AoC 2018 Day 2: Counting letters

Source: Inventory Management System

Part 1: Given a list of strings, count how many contain exactly two of a letter (a) and how many contain exactly three of a letter (b). Calculate a*b.

Bit weirdly stated. My first thought would be to use regular expressions¹, but that doesn’t actually fit supper well with this case, because you need exactly two matches. I’m not actually sure how you’d even write something like that, since you’d have to match against ’not this’ before you found the first match. Meh.

In both Racket and Python, I’m going to create a hash that maps each unique letter in a word to how many times that letter occurs:

(define (count-letters word)
  (for/fold ([count (hash)])
            ([letter (in-string word)])
    (hash-update count
                 letter
                 add1
                 0)))

def count_letters(word):
    counts = collections.defaultdict(lambda : 0)
    for letter in word.strip():
        counts[letter] += 1
    return counts

If you have that, you can count the letters and look in those hash values (we don’t actually care about keys here) for 2s and 3s:

(define-values (2s 3s)
  (for/fold ([2s 0] [3s 0])
            ([line (in-lines)])
    (define counts (count-letters (string-trim line)))
    (values (+ 2s (if (member 2 (hash-values counts)) 1 0))
            (+ 3s (if (member 3 (hash-values counts)) 1 0)))))

(* 2s 3s)

count_2s = 0
count_3s = 0

for word in fileinput.input():
    counts = count_letters(word)
    if 2 in counts.values(): count_2s += 1
    if 3 in counts.values(): count_3s += 1

print(count_2s * count_3s)

In this case, the code is pretty much the same in both the Racket and the Python case. The Racket case is half function in that we’re using for/fold to update an immutable hash, but because this is Racket (and we have the whole for family), it looks much the same as Python’s more iterative version.

If you wanted a slightly more ’traditional’ Scheme version, you could do something like this:

(let loop ([2s 0] [3s 0] [lines (port->lines)])
  (cond
    [(empty? lines) (* 2s 3s)]
    [else
     (define counts (count-letters (string-trim (first lines))))
     (loop (+ 2s (if (member 2 (hash-values counts)) 1 0))
           (+ 3s (if (member 3 (hash-values counts)) 1 0))
           (rest lines))]))

To each their own.

They all return the same value though:

$ cat input.txt | racket counting-letters.rkt

4980

$ cat input.txt | racket counting-letters-functional.rkt

4980

$ cat input.txt | python3 counting-letters.py

4980

Part 2: Given the same list of words, find the two words with exactly one letter different. Print all letters that were the same.

Not really much code you can re-use. For this, you really want a function that can count differences and another to show what is the same (you only actually need the second, but 🤷):

(define (differences word1 word2)
  (for/sum ([letter1 (in-string word1)]
            [letter2 (in-string word2)]
            #:unless (char=? letter1 letter2))
    1))

(define (shared-letters word1 word2)
  (list->string
   (for/list ([letter1 (in-string word1)]
              [letter2 (in-string word2)]
              #:when (char=? letter1 letter2))
     letter1)))

The main interesting bit here is the use of the #:when / #:unless clauses in the loops to only evaluate the body if / not if a condition is matched. An if in the body of the for would work much the same way in this case.

With that, we can use for*/first to find the first word with exactly one difference (since the puzzle spec says there will only be one) and then apply the shared_letters function to it:

(define words (port->lines))

(apply shared-letters
       (for*/first ([word1 (in-list words)]
                    [word2 (in-list words)]
                    #:when (= 1 (differences word1 word2)))
         (list word1 word2)))

And once again, I’m going to basically translate directly into Python:

def differences(word1, word2):
    count = 0
    for letter1, letter2 in zip(word1, word2):
        if letter1 != letter2:
            count += 1
    return count

def shared_letters(word1, word2):
    output = []
    for letter1, letter2 in zip(word1, word2):
        if letter1 == letter2:
            output.append(letter1)
    return ''.join(output)

words = list(fileinput.input())

for word1 in words:
    for word2 in words:
        if differences(word1, word2) == 1:
            print(shared_letters(word1, word2))
            exit()

Bit ugly to have that exit in the loop there. Sometimes I miss being able to jump out of multiple loops without leaving a function entirely (with return) in Python².

$ cat input.txt | racket near-misses.rkt

"qysdtrkloagnfozuwujmhrbvx"

$ cat input.txt | python3 near-misses.py

qysdtrkloagnfozuwujmhrbvx

Slight differences in output since I’m relying on Racket’s behavior of printing out top level values for each expression rather than explicitly printing things. So it goes.

When you have a hammer… ↩︎
Who would have thought. ↩︎