re: AoC Day 2: Inventory Management System VIEW POST

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re: Python solutions! part 1 from collections import Counter with open('input.txt', 'r') as f: twice = 0 thrice = 0 for line in f: ...
 

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[0] == pair[1], letter_pairs))
            if len(duplicates) == len(box_id_1) - 1:
                return ''.join(pair[0] for pair in duplicates)


if __name__ == '__main__':
    with open('2-input.txt') as box_file:
        all_box_ids = box_file.read().splitlines()
    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!

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