Advent of Code 2025 Day 5

— Day 5: Cafeteria —

Some ingredients are fresh and some are spoiled in the elf cafeteria.

Process the database file from the new inventory management system. How many of the available ingredient IDs are fresh?

with open('data-2025-05.txt', 'r') as f:
    inp = f.read().splitlines()

ranges = inp[:185]

ranges = [[int(r.split('-')[0]), int(r.split('-')[1])] for r in ranges]

ing = inp[186:]

ing = [int(i) for i in ing]

Loops

count_me = 0

for i in ing: 
    for r in ranges: 
        if r[0] <= i <= r[1]: 
            count_me += 1
            break 

count_me

# 888

888

— Part Two —

Look at the range of IDs to see how many ingredients could possible be fresh.

Process the database file again. How many ingredient IDs are considered to be fresh according to the fresh ingredient ID ranges?

Process sequentially after sorting the list. This should give us overlapping ranges. When there are no more overlapping ranges we can use minimum and maximum + 1 to get number of fresh ingredients.

# 
# # not overlap, take this one
# t -- t 
#        n -- n
# 
# # no special cases, just take the max of second element
# t -- t 
# n ---- n 
# 
# # 
# t -- t 
#   n -- n
# 
# 
# # 
# t -- t
#   n-n
# 
# # 
# t -- t
#   n--n 
#   

# check if they are increasing
for r in ranges: 
    if r[0] > r[1]: 
        print('bad')
        
# sort for ease of use        
ranges.sort(key=lambda x: (x[0], x[1]))

count_me = 0
this = ranges[0]

for s in range(len(ranges) - 1): 
    next = ranges[s+1]
    
    # see if they overlap
    if this[1] < next[0]: 
        # they do not overlap, this range can be added
        count_me += this[1] - this[0] + 1
        this = next[:]
    else: 
        # they do overlap, we need some memory of ranges
        this[1] = max(this[1], next[1])
        
count_me += this[1] - this[0] + 1

count_me

# 904625236266855 too high
# 344378119285354

344378119285354