# General Discussions

Discuss a wide range of topics! Questions about the Alteryx Platform should be directed to the appropriate Product discussion forum.
It's the most wonderful time of the year - Santalytics 2020 is here! This year, Santa's workshop needs the help of the Alteryx Community to help get back on track, so head over to the Group Hub for all the info to get started!

## Advent of Code - Base A Style! (Post 4 of 26)

Highlighted
14 - Magnetar

Please excuse my absence over the past few days. Fell ill with bronchitis. Still feverish, so this post may be a nonsensical ramble, but hopefully you'll respond with your coherent solutions!

Day 4 - secure container

1.4 seconds

Instructions for part 1 were good. Part 2 made no sense to me. Took lots of discussion with other colleagues before I could make sense of it.

Seemed like a straighforward problem. Generate all the numbers between two given numbers (thank you Generate Rows tool!), and then make a really ugly filter to determine that the numbers are either the same left to right or increase.

Now to determine the adjacent numbers, I used a fun little RegEx trick: REGEX_CountMatches(tostring([Num1]), "([0-9])\1")>=1 This looks for any repetition of the same number.

Part 2 was challenging for me because I had trouble understanding which numbers were ok to have more than just a pair, and which weren't.  Still not sure this was the best approach, but it did work in the end. I counted distinct numbers in each passcode that passed through part 1. and also counted how many times each number appeared in the passcode. To do this, parsed each number in each code out, and used some Summarizes, then joined back together.

An ugly formula determines if it meets the criteria for a passcode or not:

If [CountDistinct_Num1]=5 THEN "Y"
ELSEIF [CountDistinct_Num1]=4 AND [Count]>2 THEN "N"
ELSEIF [CountDistinct_Num1]=3 AND [Count]=4 THEN "N"
ELSEIF [CountDistinct_Num1]=2 AND [Count] In (3,5) THEN "N"
ELSEIF [CountDistinct_Num1]=1 THEN "N"
ELSE "Y" ENDIF

Ultimately, anything with an N doesn't meet the criteria. Here's a snapshot. Happy solving!

Cheers,

Esther

Highlighted
Alteryx Community Team

Hi @EstherB47!

Great use of REGEX_CountMatches()!

I don't use that one enough!

Other REGEX muscles got a workout though!

Highlighted
11 - Bolide

I didn't realize that numeric references were supported in POSIX regex!! Since lookarounds aren't supported, I guess I just assumed that numeric references weren't either! So, I went with the very ugly regex below to find repeating digits:

``````regex_match(
tostring([num]),
'\d*(0{2}|1{2}|2{2}|3{2}|4{2}|5{2}|6{2}|7{2}|8{2}|9{2})\d*'
)``````

For checking increasing digits, I tried several methods for splitting out the digits, including this very ridiculous combination of formulas ðŸ˜‚

I also tried:

• Using string functions to compare the characters (as @EstherB47 did)
• Splitting characters into columns (as @patrick_digan did)
• Splitting characters into rows (as @cgoodman3 did)

On my laptop, splitting the characters into rows or columns using a parsing tool was consistently faster than parsing them within the filter. I wonder if the efficiency gain is due to running the parsing operation all at once (rather than parsing and comparing each record within the filter)?

My Advent of Code solutions are on github here: https://github.com/kelly-gilbert/advent-of-code/tree/master/2019-alteryx