Advent of Code 2024 Day 25

— Day 25: Code Chronicle —

There are some locks and some keys.

Analyze your lock and key schematics. How many unique lock/key pairs fit together without overlapping in any column?

with open('data-2024-25.txt', 'r') as f:
    inp = f.read().splitlines()

Get a list of the keys and of the locks, and organise them into lists of heights.

keys = []
locks = []
prog = 0

for i in inp: 
    if prog == 0: 
        # start of the sequence
        prog += 1
        if i == '.....':
            # key
            key,lock = True,False
            this = [i]
            # print('set key')
        else:
            # lock
            lock,key = True,False
            this = [i]
            # print('set lock')
    elif i == '': 
        prog = 0
        if key: 
            keys.append(this)
            # print('append key')
        elif lock:
            locks.append(this)
            # print('append lock')
    else: 
        prog += 1
        this.append(i)
        # print('progress')
        
# added a blank line at the end of the input to get all locks and keys

Create a function that returns the columns rather than the rows

def make_things(x):
    a = [0,0,0,0,0]
    for b in x[1:]: 
        # print(b)
        for c,d in enumerate(b):
            if d == '#':
                a[c] += 1
    return a
            
def make_keys(x):
    return make_things(list(reversed(x)))

def make_locks(x):
    return make_things(x)


keysN = [make_keys(q) for q in keys]
locksN = [make_locks(q) for q in locks]

OK, good.

counter = 0

for k in keysN:
    for l in locksN:
        add = True
        for v,w in zip(k,l):
            if v+w > 5:
                add = False
        if add: 
            counter += 1


counter
# 3127

3127