**You can become a serverless blackbelt**. Enrol to my 4-week online workshop **Production-Ready Serverless** and **gain hands-on experience** building something from scratch using serverless technologies. At the end of the workshop, you should have a broader view of the challenges you will face as your serverless architecture matures and expands. You should also have a firm grasp on when serverless is a good fit for your system as well as common pitfalls you need to avoid. Sign up now and get **15% discount** with the code yanprs15!

**ps. look out for all my other solutions for Advent of Code challenges here.**

## Day 22

See details of the challenge here.

Today’s input looks like this:

root@ebhq-gridcenter# df -h

Filesystem Size Used Avail Use%

/dev/grid/node-x0-y0 87T 71T 16T 81%

/dev/grid/node-x0-y1 93T 72T 21T 77%

/dev/grid/node-x0-y2 87T 67T 20T 77%

/dev/grid/node-x0-y3 89T 65T 24T 73%

/dev/grid/node-x0-y4 93T 67T 26T 72%

…

Of the information presented to us we only need the x & y position of the node, its size and amount of space used. Everything else can either be derived from these or are not needed for this challenge.

For part 1 of the challenge, if we sort the array of nodes by the amount of available space in descending order, then for each of the elements we only need to iterate through the sorted array for as long as the available space is >= the used space of the current node. Hence reducing the average time complexity from * O(n^{2}) *to

*.*

**O(n * log n)**

### Part 2

Now that you have a better understanding of the grid, it’s time to get to work.

Your goal is to gain access to the data which begins in the node with`y=0`

and the highest`x`

(that is, the node in the top-right corner).

For example, suppose you have the following grid:`Filesystem Size Used Avail Use% /dev/grid/node-x0-y0 10T 8T 2T 80% /dev/grid/node-x0-y1 11T 6T 5T 54% /dev/grid/node-x0-y2 32T 28T 4T 87% /dev/grid/node-x1-y0 9T 7T 2T 77% /dev/grid/node-x1-y1 8T 0T 8T 0% /dev/grid/node-x1-y2 11T 7T 4T 63% /dev/grid/node-x2-y0 10T 6T 4T 60% /dev/grid/node-x2-y1 9T 8T 1T 88% /dev/grid/node-x2-y2 9T 6T 3T 66%`

In this example, you have a storage grid`3`

nodes wide and`3`

nodes tall. The node you can access directly,`node-x0-y0`

, is almost full. The node containing the data you want to access,`node-x2-y0`

(because it has`y=0`

and the highest`x`

value), contains 6 terabytes of data – enough to fit on your node, if only you could make enough space to move it there.

Fortunately,`node-x1-y1`

looks like it has enough free space to enable you to move some of this data around. In fact, it seems like all of the nodes have enough space to hold any node’s data (except`node-x0-y2`

, which is much larger, very full, and not moving any time soon). So, initially, the grid’s capacities and connections look like this:`( 8T/10T) -- 7T/ 9T -- [ 6T/10T] | | | 6T/11T -- 0T/ 8T -- 8T/ 9T | | | 28T/32T -- 7T/11T -- 6T/ 9T`

The node you can access directly is in parentheses; the data you want starts in the node marked by square brackets.

In this example, most of the nodes are interchangable: they’re full enough that no other node’s data would fit, but small enough that their data could be moved around. Let’s draw these nodes as`.`

. The exceptions are the empty node, which we’ll draw as`_`

, and the very large, very full node, which we’ll draw as`#`

. Let’s also draw the goal data as`G`

. Then, it looks like this:`(.) . G . _ . # . .`

The goal is to move the data in the top right,`G`

, to the node in parentheses. To do this, we can issue some commands to the grid and rearrange the data:

Move data from`node-y0-x1`

to`node-y1-x1`

, leaving node`node-y0-x1`

empty:`(.) _ G . . . # . .`

Move the goal data from`node-y0-x2`

to`node-y0-x1`

:`(.) G _ . . . # . .`

At this point, we’re quite close. However, we have no deletion command, so we have to move some more data around. So, next, we move the data from`node-y1-x2`

to`node-y0-x2`

:`(.) G . . . _ # . .`

Move the data from`node-y1-x1`

to`node-y1-x2`

:`(.) G . . _ . # . .`

Move the data from`node-y1-x0`

to`node-y1-x1`

:`(.) G . _ . . # . .`

Next, we can free up space on our node by moving the data from`node-y0-x0`

to`node-y1-x0`

:`(_) G . . . . # . .`

Finally, we can access the goal data by moving the it from`node-y0-x1`

to`node-y0-x0`

:`(G) _ . . . . # . .`

So, after`7`

steps, we’ve accessed the data we want. Unfortunately, each of these moves takes time, and we need to be efficient:

What is the fewest number of steps required to move your goal data to`node-x0-y0`

?

If you visualize the grid then the solution becomes pretty clear and one that we can actually work out by hand.

Here, the target node is marked as * red*, the empty node

*(which is initially at position (7, 17)), and the very large, very full nodes are marked as*

**green***. We want to move the empty node all the way to (35, 0) so we can start moving the target node at (36, 0) horizontally towards (0, 0).*

**black**We can again use the Level Order Tree Traversal approach we have seen so many times this AOC for the first part. First, let’s transform the * Node[]* into a 2D array first to make our task easier.

The trick for the traversal part of the problem is to avoid passing through the same coordinate more than once, a problem that can be easily rectified with a cache of positions that we have already passed through.

Once the empty node is at (35, 0) we can commence the second part of the problem – to move the data in the target node to (0, 0). Every time we move the data to the empty node on the left, we’ll have to move the empty node in front of it again, which means going down, left, left and then up, so that means a total of 5 moves for every step towards (0, 0). Except, when the data is at (1, 0), which requires only one move.

Which means, there’s a total of 5 * 35 + 1 move left ONCE we have move the empty node to (35, 0).

In case you’re wondering why I left out the “+ 1” on line 33, it’s because the path that’s returned by * findPaths* includes the starting position (7, 17), so really line 33 should have read

**let part2 = path.Length – 1 + 35 * 5 + 1**

## Links

- Day 22 challenge description
- Advent of Code 2015
- Solution for Day 21
- All my F# solutions for Advent of Code
- Github repo

Hi, I’m **Yan**. I’m an **AWS Serverless Hero** and I help companies go faster for less by adopting serverless technologies successfully.

Are you struggling with serverless or need guidance on best practices? Do you want someone to review your architecture and help you avoid costly mistakes down the line? Whatever the case, I’m here to help.

### Skill up your serverless game with this hands-on workshop.

My 4-week Production-Ready Serverless online workshop is back!

This course takes you through building a production-ready serverless web application from testing, deployment, security, all the way through to observability. The motivation for this course is to give you **hands-on** experience building something with serverless technologies while giving you a broader view of the challenges you will face as the architecture matures and expands.

We will start at the basics and give you a firm introduction to Lambda and all the relevant concepts and service features (including the latest announcements in 2020). And then gradually ramping up and cover a wide array of topics such as API security, testing strategies, CI/CD, secret management, and operational best practices for monitoring and troubleshooting.

If you **enrol now** you can also get **15% OFF** with the promo code “yanprs15”.

Check out my new podcast **Real-World Serverless** where I talk with engineers who are building amazing things with serverless technologies and discuss the real-world use cases and challenges they face. If you’re interested in what people are actually doing with serverless and what it’s really like to be working with serverless day-to-day, then this is the podcast for you.

Check out my new course, **Learn you some Lambda best practice for great good!** In this course, you will learn best practices for working with AWS Lambda in terms of performance, cost, security, scalability, resilience and observability. We will also cover latest features from re:Invent 2019 such as Provisioned Concurrency and Lambda Destinations. **Enrol now and start learning**!

Check out my video course, **Complete Guide to AWS Step Functions**. In this course, we’ll cover everything you need to know to use AWS Step Functions service effectively. There is something for everyone from beginners to more advanced users looking for design patterns and best practices. **Enrol now and start learning**!

Further reading

Here is a complete list of all my posts on serverless and AWS Lambda. In the meantime, here are a few of my most popular blog posts.

- All you need to know about caching for serverless applications
- Lambda optimization tip – enable HTTP keep-alive
- You are wrong about serverless and vendor lock-in
- You are thinking about serverless costs all wrong
- Just how expensive is the full AWS SDK?
- Check-list for going live with API Gateway and Lambda
- How to choose the right API Gateway auth method
- CloudFormation protip: use !Sub instead of !Join
- AWS Lambda – should you have few monolithic functions or many single-purposed functions?
- Guys, we’re doing pagination wrong
- Top 10 Serverless framework best practices
- How to break the “senior engineer” career ceiling
- My advice to junior developers

Pingback: Advent of Code F# – Day 23 | theburningmonk.com

Pingback: F# Weekly #52, 2016 – Sergey Tihon's Blog