---
title: Hash[], minmax, polymer expansion - Advent of Code 2021 - Day 14 with Ruby
slug: hash-minmax-polymer-expansion-advent-of-code-2021-day-14-with-ruby
published_at: 2021-12-19 14:00:32 +0000
updated_at: 2026-03-04 20:13:43 +0000
summary: 
description: Another puzzle that requires some data structure changes to make it fast enough to get an answer.  00:20 Part 1 11:20 Hash[] 12:47 minmax 14:14 Part 2
tags: [cjav_dev, web development tutorials, web development for beginners, vim, ruby, rails, Hash[], minmax, advent of code, advent of code 2021]
views: 320
author: CJ Avilla
url: https://www.cjav.dev/videos/hash-minmax-polymer-expansion-advent-of-code-2021-day-14-with-ruby
youtube_url: https://www.youtube.com/watch?v=sXJieqAoL8w
youtube_id: sXJieqAoL8w
embed_url: https://www.youtube.com/embed/sXJieqAoL8w
thumbnail_url: https://i.ytimg.com/vi/sXJieqAoL8w/hqdefault.jpg
type: video
---

# Hash[], minmax, polymer expansion - Advent of Code 2021 - Day 14 with Ruby

*Published: December 19, 2021*
*Views: 320*

## Watch

[Watch on YouTube](https://www.youtube.com/watch?v=sXJieqAoL8w)

[![Hash[], minmax, polymer expansion - Advent of Code 2021 - Day 14 with Ruby](https://i.ytimg.com/vi/sXJieqAoL8w/hqdefault.jpg)](https://www.youtube.com/watch?v=sXJieqAoL8w)

## Description

Another puzzle that requires some data structure changes to make it fast enough to get an answer.

00:20 Part 1
11:20 Hash[]
12:47 minmax
14:14 Part 2

## Transcript

hey what's up welcome back this is day 14 of the advent of code this exercise is called polymerization and the idea is that you get a template that is a string template that has some letters in it and then you also get sort of a mapping between pairs of characters and another character that when when you encounter that pair inside of the template you sort of like split the pair apart and insert that letter in between the two and you work your way through the entire template until you have updated all of the pairs and so at each step you sort of replace pairs of characters with the triple of characters with that new letter inserted so here where we used to have nncb nn maps to c so here we have n n and it maps to the letter c so the resulting the resulting pattern or the resulting template after step one is n c n so we're sort of like sticking in the c in between this grows super quickly so after step five we are going to have 97 characters this just shows us steps one through four and what the problem is asking us to do is figure out after we have gone some number of steps we're going to find some key that we're going to enter that's going to be our solution and our answer and the key is created by finding the most common letter and subtracting its occurrences from the least common letter so let's get into it so we're going to create a new thing called day 14 and we'll call it polymer.rb and we'll also need to create a spec day 14 spec.rb and let's jump into it so we're going to require relative day 14 polymer dot rb we're going to describe polymer and it should um replace a known pair with its like i don't know mapped letter we'll create polymer as polymer dot new and pass in the template so let's use the template actually let's make our own template so that we can control it alright so we will make our template abc i guess we'll also need to pass in okay so we're going to pass in the template but we also need to pass in the mapping so our input looks like this it's going to be the template and then a new line and then all of the mappings so our mapping is going to be a dictionary where we'll pass in pairs of letters and what those map to so we'll have a b mapped to a and bc also mapped to a and what do we get after that so i guess we can say polymer.step and we'll expect that polymer dot template to equal something what is it going to be after that first step well a b is going to be replaced by a a b and so one way we can do this is just write a b c and then what letter is going to be inserted between a and b well a is going to be inserted because that's what we mapped to and then what letter is going to be inserted between b and c a so this is what we should end up with after we make a step with the polymer so let's go solve this so class polymer is going to start off i guess we might want like a method to um parse i don't know should we have the parsing happen as part of the driver unclear so let's take in the template and let's take in the mapping and we'll just store those template is template and mapping is mapping and i'm not sure if we want to store the template as an array or what kind of how we're going to interact with it yet we also need some method step and let's run our test again undefined method template for polymer so let's just make that a reader template and as we step we'll just keep updating the underlying template object okay so now in order to take a step what we need to do is i guess what might make sense is iterating through or maybe creating yeah so if we iterate over all of the pairs so if we say template dot chars dot each cons two that will give us all of the consecutive pairs together dot i guess like do pair and we could even deconstruct this into like a and b and i guess what we want to do is say or maybe let's keep it as pair i don't know pair and we need to know what character we're going to insert so character is going to equal mapping at pair because the mapping is going to be yeah pairs of letters to a single character that's right and then what we want to ultimately return is we want to kind of like collect up the results maybe into an array results and we can just join those later so maybe if we say something like results push i guess we want to add on let's see a and b is the pair so we want to add in a then we want to add in b or char and we want to add in b and i'm not sure if that's going to work let's just see okay all right so then we want to update at template equals results.join day 14 spec close close so we were expecting a a b a c and we got a b b c so from abc we inserted something but it's not what we expected so let's let's debug this so we'll open up require debugger and or i guess by bug i don't know there's a new there's a new debugger that's getting popular again that i feel like maybe we should try out or use in one of these we'll see okay so our pair a is a and b is oops oh yeah puts b okay so then char is nil why is char nil at mapping okay so mapping of pair oh pair is going to be okay so we need to join that para.join so this should be pair.join and then we can run it okay there we go close all right so now we have a duplicate b in there somewhere so why do we have the duplicate b i don't know uh i guess why do we have a duplicate b a b b c so i think we might only want to add a if we're uh if we're at the very beginning or something so maybe results should be something like template dot chars zero and then we just don't even put a in there at all i don't know uh template okay yeah there we go so the idea there is that every time we're in every time we're looking at a pair we will have already seen like sort of the head or like the beginning of the list so we don't actually need to add that twice so we'll remove our debugger here and we can clean this up and just say results plus equals the array of char and b and now we should have cool so we're able to sort of just build the or we're able to take one step okay what is the next step here all right so after this we want to count up the occurrences after step 10 okay so we want to step basically 10 times so let's add some driver code here so polymer oh we also need to parse out the input and stuff so we're gonna read lines and uh so actually yeah let's do this so let's say file.read argv dot first and then split on two new lines and that should give us a split like right here where there is a double new line and then uh the first element that we're going to get back from that split is going to be the template and the second element is going to be the rest of all this business so we can say template and then lines i guess or something and then we can say lines dot map so we want to split the lines on that arrow i guess yeah so lines dot map line line dot split on this arrow and then that will give us the pairs of sort of the left-hand side and the right-hand side and that might be close so let's just print those out and let's add some example input here so we're going to just grab this example input and then we'll run this ruby day 14. and when we run it we see undefined method map for string why oh that's because we need to split on newline okay so this is what we get back pairs of okay so we have these all these pairs and then i think how do we get that into yeah that should just be i guess okay so there's there's this let's see if this works okay so so the square bracket operator on the class itself is kind of another another hack or another trick um that is kind of interesting so hash the hash class has like a square bracket operator so we're not actually calling hash.new here we're calling hash square bracket and passing in uh this array that has arrays that are all tuples and the first value is going to become the key and the second value is going to become the value that's like there is a like if you do hash.2a or sort or something it spits out like an array of arrays like this array of tuples and then this is kind of the way to get it back into the other the other format so this is our mapping and we want to pass in both the template and the mapping and then we want to say 10 times polymer dot step and then we want to say uh i guess we need to okay so we need to count up or we need to figure out right yeah okay so we need to figure out what the most common or the value for the most common letter and the value for the least common letter so let's say our score here is going to be um i guess okay so there's a there's a there's a couple like really cool things here that i was hoping to use so one is called min max so i think what we can do is we can say at template.chars.tally and that should we we looked at tally in the last episode it's really cool and that should give us i believe that should give us if we have something like a is a b c a b a b a b a b c c c c c c c and we say a dot chars dot tally then we get back the counts of each of those and what we want to do then is say values and then dot min max and that should give us back the pair of the minimum and the maximum and then we can inject uh oh gosh this is fun uh inject minus dot absolute value also this looks so funny inject minus is like a smiley face so i love that and that is sort of like the answer uh this is i don't know it's kind of fun tally values min max inject absolute value i don't know how clear it is but it's it's definitely fun so let's let's try that so we're going to tally then we're going to say values then we're going to say min max then we're going to inject minus to give us back a score and that is what we actually want to print out so puts polymer.score and let's just run it against the example input again 1588 and the answer here was 1588 so let's get our puzzle input copy it come back over here and say input and drop it in and then we can run this against our input and we get back three four zero eight three four zero eight all right so that is the end of part one for day fourteen of the advent of code pretty simple stuff so far now the next part gets a little gnarly so part 2 you want to do a total of 40 steps so if you imagine like it's growing this quickly uh if you just turn your head a little bit that's the kind of business you want is one that just kind of goes like a hockey stick up into the right but the result is going to be that if we try to run this 40 steps i don't think it's going to work because it was even slow for 10 steps so let's just run it and it's going to keep going and it's not going to finish because it gets way too big like the the using the approach of uh can like maintaining the string the string value itself is going to get too big so it just yeah it's uh not not a workable approach so if we look at sort of this example down below we see that the most common element is b and it occurs this many times so what is that uh so 2 trillion times or something and then h occurs i don't know 3 billion times and subtracting those ends up with this massive number and so if you imagine the string that is like 2 trillion characters long it's it's gonna just like fill up memory in fact i'm going to kill this so okay what we need to do is find again an alternative representation for storing these values and we did this before with with the ages of the fish in the school i can't remember what day it was but we went from using an array where we kept track of the ages to a dictionary and so if we think about ways that we might be able to use a dictionary to solve this particular problem then you might come again upon a dictionary as a solution now let's talk through it so how how do we want to actually store the keys well if every time we're kind of considering these pairs and what those pairs are mapping to then for one single iteration what we're kind of doing is we're looking at each pair and we're splitting it apart and when it splits apart it actually becomes two new pairs one pair where the first element is paired with the center replacement and the other pair where the center replacement is paired with the second element right and so what i think we want to do is create a hash where the keys are each of the possible pairs and the values are the number of times that those pairs appear in the string and what we can do is then we can every time that we want to take a step we can just look through all of the keys that have any values and then that many times we'll sort of split the key so we will we'll need to like subtract the ones that are being removed from the string we'll need to add ones that are being added to the string so it might be a little tricky but we'll figure it out so let's change our representation polymer or day 14 spec i think this is we're not actually using the test very much but and we don't actually want to care about the template anymore i guess in fact like the template is going to become a dictionary that will be something like a b points at a number and then actually so yeah let's let's let's keep this around for a second because this is going to be this string value will be interesting so we want to go through that string value and say that like the result should be something like a a points at 1 a b points at one b a points at one and a c points at one and then we're also probably going to have a b points at zero b or no a b is in there okay so or no bc points at zero so bc was one of the original pairs but there is no bc anymore in this list so that should be a zero so that's kind of like the new representation that we want to have and i missed a comma somewhere yeah okay this is what it is now but we're going to change how that works so for our template instead of storing it as a string what we want to do is we want to store it as a dictionary so let's use hash dot new 0 because we're going to use a counter of those pairs and then we want to say template.chars.each cons 2. so each consecutive pair of 2 again we're kind of using the same situation here do pair and we're going to say at template at pair plus equals 1. that should give us the count i think well that should give us uh let's actually not step well okay so let's uh confirm that after we first create our situation here we end up with a b points at one and then b c points at one that's like right after we initialize that's what we should have okay so those pairs did not again we have to join i thought i wonder if we can map to and join here both block and arg actual given oh and then dot each okay i wonder if we can just tally at template equals this thing dot tally does that work whoa sweet okay so that's totally what we want tally gosh that is such a cool thing i learned that from colin gilbert and roman from rubyconf tally is a fun that's a fun one okay so at this point we have our template we have our mapping and we need to update our step logic so now what we want to do is we want to go through each pair of each yeah each pair of letters inside of the keys for the template so template.keys.each do in fact yeah keys is template.keys and then we want to say keys that each do key and first we have to let's see subtract key value i don't know so inside of inside of template so at template at key oh you know what is going to be bad about doing this doing it this way is that we're not going to have a default we're not going to have a default a hash default which we do want so let's actually let's go back to what we had before um yeah this this is actually like uh superior here so we can say dot each do pair again okay i see i see all right so instead of chars we want to say keys so keys is template.keys the reason why i'm i'm pulling off template.keys here is that while we're iterating we're going to be modifying the template and i don't want the template to be changing underneath us while we're iterating over it i don't know if that makes any sense but hopefully it does okay so here what we're going to do is we're going to say keys.each do key now what we want to do is we want to look at the template at the key that's going to give us some value and that value is important because that's how many times we want to split the key so the key dot chars we also i guess we also want the mapping so char is at mapping at key and now what we want to do is we want to split the key so we're going to have a we're going to have b and that's going to be key dot chars and we want to combine a with char and b with char and increment the occurrences inside of template the number value number of times so what we want to do is we want to say at template at a char.join plus equals value and at char b dot join plus equals value and then we want to say template at key minus equals value because we're in the in the next step of the string we've sort of split apart that we've split apart that key so it should no longer be in there but as we're iterating through the keys we might encounter one where it's when it's being combined with something else it ends up adding values to this so those numbers are going to kind of like shift over time okay so what does this actually look like i don't know if we want to join any results or anything so let's just let's step and then see what we end up with no method chars for hash do we call chars somewhere oh down in score or we're not calling score anywhere are we okay so let's see oh this is this i don't think we want that anymore okay we do need to somehow figure out how to deal with that first character again but we'll come back and try to figure that out later okay so that's great so now we have now we have something that looks somewhat like it's working okay and i hope i hope this makes sense we're kind of like yeah we're incrementing and we're splitting apart and then we're putting it all back together and we end up with this this big template that should should have something for us okay so then the next thing was that we wanted to run it 40 times and what do you get if you take the last quantity of the most common element and subtract the quantity of the least common element so we're getting we're doing the same thing again where we need to get the quantities now the problem with this is that when you're looking at the template like abc right if this was the template the original template our new template is going to be this dictionary that has uh like a b points at some number and then bc points at some number actually this is going to be one and one right and so one thing that we could do is we could say all right show me all of the letters the individual letters as like a we're going to kind of just like look through every single key and if a appears in that key then give us the value for that so we could say a points at 1 b points at 2 and c points at 1 right because b is in this one and it's in this one and then we'll just be off by one for the first letter and for the last letter or maybe just the first letter uh if we take all of these and then we divide by two i think we might be in a good place so what we wanna do is we wanna iterate over the template keys again so we're gonna say something like at uh yeah at template dot keys that each do actually we want to iterate over both the key and the value so we'll say template dot each do key and value and we're gonna kind of keep track of a new dictionary of sort of uh frequencies and it's going to be hash.new of zero against because we're going to use this sort of counter style uh and in fact actually yeah like freak the frequency i think we're going to want frequency.valuesmin max inject again that whole thing okay so freak is a hash of values we're going to have a single letter pointing at its value so now we're going to actually we could kind of well yeah okay so here we want to split the key into a and b and that's going to be k.chars actually let's make this a little more clear okay i don't know if that is more clear or not okay so then we want to say freak at a plus equals value and freak at b plus equals value and then i guess yeah we want the min max min max is equal to this and then we're going to say the return value is going to be max divided by 2 minus the min divided by two because i think yeah i think if we divide by two we should be okay maybe we can yeah let's let's do let's do it that way let's see what happens uh all right so we have do we have an example we have an example so from from the example input i think we're supposed to get back this giant number so if we just run against example input we get a giant number let's see if it's the right one it is not the right one okay so uh let's make sure actually should we just do min minus or max minus min and then divide the whole thing by two i don't know let's see same number okay uh math gosh transitive properties i don't know something something property something something okay so in in the first example what we did was we only incremented it uh for gosh okay so what is going on with our logic here all right so we've got we're stepping we think that the step is increasing as expected why don't we step again polymer.step again and now we're going to have to we're gonna have to build out our map a little bit more because we don't actually have a a a let's make that also go to a just to like the more we can collapse on a the better uh so this is gonna end up with so yeah our our old one was a uh gosh it's so much it's so much trickier so we had yeah we had all eight except again we're gonna get aabac and this one is gonna become a a b a c a right so now we should have one two three four five aas and we should have one a b and we should have one ac and we should have yep one ba and i think okay so i think that's right but that's not what we're getting we're getting six aas because again i wonder if we need to do this situation here with the the very first time through i don't know if we want to actually increment that a thing so i equals 0 i plus equals actually yeah we can just do each with index with index and say this is i and then say like if i is greater than zero or something i don't know like uh that's a different number but it's not the same number and okay so does that work that does not work and we end up with the wrong template there's no aas how is that possible oh because we're not we're not ever adding it right okay okay so i think there's a bug here where we are we're reading out this value but then we're like updating the value directly as part of the step and so what i think we want to do is actually split out and create like a temporary template and then use that template um as like a two-step process where first we kind of like update the temporary template temp template is going to be new of xero and then we will update the temporary template temp template and then after we've gone through and updated the temporary template after we've gotten like all of our numbers based on the current values that are in the template then later we'll go in a second step and we'll update these values because what we're doing right now is we were overriding what we were reading and that was messing us up so keys let's see i bet there's some sort of merge thing we can do here but let's just do it the hard way because we don't know actually yeah let's iterate through temp template and we don't need to index we just want key and value and now we're going to say at template at key plus equals v and it's possible we're going to get negative values and that's okay for the temp template so let's run this again or let's let's run our test again okay now our test is passing now let's run it against the example input again okay that's a different number and that looks promising and oh it's so close two one eight eight one eight nine six nine three five two eight wow we're just off by one okay maybe this needs to be like dot ceiling or something uh or like yeah make this a float and make it the ceiling uh i don't know maybe that's gonna be that's cheating but we're close we're super close all right so now let's do 40 steps of our own input and see what we end up with and okay 372 okay so that was my puzzle answer so that was the answer all right so if we come back if we come back through this so the bug was that we were overriding while we were iterating and what's really cool is that old solution would have taken a super super long time to finish and this solution is lightning fast again because we're using hashes instead of arrays and or instead of strings in this case because we don't need to build up the entire string we can just kind of like keep counts of things and shift them around that is it for day 14 of the advent of code with ruby thanks so much for watching and we'll see in the next [Music]

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