 I solved day 4 much faster than day 3, probably because it moved away from spacial reasoning and just into input validation. I’m given a range of 6-digit numbers, and asked to pick ones that meet certain criteria.

## Challenge

The puzzle can be found here. I’m given a starting and ending point, both 6-digit numbers. For part 1, I need to count the number of numbers in that range that have only increasing (or staying the same) digits left to right, and contain two matching digits at some point in the password. For part 2, I add the complication that a grouping of 3 repeating digits doesn’t count towards the pair requirement.

## Solution

### Part 1

There are so many ways to approach this. I considered recurrsively building the password rather than looping over a range because I could avoid even checking many of the number. For example, for 5678, I’d only have to chec 567888, 567889, 567899, and not the other 97 possible two digit combinations.

But then I decided to try first just looping over the entire range, and while it isn’t instant, it still took less than a second.

For part one my code is:

#!/usr/bin/env python3

import re
import sys

count = 0
for passwd_int in range(int(sys.argv), int(sys.argv) + 1):
passwd = f"{passwd_int}"
if re.search(r"(\d)\1", passwd) and all(
int(i) <= int(j) for i, j in zip(passwd, passwd[1:])
):
count += 1

print(f"Part 1: {count}")


This loops over the numbers in the given range, and checks first if there’s a repeating digit using re.search, and then if for each character in the password, the next one is greater than or equal. zip is smart enough to take the six characters in passwd and the five characters in passwd[1:] and return five tuples.

This runs in less than a second, and returns the right answer:

$time ./day4.py 231832 767346 Part 1: 1330 real 0m0.848s user 0m0.840s sys 0m0.008s  ### Part 2 In part two, when checking for consecutive characters, there must be a group of only two for it to be valid. This is tricky because while 123444 is no longer valid, 111122 still is, because while the four 1s don’t count, the pair of 2s do. I decided to add a check after incrementing count to look for this. I’m going to do a findall on the consecutive characters and then look for ones that are two in length. This gets a bit tricky with the \1 capture group reference. I started with something like this: >>> re.findall(r'(\d)\1+', '111122') ['1', '2']  It’s finding what I want it to find, but only returning what’s in (). I tried changing it to: >>> re.findall(r'((\d)\1+)', '111122')  But that crashes, because the \1 is referencing a capture group that isn’t yet closed. The trick then is to use \2, since the (\d) is now the second capture group. This worked: >>> re.findall(r'((\d)\2+)', '111122') [('1111', '1'), ('22', '2')]  I can use a list comprehension to get just the part I care about: >>> [x for x in re.findall(r'((\d)\2+)', '111122')] ['1111', '22']  Now I can change that into a check for length: >>> [len(x) == 2 for x in re.findall(r'((\d)\2+)', '111122')] [False, True]  I’ll add this to my code, and when I run this, it solves both in less than a second: $ time ./day4.py 231832 767346
Part 1: 1330
Part 2: 876

real    0m0.644s
user    0m0.644s
sys     0m0.000s


## Final Code

#!/usr/bin/env python3

import re
import sys

count = 0
count2 = 0
for passwd_int in range(int(sys.argv), int(sys.argv) + 1):
passwd = f"{passwd_int}"
if re.search(r"(\d)\1", passwd) and all(
int(i) <= int(j) for i, j in zip(passwd, passwd[1:])
):
count += 1
if any([len(x) == 2 for x in re.findall(r"((\d)\2+)", passwd)]):
count2 += 2

print(f"Part 1: {count}")
print(f"Part 2: {count2}")