Advent of Code 2025 Day 3

— Day 3: Lobby —

Fix the escalator with some battery power.

There are many batteries in front of you. Find the maximum joltage possible from each bank; what is the total output joltage?

import re

with open('data-2025-03.txt', 'r') as f:
    inp = f.read().splitlines()

Can I just loop backwards from 99 and see what I can find? A little regex?

Alternatively: Find the highest number in the sequence that’s not at the end, and then find the highest number after that. Which is really just two loops instead of one.

summer = 0

for b in inp:
    for i in range(99, 9, -1): 
        # print(i)
        j = str(i)[0]
        k = str(i)[1]
        patt = '[' + j + '].*[' + k + ']'
        # print(patt)
        if re.search(patt, b) is not None: 
            summer += int(j+k)
            break
    
# 17412 

— Part Two —

The battery’s joltage is actually 12 digits, not two.

What is the new total output joltage?

This feels like the previous alternative is actually the right suggestion.

Find the highest number in the first 89 characters (lines are 100 characters long).

If the 89th character is a 9, and there are no 9s before that, then whatever comes after that is the largest number possible.

Need to find the position of the maximum, and then the next maximum from that position until L-10, and then the next from L-9, etcetera.

max() of a string works, that’s nice

summer = 0

for b in inp: 
    # current list of digits, make this an integer later
    curr = [] 
    # search from this position until the end, update each time
    pos = 0 
    # work backwards to find the digits
    # we can search up until this position
    for z in range(11, -1, -1): 
        # find the next number
        if z == 0: 
            nxt = max(b[pos:])
        else: 
            nxt = max(b[pos:-z])
        # find the position of that number
        pn = b[pos:].find(nxt) 
        # update our search parameters
        pos = pos + pn + 1
        # checks
        # print(nxt, pos, pn)
        # append
        curr += [nxt]
    # add to our count
    summer += int(''.join(curr))
        
summer

# 172681562473501 

172681562473501