Discussion on: Advent of Code 2019 Solution Megathread - Day 4: Secure Container

[Sarah]

It's not as good as I wanted it to be, am pretty sure it was possible to solve in a nicer way with regex. But had to do it quickly between meetings so 🤷‍♀️ Am satisfied though

matches = 0

for number in range(min, max + 1):
    is_increasing = True
    has_double = False

    previous = []

    for character in str(number):
        if len(previous) > 0:
            if character < previous[-1]:
                is_increasing = False
                break
            if previous.count(character) > 0:
                has_double = True

        previous.append(character)

    if is_increasing and has_double and 2 in Counter(previous).values():
        matches += 1


print(matches)