] ll] (when (seq line) (or (some (partial nearly= line) xlines) (recur xlines))))) (defn star2 [input-file] (compare-lines (lines input-file)))  The nearly= function now returns a string, rather than the array of characters, but hasn't changed much. I was still unsatisfied with it not terminating the test early, so I rewrote it to be faster. However, the resulting code lacks any kind of elegance, and I'm not convinced that it's worth the effort: (defn nearly= [left right] (loop [[l & xl] left [r & xr] right diffs 0] (if (nil? l) (apply str (filter identity (map #(#{%1} %2) left right))) (if (not= l r) (when (zero? diffs) (recur xl xr 1)) (recur xl xr diffs)))))  Hopefully I'll get some sleep before attempting day 3. 😊 A bit late to the party, here are my solutions on Ruby: Part 1: twos = 0 threes = 0 File.open('day-02_input.txt').each_line do |line| counter = {} line.each_char do |char| next if line.count(char) <= 1 counter[char] = line.count(char) end twos += 1 if counter.value? 2 threes += 1 if counter.value? 3 end puts twos * threes  Part 2: boxes = [] File.open('day-02_input.txt').each_line { |box| boxes << box.chomp! } def find_distance pair, distance = 0 pair.each_char.with_index do |char, index| distance += 1 if char != pair[index] end distance end boxes.combination(2).each do |pair| next unless find_distance(pair) == 1 str = '' pair.chars.each_with_index do |_, i| str << pair[i] if pair[i] == pair[i] end puts str end  I learnt about the combination method :) My solution in JavaScript / Node 11, using the readline interface: readLines.js const fs = require('fs'); const readline = require('readline'); const readLines = (file, onLine) => { const reader = readline.createInterface({ input: fs.createReadStream(file), crlfDelay: Infinity }); reader.on('line', onLine); return new Promise(resolve => reader.on('close', resolve)); }; const readFile = async file => { const lines = []; await readLines(file, line => lines.push(line)); return lines; } module.exports = { readLines, readFile };  02a.js const { readFile } = require('./readLines'); (async () => { const lines = await readFile('02-input.txt'); let twoLettersCount = 0; let threeLettersCount = 0; for (let line of lines) { const frequencyMap = {}; for (const char of line.split('')) { frequencyMap[char] = (frequencyMap[char] || 0) + 1; } const hasTwoLetters = Object.values(frequencyMap).some(frequency => frequency === 2); const hasThreeLetters = Object.values(frequencyMap).some(frequency => frequency === 3); twoLettersCount += +hasTwoLetters; threeLettersCount += +hasThreeLetters; }; const checksum = twoLettersCount * threeLettersCount; console.log(The checksum is${checksum});
})();


02b.js

const { readFile } = require('./readLines');

// Compares two strings to see if they differ by one char and which one
function compare(string1, string2) {
const length = string1.length;
let differentChars = 0;
let differIndex;
for (let i = 0; i < length; i++) {
if (string1.charAt(i) !== string2.charAt(i)) {
differentChars++;
differIndex = differentChars === 1 ? i : undefined;
}
}

return {
differByOneChar: differentChars === 1,
differIndex
};
}

// Compare each strings to every other string
// and get the common letters in case the differByOneChar is true
function getCommonLetters(ids) {
const idsCount = ids.length;
for (let i = 0; i < idsCount; i++) {
const id = ids[i];
for (let j = i + 1; j < idsCount; j++) {
const { differByOneChar, differIndex } = compare(id, ids[j]);
if (differByOneChar) {
return id.slice(0, differIndex) + id.slice(differIndex + 1);
}
}
}
}

(async () => {

const commonLetters = getCommonLetters(lines);
console.log(The common letters are ${commonLetters}); })();  My solutions for part 1 Python def checksum(inputs): ids = inputs.read().splitlines() twos = 0 threes = 0 for row in ids: seen = {} for letter in row: seen[letter] = seen.get(letter, 0) + 1 if 2 in seen.values(): twos += 1 if 3 in seen.values(): threes += 1 return twos * threes checksum(open('input.txt', 'r'))  Go package main import ( "strings" "io/ioutil" ) func containsVal(m map[string]int, v int)(bool) { for _, x := range m{ if x == v { return true } } return false } func checksum(x string)(int) { two := 0 three := 0 ids := strings.Split(x, "\n") for i := 0; i < len(ids); i++ { counter := make( map[string]int ) s := strings.Split(ids[i], "") for _, l := range s { counter[l]++ } if containsVal(counter, 2) { two++ } if containsVal(counter, 3) { three++ } } return two * three } func main() { f, err := ioutil.ReadFile("input.txt") if err != nil { panic(err) } checksum(string(f)) }  Benchmark difference python3 2_1.py 100 times real 0m4.744s user 0m3.129s sys 0m0.967s go run 2_1.go 100 times real 0m37.649s user 0m28.807s sys 0m14.105s go build ./2_1 100 times real 0m0.709s user 0m0.327s sys 0m0.280s  My solution to day 2, in Elixir. The double for-loop in part 2 is certainly not optimal, but it works. The available time was short today. :) Part one: defmodule AoC.DayTwo.PartOne do alias AoC.DayTwo.Common def main() do "lib/day2/input.txt" |> Common.read_input() |> Enum.map(&count_characters/1) |> Enum.map(&Map.values/1) |> get_multiplicants() |> Enum.reduce(1, fn x, acc -> acc * x end) end defp count_characters(box_id) do box_id |> String.graphemes() |> Enum.reduce(%{}, fn x, acc -> Map.put(acc, x, (acc[x] || 0) + 1) end) end defp get_multiplicants(occurrences_lists) do twos = get_multiplicant(occurrences_lists, 2) threes = get_multiplicant(occurrences_lists, 3) [twos, threes] end defp get_multiplicant(occurrences_lists, count) do multiplicant = Enum.reduce(occurrences_lists, 0, fn x, acc -> if Enum.member?(x, count), do: acc + 1, else: acc end) end end  Part two: defmodule AoC.DayTwo.PartTwo do alias AoC.DayTwo.Common def main() do "lib/day2/input.txt" |> Common.read_input() |> find_correct_boxes() receive do {x, y} -> get_common_characters(x, y) end end defp find_correct_boxes(box_ids) do for x <- box_ids do for y <- box_ids do zipped = Enum.zip(String.graphemes(x), String.graphemes(y)) differences = Enum.reduce(zipped, 0, fn {z1, z2}, acc -> if z1 == z2, do: acc, else: acc + 1 end) if(differences == 1) do send(self(), {x, y}) end end end end defp get_common_characters(x, y) do Enum.zip(String.graphemes(x), String.graphemes(y)) |> Enum.reduce("", fn {x1, x2}, acc -> if x1 == x2, do: acc <> x1, else: acc end) end end  Common: defmodule AoC.DayTwo.Common do def read_input(path) do path |> File.stream!() |> Stream.map(&String.trim_trailing/1) |> Enum.to_list() end end  Giving it a go in good ol' JavaScript. Part 1 First creates an Object to track the # of times a letter appears in a string. That gets converted to a Set to filter out any duplicate values to account for situations where 2 or more letters appeared twice (as the string only gets counted once for the check sum). const input = text.split("\n").map(d => d.split('')) let x2 = 0 let x3 = 0 input.forEach(line => { let counted = [...new Set(Object.values(line.reduce((allLetters, letter) => { if (letter in allLetters) { allLetters[letter]++ } else { allLetters[letter] = 1 } return allLetters }, {})))] x2 += counted.filter(d => d === 2).length x3 += counted.filter(d => d === 3).length }) let checksum = x2 * x3  Part 2 I struggled with this one, so I'm sure there's a cleaner/more efficient way to do this. This takes each line of the input and compares it to the other lines, checking the characters and tracking the # of differences and the position of the last accounted for difference. The loops are set to break when a result with only 1 difference is found to prevent unnecessary looping. const input = text.split("\n") let check = true for (let line of input) { const comparisonLines = input.filter(d => d !== line) for (let entry of comparisonLines) { let result = findDifferences(line, entry) if (result.numDiff === 1) { let string = line.slice(0, result.posDiff) + line.slice(result.posDiff + 1) console.log(string) check = false break } } if (!check) break } function findDifferences(string1, string2) { if (string1 === string2) return const length = Math.max(string1.length, string2.length) let numDiff = 0 let posDiff = 0 for (let i = 0; i <= length; i++) { if (string1[i] !== string2[i]) { numDiff++ posDiff = i } } return {numDiff: numDiff, posDiff: posDiff} }  ### Part 1 in Ruby: have_two_letters = 0 have_three_letters = 0 File.open('./input.txt', 'r').each do |str| freq = str.split('').group_by(&:itself).map {|k,v| [v.size, k] }.to_h have_two_letters += 1 if freq have_three_letters += 1 if freq end puts have_three_letters * have_two_letters  I enjoyed playing around with the one liner str.split('').group_by(&:itself).map {|k,v| [v.size, k] }.to_h  which produces a frequency chart something like this: {1=>"j", 4=>"f", 2=>"s"}  i.e. there are exactly 2 occurrences of the letter s in str ### Part 2 in Ruby lines = File.read('./input.txt').split("\n"); def get_shared_letters(str1, str2) differences = 0 same_letters = nil str1.split('').each_with_index do |letter, i| if letter != str2[i] differences += 1 same_letters = str1.dup same_letters.slice!(i) end end differences === 1 ? same_letters : nil end lines.each_with_index do |str1, i1| lines.each_with_index do |str2, i2| next if i2 === i1 shared = get_shared_letters str1, str2 puts shared if shared end end  Not happy about the double loop through the input file... but also can't think of a way to avoid it! It occurred to me that sorting the list first might improve the chances of finding a match earlier in the loop since all similar strings would be together, but thinking about it, perhaps there's equal chance that the similar strings would end up being sorted to the bottom of the list and it wouldn't help at all :/ Another thought - many of the input strings are v. similar, maybe they could be grouped into sets early on (e.g. based on the first 4 chars or something) and then you only look for similar strings in the same set? Going to read into hamming distances now! I lost about 10 minutes to my son stalling getting into bed. ### Kotlin Solution #### Answer 1 Once again a fairly straightforward opener. Just run through, do some simple frequency checking and spit back the results. I think this is technically O(n2) but it moved fast enough for me. (And in a more lazy language, it ends up being closer to O(n) anyway) fun String.frequencies(): Map<Char, Int> = groupBy { it }.mapValues { (_, v) -> v.count() } fun answer1(input: List<String>) = input.map { val freqs = it.frequencies() freqs.anyValue(2::equals) to freqs.anyValue(3::equals) } .unzip().toList() .map { bs -> bs.count { it } } .fold(1, Int::times)  #### Answer 2 As Ryan said, this is just Hamming distance with the added wrinkle that you need to throw away the differences while counting them. Lots of optimization room here, but once again, we shave off just under half the possibilities by only doing unique combinations and going from a raw n^2 to (n*(n+1))/2. At around 10 ms (calculated by shuffling my input 1000 times), I don't think there's an easy way to make this noticeably faster at this point without a more esoteric method. fun <A, B> List<A>.cartesian(transform: (A, A) -> B): Sequence<B> { return init.asSequence().mapIndexed { i, a -> drop(i + 1).asSequence().map { b -> transform(a, b) } }.flatten() } fun <A> Pair<A, A>.same() = first == second fun <A> Pair<A, A>.different() = first != second fun answer2(input: List<String>) = input.cartesian { s1, s2 -> s1.zip(s2) }.find { // find short circuits on Sequence it.count(Pair<Char, Char>::different) == 1 }?.filter { it.same() }?.joinToString("") { it.first.toString() }  ## Node.js Common async generator. Read the file in chunk by chunk and yield each product ID based on new lines. // Generator // "abcdef\n+bababc\n" yields -> abcdef -> bababc async function* streamToProdctId(stream) { let previous = ""; for await (const chunk of stream) { previous += chunk; let eolIndex; while ((eolIndex = previous.indexOf("\n")) >= 0) { // productId excludes the EOL const productId = previous.slice(0, eolIndex); yield productId; previous = previous.slice(eolIndex + 1); } } if (previous.length > 0) { yield previous; } }  Part 1 was fun with some ES6 array -> Set -> Map to get the value counts // ['b', 'a', 'b', 'a', 'b', 'c'] returns Map {3 => 'b', 2 => 'a', 1 => 'c} const count = array => { return new Map( [...new Set(array)].map(x => [array.filter(y => y === x).length, x]) ); }; const checksum = async stream => { let idsWith2MatchingLetters = 0; let idsWith3MatchingLetters = 0; for await (const productId of streamToProductId(stream)) { const valueCounts = count([...productId]); if (valueCounts.has(2)) { idsWith2MatchingLetters++; } if (valueCounts.has(3)) { idsWith3MatchingLetters++; } } return idsWith2MatchingLetters * idsWith3MatchingLetters; };  Part 2 got interesting. I needed to generate all pairs for every product ID. I made my Hamming Distance function also return the common letters. Then tied it all together by running each pair through the Hamming Distance function and getting the lowest. const generatePairs = array => { return array.reduce( (acc, _, i1) => [ ...acc, ...new Array(array.length - 1 - i1) .fill(0) .map((v, i2) => [array[i1], array[i1 + 1 + i2]]) ], [] ); }; const hammingDistance = (stringOne, stringTwo) => { let distance = 0; let commonLetters = ""; for (let i = 0; i < stringOne.length; i++) { if (stringOne[i] !== stringTwo[i]) { distance += 1; } else { commonLetters = commonLetters.concat(stringOne[i]); } } return [distance, commonLetters]; }; const findLowestPairAndRemoveDifferences = pairs => { let lowestDistance = Infinity; let lowestPair; pairs.forEach(pair => { const [distance, commonLetters] = hammingDistance(...pair); if (distance < lowestDistance) { lowestDistance = distance; lowestPair = commonLetters; } }); return lowestPair; }; const productIds = async stream => { const ids = []; for await (const productId of streamToProductId(stream)) { ids.push(productId); } return ids; }; const findCommon = async stream => { return findLowestPairAndRemoveDifferences( generatePairs(await productIds(stream)) ); };  Putting it all together: const productIdStream = () => { return fs.createReadStream(__dirname + "/input.txt", { encoding: "utf-8", highWaterMark: 256 }); }; const main = async () => { try { const part1 = await checksum(productIdStream()); console.log({part1}); const part2 = await findCommon(productIdStream()); console.log({part2}); } catch (e) { console.log(e.message); process.exit(-1); } };  Full code here: github.com/MattMorgis/Advent-Of-Co... Making hamming distance return common letters is a slick way to go. I was looking at the duplication between the hamming distance and common letters functionality in my solution and was a little bummed about it, but I couldn't figure out a good way to do it. I like this! Parts 1 and 2 in Ruby: require 'set' box_ids = DATA.read.split.map(&:each_char) counts = Hash.new(0) box_ids.map do |box_id| char_counts = box_id.group_by(&:itself).values.map(&:count) Set.new(char_counts).each { |n| counts[n] += 1 } end puts counts * counts # 6474 pair = box_ids.combination(2).map { |a1, a2| a1.zip(a2) }.lazy.find { |zipped| zipped.count { |c1, c2| c1 != c2 } == 1 } # mxhwoglxgeauywfkztndcvjqr puts pair.map { |c1, c2| c1 if c1 == c2 }.join __END__ myhposlqgeauywfikztndcvrqr mbhposlxfeauywoikztndcvjqi mbhpoulxgeagywfikytndcvjqr jbhposlxgeauywdikztndcvjqk mbhpsslxueauywfikzfndcvjqr mbhposnxgeauzyfikztndcvjqr # data section abbreviated  Rust Part 1 use std::collections::HashMap; fn contains_repeats(input: &str) -> (bool, bool) { let mut count = HashMap::new(); input.chars().for_each(|c| { let n = count.entry(c).or_insert(0); *n += 1; }); let two = count.iter().any(|(_, &v)| v == 2); let three = count.iter().any(|(_, &v)| v == 3); (two, three) } pub fn checksum(input: &[&str]) -> usize { let (twos, threes): (Vec<_>, Vec<_>) = input.iter().map(|&s| contains_repeats(s)).unzip(); let two = twos.iter().filter(|&x| *x).count(); let three = threes.iter().filter(|&x| *x).count(); two * three }  Part 2 I'm still searching for a nice iterator adapter to replace the nested loop. I finally used the itertools crate and it's AMAZING!!! use itertools::Itertools; fn hamming(input1: &str, input2: &str) -> usize { input1 .chars() .zip(input2.chars()) .filter(|(a, b)| a != b) .count() } fn common(input1: &str, input2: &str) -> String { input1 .chars() .zip(input2.chars()) .filter_map(|(a, b)| match a == b { true => Some(a), false => None, }) .collect() } pub fn find_common(input: &[&str]) -> String { input .iter() .tuple_combinations() .find_map(|(first, second)| match hamming(first, second) { 1 => Some(common(first, second)), _ => None, }) .unwrap() }  I am using Advent of Code to learn Golang, and here is the solution I came up with. Suggestions for improvements are always welcome! Part 1: package main import ( "bufio" "fmt" "os" "strings" ) // readLines reads a whole file into memory // and returns a slice of its lines. func readLines(path string) ([]string, error) { file, err := os.Open(path) if err != nil { return nil, err } defer file.Close() var lines []string scanner := bufio.NewScanner(file) for scanner.Scan() { lines = append(lines, scanner.Text()) } return lines, scanner.Err() } func main() { boxes, err := readLines("input") if err != nil { panic(err) } var twos, threes int for _, box := range boxes { // Iterate through each box letter-by-letter and check if letters appear // two or three times: two, three := false, false for _, letter := range box { if !two && strings.Count(box, string(letter)) == 2 { twos++ two = true } else if !three && strings.Count(box, string(letter)) == 3 { threes++ three = true } if two && three { // We already found the maximum number of appearing letters we count break } } } checksum := twos * threes fmt.Printf("Checksum is: %d\n", checksum) }  Part 2: package main import ( "bufio" "fmt" "os" ) // readLines reads a whole file into memory // and returns a slice of its lines. func readLines(path string) ([]string, error) { file, err := os.Open(path) if err != nil { return nil, err } defer file.Close() var lines []string scanner := bufio.NewScanner(file) for scanner.Scan() { lines = append(lines, scanner.Text()) } return lines, scanner.Err() } func findSimilar(boxes []string) string { // For each box name we remove the i'th element and store the rest of the // name inside a map: visited := make(map[int]map[string]bool) for _, box := range boxes { // Go through each box: for i := range box { // Remove i'th character from box name: subname := box[:i] + box[i+1:] // Check if we have already visited a similar box name: _, ok := visited[i][subname] if !ok { // If we have never encountered a similar box name, add it: _, ok := visited[i] if !ok { sub := make(map[string]bool) visited[i] = sub } visited[i][subname] = true } else { // We have encounter a similar box name, return it: return subname } } } return "" } func main() { boxes, err := readLines("input") if err != nil { panic(err) } subname := findSimilar(boxes) fmt.Println(subname) }  My idea was to use a dictionary and store the subnames and see if we encounter one we have already visited. Since there should only be one match we can immediately return it. I learned a lot about using maps in Golang! I am also using Python that I have more experience with to cross check solutions. I have tried to implement it with readability in mind, not performance. Part 1: with open('input') as f: boxes = f.readlines() twos, threes = 0, 0 for box in boxes: twos += any([box.count(letter) == 2 for letter in set(box)]) threes += any([box.count(letter) == 3 for letter in set(box)]) checksum = twos * threes print('Checksum {}'.format(checksum))  Part 2: with open('input') as f: boxes = f.readlines() def closest_boxes(): visited = {} for box in boxes: for i in range(len(box)): if i not in visited: visited[i] = set() subname = box[0:i] + box[i+1:] if subname in visited[i]: return subname else: visited[i].add(subname) print(closest_boxes())  Late to the party! Still digging F#, but I'm definitely hindered by my lack of familiarity with what's available in .NET namespace Day2 open Microsoft.FSharp.Collections module util = type FreqMap = Map<char, int> let getBoxIds fileName = let lines = System.IO.File.ReadLines(fileName) lines |> Seq.toList let tickFreq c (freqs: FreqMap) = match (freqs.TryFind c) with | Some x -> Map.add c (x + 1) freqs | None -> Map.add c 1 freqs let rec goGetFreqs (freqs: FreqMap) boxId = let len = String.length boxId if len = 1 then tickFreq boxId. freqs else goGetFreqs (tickFreq boxId. freqs) boxId.[1..len - 1] let getFreqs boxId = goGetFreqs (new Map<char, int> (Seq.empty)) boxId let countOfFreq num (freqMap: FreqMap) = freqMap |> Map.filter (fun _ v -> v = num) |> Map.toList |> List.length module part1 = let getTwosAndThrees boxId = let freqs = util.getFreqs boxId let twos = util.countOfFreq 2 freqs let threes = util.countOfFreq 3 freqs (twos, threes) let rec getTotals twos threes l = match l with | [] -> (twos, threes) | (two, three)::tail -> let newTwos = if two > 0 then twos + 1 else twos let newThrees = if three > 0 then threes + 1 else threes getTotals newTwos newThrees tail let execute fileName = let boxIds = util.getBoxIds fileName let twosAndThrees = List.map (fun bid -> getTwosAndThrees bid) boxIds let (twos, threes) = getTotals 0 0 twosAndThrees twos * threes module part2 = let checkTwo id id' = let s = (Seq.map2 (fun c c' -> if c <> c' then '!' else c) id id') |> Seq.filter (fun c -> c <> '!') |> Seq.toArray |> System.String if (String.length id) - 1 = String.length s then Some s else None let findWinner boxIds = List.map (fun id -> List.map (fun id' -> if id = id' then None else checkTwo id id') boxIds) boxIds |> List.map (List.filter (fun el -> el.IsSome)) |> List.filter (fun el -> List.length el > 0) |> List.concat |> List.distinct let execute fileName = let boxIds = util.getBoxIds fileName findWinner boxIds  I'm not super thrilled with my Day 2 code, but I haven't really had the time to tweak it with everything going on at work currently. Swift Solutions Part 1 This one was fairly simple. Just count how many times each letter appears in each String and act appropriately. I do like the fact that a Swift String type is really an Array of Characters. // Part 1: Find the checksums func calculateChecksum(_ idCodes: [String]) -> Int { var doubles = 0 var triples = 0 idCodes.map { (boxID) in var doubleTrue = false var tripleTrue = false for char in boxID { let count = boxID.filter {$0 == char }.count
if count == 2 {
doubleTrue = true
}
if count == 3 {
tripleTrue = true
}
}
doubles += doubleTrue ? 1 : 0
triples += tripleTrue ? 1 : 0
}
return doubles * triples
}

let checksum = calculateChecksum(boxIDs)
print("Boxes checksum is: \(checksum)")


Part 2
This one feels clunky, if I get a chance I'll revisit it.

I break on the first hit for a solution to short circuit everything and return the answer, this can help a lot with so many Characters to test.

I use zip(_:_:) with a filter and count to quickly test how many differences there are in the same positions. When I see two strings that differ by one character in the same position I move to the next step.

In the second part I cast the Arrays into Set types so that I can use the Collection Algebra features to quickly find the actual character to remove by subtracting one collection from the other. With that done I can just remove it from the source Array and return what's left.

// Part 2: Box finder
func findTheBoxes(_ idCodes: [String]) -> String {
var result = ""
var differceCount = 0

for boxID in idCodes {
if differceCount == 1 {
break
}
for code in idCodes {
differceCount = zip(boxID, code).filter{$0 !=$1}.count
if differceCount == 1 {
let diff = Set(boxID).subtracting(code)
if let charToRemove = diff.first {
result = boxID
if let foo = result.index(of: charToRemove) {
result.remove(at: foo)
break
}
}
}
}
}
return result
}

let theBoxes = findTheBoxes(boxIDs)
print("Matching box ID is: \(theBoxes)")


Normally I would import Foundation here so that I could just use NSOrderedSet and skip a few steps. I wanted to try and keep it all in the Swift Standard Library though, so I didn't!

Here's my C# .Net solution:

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.IO;

namespace Day2Part1
{
class Program
{
static void Main(string[] args)
{
List<int> TwoMatchList = new List<int>();// List of id's to find box in the input array
List<int> ThreeMatchList = new List<int>();// list of id's to find box in the input array
foreach (string boxID in inputTxt)
{
int id = Array.IndexOf(inputTxt, boxID);
var distinctLetters = boxID.Select(x => x).Distinct().ToList();
foreach (char letter in distinctLetters)
{
var matches = boxID.Where(x => x == letter).ToList();
switch (matches.Count)
{
case 2:
//if there's exactly 2, and it's not already in the list then add it to the TwoMatchList
if (!TwoMatchList.Contains(id))
{
}
break;
case 3:
//if there's exactly 3, and it's not already in the list then add it to the ThreeMatchList
if (!ThreeMatchList.Contains(id))
{
}
break;
default:
break;
}//end switch
}//end foreach letter
}//end foreach boxID
int checksum = TwoMatchList.Count * ThreeMatchList.Count;
Console.WriteLine("Number of boxes with exactly two matching letters {0}", TwoMatchList.Count);
Console.WriteLine("Number of boxes with exactly three matching letters {0}", ThreeMatchList.Count);
Console.WriteLine("Checksum: {0}", checksum);
Console.WriteLine();

Console.WriteLine("Now for part two!");

//empty Dictionary to hold any boxes we find and the index of the letter that's different
Dictionary<string, int> savedIDsAndIndexes = new Dictionary<string, int>();

//empty string to hold the answer

//Loop through to test and find correct boxes ==>> this could be a foreach loop
for (int i = 0; i < inputTxt.Length; i++)
{
string boxID = inputTxt[i];
for (int j = inputTxt.Length - 1; j >= 0; j--)
{
List<int> mismatchIndexes = new List<int>();
string testBoxID = inputTxt[j];
for (int k = 0; k < boxID.Length; k++)
{
if (boxID[k] != testBoxID[k])
{
}
}
if (mismatchIndexes.Count == 1)
{
}
}
}

Console.WriteLine("number of correct boxes found: {0}", savedIDsAndIndexes.Count);

+ savedIDsAndIndexes.Keys.First().Substring(savedIDsAndIndexes.Values.First() + 1);

Console.WriteLine("here are the ids:");
foreach (var id in savedIDsAndIndexes)
{
Console.WriteLine(id.Key);
}

Console.WriteLine("saved index: {0}", savedIDsAndIndexes.Values.First());
}
}
}


I didn't see any C# glancing through:

public class InventoryManagement
{
public int GetCheckSum(IEnumerable<string> inventory)
{
return GetCheckSum(GetCheckSumCandidates(inventory));
}

public int GetCheckSum(IEnumerable<InventoryChecksumCandidates> candidates)
{
var checksumPieces = candidates
.GroupBy(gb => gb.DuplicateCount)
.Select(s => new
{
s.Key,
Count = s.Count()
})
.ToList();

Debug.Assert(checksumPieces.Count() == 2);

return checksumPieces.Count * checksumPieces.Count;
}

public IEnumerable<InventoryChecksumCandidates> GetCheckSumCandidates(IEnumerable<string> inventory)
{
List<InventoryChecksumCandidates> list = new List<InventoryChecksumCandidates>();

foreach (var item in inventory)
{
var checksumCandidate = item
// make everything same case (just in case)
.ToLower()
// where it's a letter
.Where(char.IsLetter)
// Group by the default (letter)
.GroupBy(gb => gb)
// Project the found values into their new type
.Select(s => new InventoryChecksumCandidates()
{
DuplicateCharacter = s.Key.ToString(),
DuplicateCount = s.Count()
});

if (checksumCandidate != null)
{
checksumCandidate.FirstOrDefault(f => f.DuplicateCount == 2)
);
checksumCandidate.FirstOrDefault(f => f.DuplicateCount == 3)
);
}
}

return list;
}
}


(Note AddIfNotNull is from my extension methods package nuget.org/packages/Kritner.Extensi...

Part 2:

public class PrototypeFabricFinder
{
public string GetProtoTypeFabric(IEnumerable<string> inventory)
{
var inventoryPermutations = SwapCharForEachInventoryPermutation(inventory);
var foundDuplicateish = inventoryPermutations
// Group by default
.GroupBy(gb => gb)
// We only want the instance where there's more than
// one in the group by (the prototype fabric)
.Where(w => w.Count() > 1)
.Select(s => new
{
s.Key
})
.FirstOrDefault();

// Return the prototype fabric, minus the "different" single character
return foundDuplicateish.Key.Replace("*", "");
}

/// <summary>
/// Builds and returns a <see cref="IEnumerable{string}"/>
/// containing every permutation of iventory items, with
/// one character (*) swapped in at a differing index.
///
/// I have no idea if that makes sense.
/// </summary>
/// <param name="inventory">each inventory id.</param>
/// <returns></returns>
private IEnumerable<string> SwapCharForEachInventoryPermutation(
IEnumerable<string> inventory
)
{
List<string> list = new List<string>();

foreach (var item in inventory)
{
// Create new strings and add them to the list.
// The new strings will be a copy of the original,
// with a single character (the current index) swapped in with "*"
for (var index = 0; index < item.Length; index++)
{
var charArrayOfItem = item.ToCharArray();
charArrayOfItem[index] = '*';

}
}

return list;
}
}


I dunno how I'm gonna keep up during the week, but putting my solutions in the repo github.com/Kritner/Kritner.AdventO...

Hosting my solutions on my github...

## MustafaHaddara / advent-of-code-2018

I'm trying to learn Julia so I'm using AoC to force myself to learn.

I'm pretty sure part 2 has to be O(n2) but you can cut down on the total number of iterations by only looking forward...here's my solution (implemented in Julia)

function inventory_management_diff(input)
word_idx = 0
while word_idx < length(input)
word_idx += 1
word1 = input[word_idx]

for word2 in input[word_idx+1:end]
ch_diff = one_char_diff(word1, word2)
if ch_diff != nothing
return string(word1[1:ch_diff-1], word1[ch_diff+1:end])
end
end
end
end

function one_char_diff(word1, word2)
found_diff = nothing
ch_idx = 0
while ch_idx < length(word1)
ch_idx += 1
if word1[ch_idx] != word2[ch_idx]
if found_diff == nothing
found_diff = ch_idx
else
return nothing
end
end
end
return found_diff
end

function main()
filename = ARGS  # julia arrays are 1-indexed!
test_input = ["abcde","fghij","klmno","pqrst","fguij","axcye","wvxyz"]

println(inventory_management_diff(input))
end

main()


Julia's a weird language...it's so close to python that I forget it does weird things like "Arrays are 1-indexed", and it spells None/null as nothing

I agree that Julia is weird, but I actually love it because it adds some functional stuff that I miss in python like the pipe operator!

We need better cover images 😂

I was kind of hoping that the magical resizing was an automated resizing thing that happens and not one of you guys fixing it for me. I can actually put a background on it and scale it. What are the optimal dimensions? Is the white text on black ok or should I come up with something more fancy? I have very little aesthetic skill, so I’d appreciate any suggestions you or anyone else has.

Ryan, do you want me to design something? The only hard part is that I probably won't be up at midnight most nights to add specifics. Do you have any design software? If so, I can transfer you a file! We could also use Canva.

That would be amazing! I have a Figma account, but that’s it. I’ve never heard of Canva and pretty sure I don’t have any design software, but if you tell me the best way to handle it, I’ll happily do whatever you suggest. I’m happy to learn.

Cool -- sending you a DM via /connect!

This one was difficult for me, but I eventually got it! I'm also not happy about that double for loop in part 2, but I think I sped it up by removing the elements as I compared them?  