# Advent of Code 2019 Day 4

This year I have decided to try and do the code challenges on the Advent of Code website in Scala and possibly Spark if needed (or an interesting solution arises).
These are simple little coding challenges given once per day like an Advent Calendar before Christmas.

## Day 4: Part 1

This challenge is relatively short so I will include the whole thing below:

— Day 4: Secure Container —

You arrive at the Venus fuel depot only to discover it’s protected by a password. The Elves had written the password on a sticky note, but someone threw it out.

• It is a six-digit number.
• The value is within the range given in your puzzle input.
• Two adjacent digits are the same (like `22` in `122345`).
• Going from left to right, the digits never decrease; they only ever increase or stay the same (like `111123` or `135679`).

Other than the range rule, the following are true:

• `111111` meets these criteria (double `11`, never decreases).
• `223450` does not meet these criteria (decreasing pair of digits `50`).
• `123789` does not meet these criteria (no double).

How many different passwords within the range given in your puzzle input meet these criteria?

Your puzzle input is `136760-595730`.

So we need to crack that password! Or at least work out how many combinations there are.

This is a nice and simple thing to do in Scala:

``````val min = 136760
val max = 595730

val fullRange = min to max
``````

First we define the minimum and maximum and create a range between them.

Next I want to extract each digit inside each item in the range into a single number. I actually use a bit of a short-cut to do this:

``````def charToInt(char: Char): Int = char.toInt - '0'
``````

This method will take a character and assuming it is a number character will convert it into a matching integer. Combined with a string version of a candidate password this lets me produce an array of digits with ease like so:

``````fullRange
.map(n => n.toString)
.map(string => string.map(char => charToInt(char)))
``````

Now all we need to do is filter down this big collection of digits to match the criteria described:

First lets find all the combinations with repeating digits:

``````def hasRepeatedDigit(number: IndexedSeq[Int]): Boolean = {
for (index <- 0 until number.size - 1) {
val digit = number(index)
val nextDigit = number(index + 1)
if (digit == nextDigit) {
return true
}
}
false
}
``````

That’s pretty simple and easy.

Next let us filter to just those digits with incrementing or remaining the same digits:

``````def isIncrementingOrSame(number: IndexedSeq[Int]): Boolean = {
var index: Int = 0
while (index < number.size - 1) {
val digit = number(index)
for (i <- index + 1 until number.size) {
val testDigit = number(i)
if (testDigit < digit) {
return false
}
}
index += 1
}
true
}
``````

A little more complex but not hard.

Putting these together like so:

``````val validPasswords = fullRange
.map(n => n.toString)
.map(string => string.map(char => charToInt(char)))
.filter(hasRepeatedDigit)
.filter(isIncrementingOrSame)

``````

Will print out the amount of valid values asked for in part 1!

## Day 4: Part 2

Now part 2 modifies one of the conditions slightly:

— Part Two —

An Elf just remembered one more important detail: the two adjacent matching digits are not part of a larger group of matching digits.

Given this additional criterion, but still ignoring the range rule, the following are now true:

• `112233` meets these criteria because the digits never decrease and all repeated digits are exactly two digits long.
• `123444` no longer meets the criteria (the repeated `44` is part of a larger group of `444`).
• `111122` meets the criteria (even though `1` is repeated more than twice, it still contains a double `22`).

How many different passwords within the range given in your puzzle input meet all of the criteria?

Now we can add an extra filter to cover this.

There’s a few ways of writing this filter, one slightly hacky way is to convert the digits back to a string and use a Regular Expression to find all the repeating digits:

``````val pattern = Pattern.compile("(?<=(.))(?!\\1)")
def repeatDigitsNotPartOfLargerGroup(number: IndexedSeq[Int]): Boolean = {
val asString = number.map(digit => digit.toString).mkString
val repeatedDigits = pattern.split(asString).toSeq
repeatedDigits.exists(repeat => repeat.length == 2)
}
``````

The pattern does a positive lookbehind and negative lookahead. Kind of hard to understand unless you use Regular Expressions a lot.
You could do a similar thing with a Java `Scanner` too.

But if we wanted to do this properly without converting to a string we really only need 2 nested loops to perform the same logic on the digits:

``````def repeatDigitsNotPartOfLargerGroup(number: IndexedSeq[Int]): Boolean = {
val groupCounts = mutable.Buffer[(Int, Int)]()

var start = 0
while (start < number.length - 1) {
val digit = number(start)
var count = 1
var i = start + 1
var changed = false
while (i < number.length && !changed) {
val nextDigit = number(i)
if (digit != nextDigit) {
changed = true
} else {
count += 1
i += 1
}
}
val group = (digit, count)
groupCounts += group
start += count
}

groupCounts.exists(group => group._2 == 2)
}
``````

And with either of these filters added we get our result for part 2!

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