Advent of Code 2021 in Kotlin, Day 3: Binary Diagnostic

Solve some problems, learn some Kotlin lessons – and win some Kotlin swag this holiday season!

The "Advent of Code 2021 in Kotlin, Day 3: Binary Diagnostic" blog post:

[cisimon]

By default, I use map or forEach and not for-loops, although the earlier sometimes do not allow to use break, then I switch to for-loop

Making these videos probably not so easy and takes time, but I hope to see all the solutions to this years aoc on this channel. I'd like to compare your approaches with mine. Thanks seb :)

[vitaliiplagov]

I learned about the `removeIf { }` function from this video. I also used mutable lists (would prefer to use immutable code but didn't know how to do in this task :D).

[EricMesa]

I like the idiomatic Kotlin idea of data classes. Reminds me of named Tuples in Python. Maybe I'll start using that more often for clarity

[jonathankolberg]

I'm not to pleased with the mutabilities. Also I have to say I computed with the Ints themself not the strings.

[MaxrX]

Instead of iterating over the binary string to invert it, it should also be possible to do the same thing (likely much faster) with some math:

Create a bitmask of the same length as the input with (1 << (length + 1) - 1), then use inv() on the gammaRate integer, and return the result as a binary and of bitmask and inverse.

Less Kotlin specific features though, thus not as interesting. 😄

[dazraf]

The functional form is much more concise and clear of intent than the imperative form. I also really like to avoid using mutating variables if possible.

[skarloti]

Complexity of your solution is O(n^2) but this is very slow, instead O(n)
fun part1(sequence: Sequence<String>) = sequence
.withIndex()
.map { (index, value) -> index to value.toCharArray().map { it.digitToInt() } }
.reduce { (indexAcc, acc), (index, arr) -> index to List(arr.size) { acc[it] + arr[it] } }
.let { (index, arr) -> List(arr.size) { if (arr[it] > index / 2) 1 to 0 else 0 to 1 } }
.unzip()
.let { (a, b) -> a.joinToString("").toLong(2) * b.joinToString("").toLong(2) }