Advent of Code 2023: Day #1

Advent of code is a 25 day challenge occurring from December 1^st to Christmas day on the 25^th. Each day a new challenge is released consisting of two parts; the first part being to familiarize yourself with the problem and the second part being a more difficult twist. The first part involves finding the first and last digit in a string and taking them to be the tens and ones place value of a two digit number. They can be the same number if there is only one digit in the string. Adding all of the resulting numbers for each string in the input gives the answer. An example would be:

1aaljio3jl
hello1goodbye

Here the first line would be 13 and the second 11. Adding them up gives the solution of 24.

To solve this problem I took each input line and created a reversed version.

values = []

file1 = open('p1_input.txt','r')
#file1 = open('p1_input.example','r')
Lines = file1.readlines()

for line in Lines:
    liner = line[::-1]

Now all that needs to be done is iterate over each character in both the original string and reversed string until a digit is found, save that digit, and break from the loop.

    '''Find first digit'''
    for char in line:
        if char.isdigit():
            value += int(char)*10
            break

    '''Find second digit'''
    for char in liner:
        if char.isdigit():
            value += int(char)
            break

Adding all the values up using the complete python script gives a result of 55488 using my given input for part one. Here is the full code:

value = 0
file1 = open('p1_input.txt','r')
#file1 = open('p1_input.example','r')
Lines = file1.readlines()

for line in Lines:
    liner = line[::-1]
    '''Find first digit'''
    for char in line:
        if char.isdigit():
            value += int(char)*10
            break

    '''Find second digit'''
    for char in liner:
        if char.isdigit():
            value += int(char)
            break

print(value)

Part 2

The twist for part two is the numbers spelt out are also valid digits. For example:

three1hello5

would give a value of 35 instead of 15 as the first digit is the word ‘three’. For this second part I used the regular expression python module, regex, to search through each string to find all valid digits. This module does need to be installed via anaconda or pip as it’s not provided with python by default.

import regex as re

p = re.compile(r'one|two|three|four|five|six|seven|eight|nine|\d',re.IGNORECASE)

This regular expression will recognize the digits written as words as well as \d being a representation to [0-9]. Putting in the vertical slashes between each word acts as or allowing the regular expression to match any of them. Here instead of reversing the string we can simply search for all matches in the string as:

file1 = open('p1_input.txt','r')
#file1 = open('p2_input.example','r')
Lines = file1.readlines()

for line in Lines:
    tmp = re.findall(p,line,overlapped=True)

And then index into the first and last element of the array and multiply by the first instance by 10 as:

value += to_digit(tmp[0]) * 10
value += to_digit(tmp[-1])

Here I have also defined a function which will convert the string values correctly into their digit form. The function uses a predefined dictionary for conversion if it is represented as a word or the int cast if it is a character.

words = {"one":1,"two":2,"three":3,"four":4,"five":5,"six":6,"seven":7,"eight":8,"nine":9}

def to_digit(string):
    if string in words:
        return words[string]
    else:
        return int(string)

Putting it all together and running it on my given input gives 55614. Below is the code I used for part 2.

import regex as re

p = re.compile(r'one|two|three|four|five|six|seven|eight|nine|\d',re.IGNORECASE)

value = 0
words = {"one":1,"two":2,"three":3,"four":4,"five":5,"six":6,"seven":7,"eight":8,"nine":9}

file1 = open('p1_input.txt','r')
#file1 = open('p2_input.example','r')
Lines = file1.readlines()

def to_digit(string):
    if string in words:
        return words[string]
    else:
        return int(string)

for line in Lines:
    tmp = re.findall(p,line,overlapped=True)
    value += to_digit(tmp[0]) * 10
    value += to_digit(tmp[-1])

print(value)