day 2

2a.py

# Each line gives the password policy and then the password.
# The password policy indicates the lowest and highest number of times a given letter
#   must appear for the password to be valid.
# For example, 1-3 a means that the password must contain a at least 1 time and at most 3 times.
# input
with open('2.txt', 'r') as file:
    input = file.read()
# turn the input into a list; each entry is '1-3 a: abcde'
input_list = list(input.split('\n'))
# break it up into tuples
password_rules = []
for str in input_list:
    tmp_tuple = str.split(' ') # ('1-3', 'a:', 'abcde')
    range = tmp_tuple[0].split('-') # ('1', '3')
    letter = tmp_tuple[1][:-1] # 'a' (removes last character) 
   password_rules.append((int(range[0]), int(range[1]), letter, tmp_tuple[2])) # (1, 3, 'a', 'abcde')
# check each string against the rules
good_pw = [] # these pass the check
bad_pw = [] # these fail the check
for tuple in password_rules:
    min = tuple[0]
    max = tuple[1]
    letter = tuple[2]
    pw_temp = password = tuple[3]
    is_good = False
    let_count = 0
    while len(pw_temp) > 0: # while the string is nonempty
        pw_let = pw_temp[0] # first character
        if pw_let == letter: # if the pw_let is the letter you're checking for
            let_count += 1
        pw_temp = pw_temp[1:] # removes first character from string
    if let_count >= min and let_count <= max: # it satisfies the conditions
        good_pw.append(password) # add to good password list
    else: # it does not satisfy the conditions
        bad_pw.append(password) # add to bad password list
# the passwords are in good_pw and bad_pw lists
print(len(good_pw))

2b.py

# Each line gives the password policy and then the password.
# Each policy actually describes two positions in the password, where 1 means the first character,
#   2 means the second character, and so on.
# (Be careful; Toboggan Corporate Policies have no concept of "index zero"!)
# Exactly one of these positions must contain the given letter.
# Other occurrences of the letter are irrelevant for the purposes of policy enforcement.
# For example, 1-3 a means that the password must contain a at position 1 (0) xor at position 3 (0).
# input
with open('2.txt', 'r') as file:
    input = file.read()
# turn the input into a list; each entry is '1-3 a: abcde'
input_list = list(input.split('\n'))
# break it up into tuples
password_rules = []
for str in input_list:
    tmp_tuple = str.split(' ') # ('1-3', 'a:', 'abcde')
    range = tmp_tuple[0].split('-') # ('1', '3')
    letter = tmp_tuple[1][:-1] # 'a' (removes last character)
    password_rules.append((int(range[0]), int(range[1]), letter, tmp_tuple[2])) # (1, 3, 'a', 'abcde')
# check each string against the rules
good_pw = [] # these pass the check
bad_pw = [] # these fail the check
for tuple in password_rules:
    i = tuple[0]
    j = tuple[1]
    letter = tuple[2]
    password = tuple[3]
    letij = 0 # how many times the letter appears in position i or j
    if password[i-1] == letter:
        letij += 1
    if password[j-1] == letter:
        letij += 1
    if letij == 1: # if the letter appears in i xor j ((i and ¬j) or (¬i and j))
        good_pw.append(password) # it is a good password
    else: # the letter is in neither or both positions
        bad_pw.append(password) # it is a bad password
# the passwords are in good_pw and bad_pw lists
print(len(good_pw))

overall idea for this was to format the list into tuples so that i could easily refer to the variables (min/max number of letters, the letter i was checking for, the password itself) - for the first challenge i just took letters off the string one by one, counted it if it was the right letter, and then added the password to the good list if this count was in the right range, and for the second i.. never bothered looking up if i could use xor in python, so i just counted the number of times that letter appeared if i took just the i^th and j^th characters and added it to the good list if this was exactly 1.