Advent of Code 2023 Day 12

— Day 12: Hot Springs —

For each row, count all of the different arrangements of operational and broken springs that meet the given criteria. What is the sum of those counts?

from itertools import combinations, permutations, product

with open('data-2023-12.txt', 'r') as f:
    inp = f.read().splitlines()

This is the first time I don’t immediately have an approach and am worried about not getting finished.

Update: create all the possible options

# let us start by getting things organised
spr = [[x.split()[0], list(map(int, x.split()[1].split(',')))] for x in inp]

dothash = ['.', '#']

# somewhat brute force to create all options
def create_options(w, aim):
    # how many questions marks, means how many replacements
    q = w.count('?')
    # print(q)
    # print(w)
    # this provides binary "combinations" 
    # (not the combinations or permutations function)
    combo = list(product(dothash, repeat=q))
    # this is what we are collecting, possible arrangements
    arr = []
    # loop through our different dothash combos
    for c in combo: 
        # print(c)
        # current possibility
        this = []
        # position of replacement
        it = 0
        # original string of characters
        for d in w: 
            # print(d)
            # make a replacement, iterate the replacer
            if d == '?':
                # print(c[it])
                this.append(c[it])
                it += 1
            # take the original char
            else: 
                this.append(d)
        # once used up the combos, assign back
        # print(this)
        if this.count('#') == aim: 
            arr.append(''.join(this))
    return arr
    
  
# check if our things match the thing we want
def verify_arr(opt, v):
    num = 0
    # loop through opt and count the hash to see if they match v
    for o in opt: 
        # collect the current count
        curr = []
        # set the count to zero
        cnt = 0
        # characters in current option
        for c in o: 
            if c == '.':
                if cnt > 0: 
                    curr.append(cnt)
                cnt = 0
            else: 
                cnt += 1
        if cnt > 0: 
            curr.append(cnt)
        # add to our tally for correct sequences
        if curr == v: 
            num += 1
    return num
  
  
summer = 0

for s in spr: 
    aim = sum(s[1])
    opt = create_options(s[0], aim)
    cnt = verify_arr(opt, s[1])
    summer += cnt
    
summer

# 7916

7916

— Part Two —

Actually, each record is folded up 5 times.

Unfold your condition records; what is the new sum of possible arrangement counts?

spr2 = []

for s in spr:
    spr2.append([s[0]*5, s[1]*5])

dothash = ['.', '#']

# somewhat brute force to create all options
def create_options(w, aim):
    # how many questions marks, means how many replacements
    q = w.count('?')
    # print(q)
    # print(w)
    # this provides binary "combinations" 
    # (not the combinations or permutations function)
    combo = list(product(dothash, repeat=q))
    # this is what we are collecting, possible arrangements
    arr = []
    # loop through our different dothash combos
    for c in combo: 
        # print(c)
        # if we don't have the right number don't entertain the possibility
        if ''.join(c).count('#') + w.count('#') != aim: 
            continue        
        # current possibility
        this = []
        # position of replacement
        it = 0
        # original string of characters
        for d in w: 
            # print(d)
            # make a replacement, iterate the replacer
            if d == '?':
                # print(c[it])
                this.append(c[it])
                it += 1
            # take the original char
            else: 
                this.append(d)
        # once used up the combos, assign back
        # print(this)
        if this.count('#') == aim: 
            arr.append(''.join(this))
    return arr
    
  
# check if our things match the thing we want
def verify_arr(opt, v):
    num = 0
    # loop through opt and count the hash to see if they match v
    for o in opt: 
        # collect the current count
        curr = []
        # set the count to zero
        cnt = 0
        # characters in current option
        for c in o: 
            if c == '.':
                if cnt > 0: 
                    curr.append(cnt)
                cnt = 0
            else: 
                cnt += 1
        if cnt > 0: 
            curr.append(cnt)
        # add to our tally for correct sequences
        if curr == v: 
            num += 1
    return num
  
  
summer = 0

for s in spr: 
    aim = sum(s[1])
    opt = create_options(s[0], aim)
    cnt = verify_arr(opt, s[1])
    summer += cnt
    
summer

7916