--- Day 4: Secure Container ---

Python, 1042/3178

Well after looking at this forever because of how many times I got part 2 wrong, I noticed two things

a) If you simply sort digit-wise and compare and they're not the same, that eliminates it

b) If it wasn't eliminated in a), that means that you just need to count the occurrences of the digits to see if it's still legit

Although this isn't my actual code, it's just the super short version with a comment for part 1/2, here we are:

inp = open("inputs/day04.txt")
low, high = [int(num) for num in inp.read().split("-")]

ans = 0
for i in range(low, high):
    password = [int(j) for j in str(i)]
    if password != sorted(password):
        continue

    for digit in password:
        # password.count(digit) == 2 for part 2
        if password.count(digit) >= 2:
            ans += 1
            break

print (ans)

It's super clean and really easy to see what exactly is going on, though it took me a while to figure it out, I'm happy with it. Very clever puzzle.