Yo, Happy New Year, peeps!

Last year, I was on that grind, tackling the 2023 Advent of Code bit by bit. So, I’m droppin’ my solutions and some quick notes here, ya dig?

I’m coding all this in Python, straight up.

I’m thinkin’ of droppin’ the whole thing on GitHub when I’m done, but no promises on when that’ll be, ya feel me?

Aight, let’s jump into Day 1’s solution, no time to waste.

Part A: The Lowdown

  • What’s the Deal?

    We got strings mixed with numbers and letters, a real word salad. For each line, we gotta snag the first and last digit and mash ’em together into a number.

    Peep these examples: #+END_SRC 1abc2 → 12 pqr3stu8vwx → 38 a1b2c3d4e5f → 15 treb7uchet → 77 #+END_SRC

    The challenge? Add up all these “calibration values” from the whole input file.

    In this example, we add up 12 + 38 + 15 + 77, and boom, we get 142.

  • My Solution, My Flow

    from fastcore.utils import L
from aocd import get_data, submit
import re, regex
from regex import findall

def get_calibration_value(x:str):
    digits = re.findall(r"\d", x)
    return int(digits[0]+digits[-1])

def solve1a(inp:str):
    lines = L(inp.splitlines())
    return lines.map(get_calibration_value).sum()

Part B: Keepin’ It Real

  • The Twist

    Part 2 throws a curveball. Now, some numbers are spelled out as words, sneaky, huh?

    Examples, yo: #+END_SRC two1nine → 29 eightwothree → 83 abcone2threexyz → 13 #+END_SRC

    So, we gotta treat both digits (“1”, “2”, etc.) AND words (“one”, “two”, etc.) as numbers. Find the first and last on each line, same as before.

  • My Solution, Keepin’ It 100

    d = {str(i):i for i in range(10)}
number_words = ["one", "two", "three", "four", "five", "six", "seven", "eight", "nine"]
d |= {o:i+1 for i,o in enumerate(number_words)}

pat = "|".join(d.keys())

def get_calib_with_letters(x):
    m = findall(pat, x, overlapped=True)
    return d.get(m[0])*10 + d.get(m[-1])

def solve1b(inp):
    lines = L(inp.splitlines())
    return lines.map(get_calib_with_letters).sum()

Thoughts & Notes, Straight from the Source

I tripped up a bit on Part B cuz I didn’t fully grasp how the standard library re works.

Like, check this:

x = "7pqrstsixteen"
print(re.findall(pat, x))

['7', 'six']

That works, no problem

But peep this:

x = "oneight"
print(re.findall(pat, x))

['one']

I wanted ['one', 'eight'], but it only grabbed ['one']. What gives?!

It’s cuz the “e” in “one” is overlapping with the “e” in “eight”. The standard re ain’t built for that.

To fix that, I used the regex library, which wraps the standard re. This library’s findall has an overlapped argument. Set it to True, and it handles overlapping matches like a champ.

Check it out, homie.

x = "oneight"
print(regex.findall(pat, x, overlapped=True))

['one', 'eight']

The regex library has a bunch of other dope features, too. I’m tryna master that, stay tuned.

Aight, that’s it for Advent of Code 2023 Day 1. Peace out!