## Advent of Code 2020 Day 10

## Try the simulator for Part 2 using NullDev's algorithm

##
Task: Solve for `X`

where `X =`

- The product of the count of 1- and 3-integer differences between sorted jolt measures
- The total number of distinct ways you can arrange the adapters

### Example input

```
16
10
15
5
1
11
7
19
6
12
4
```

It represents:

- A collection of adapters
- Each of their output joltage

## Part 1

- Confusion, then clarity, then confidence
- Writing a working algorithm

### Confusion, then clarity, then confidence

- Confusion: The introductory sentences of this puzzle made me concerned that I'd have no idea where to begin
- Clarity: The walkthrough using the example input helped me understand what the rules of this puzzle were
- Confidence: I knew immediately how I would solve it algorithmically - and thankfully my algorithm worked!

### Writing a working algorithm

```
Process the input into a sorted array of numbers
Insert two values: one at the beginning and another at the end
Accumulate a 3-element array representing tallies for each 1-, 2- and 3-integer difference between each subsequent value
Return the product of the integers tallying all 1- and 3-integer differences
```

Here's a visualization of my algorithm:

## Part 2

- I should have seen that coming
- I thought I figured it out
- I knew it was that simple!
- I had to see how it worked

### I should have seen that coming

- When Part 1 feels easy - to me! - it is often a sign that Part 2 is a test of speed...which I'm often unequipped to solve
- Given the 1, 2 and 3-integer differences, it should have been clear that Part 2 would be a test of 'how many possible arrangements?'

### I thought I figured it out

Using the first example input as reference, the numbers in order are:

```
1, 4, 5, 6, 7, 10, 11, 12, 15, 16, 19
```

My faulty logic followed this path toward a solution:

```
Legend:
0 numbers in-between = 1
1 number in-between = 2
2 numbers in-between = 4
1 - 4: 0 -> 1
4 - 7: 2 -> 4
7 - 10: 0 -> 1
10 - 12: 1 -> 2
12 - 15: 0 -> 1
15 - 16: 0 -> 1
16 - 19: 0 -> 1
1 * 4 * 1 * 2 * 1 * 1 * 1
4 * 2
8
```

Sadly, using this formula on the second example produced ~2000, not 19208.

### I knew it was that simple!

Thankfully, my go-to JavaScript solver, NullDev, wrote an elegant, sub-5-line solution that I felt compelled to fully understand...using simulation!

```
Create step, a 1-element array containing the value 1
For i from 0 to the maximum value in the sorted array of numbers from the input
Set j to i + 1
While the number at location j in the sorted array is less than or equal to i + 3
Set the item at location j in step equal to the sum of the item at location i in step and either the item at location j in step or 0
Increment j
Return the last value in step
```

I was confused about how NullDev's algorithm worked when reasoning about it in my head.

Even after using `console.log`

at the end to show the filled `step`

array, I still didn't see what it was doing in each iteration.

I knew I needed to simulate each iteration.

### I had to see how it worked

- Re-factoring NullDev's code into
`if...else`

clauses instead of combination of`for`

and`while`

loops wasn't too difficult - The result is a simulator that renders a list which grows taller and always shows the last value at the top
- It finally helped me realize how NullDev's algorithm goes '3 steps forward, 2 steps back' in a way to continually double each value

Try the simulator for Part 2 using NullDev's algorithm

Seeing NullDev's solution showed me that I was 'on to something' in my 1-2-4 doubling logic. Though, clearly my approach was too shallow.

I may not have solved Part 2, but I had fun noodling around with this puzzle and simulating another solver's algorithm.

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