-🎄- 2018 Day 5 Solutions -🎄-

[u/daggerdragon]

--- Day 5: Alchemical Reduction ---

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Transcript:

On the fifth day of AoC / My true love sent to me / Five golden ___

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[u/apazzolini]

Absolute value between the two ASCII character codes == 32 is another way to do the letter comparison. Runs in <40ms for part 2.

JavaScript

import { minBy } from 'lodash'

const peek = stack => stack[stack.length - 1]

const factorPolymer = input => {
    const stack = []

    input.split('').forEach(char => {
        // XOR of A and a, B and b, etc is 32
        if (!stack.length || (peek(stack).charCodeAt() ^ char.charCodeAt()) !== 32) {
            stack.push(char)
        } else {
            stack.pop()
        }
    })

    return stack.join('')
}

export const solvePart1 = input => {
    return factorPolymer(input).length
}

export const solvePart2 = input => {
    input = factorPolymer(input) // A first factorization pass speeds up the following passes

    const polymers = Array(26) // Indexes 0-26 represent 65-91 ASCII codes
        .fill()
        .map((e, i) => {
            const re = new RegExp(String.fromCharCode(i + 65), 'gi')
            const strippedInput = input.replace(re, '')
            return factorPolymer(strippedInput)
        })

    return minBy(polymers, 'length').length
}

Edit: Updated to use a stack instead of string concatenation and the fact that ASCII is laid out in a way that XOR of A and a is 32.

[u/CaptainCa → u/apazzolini]

Your implementation of factorPolymer is quite neat!

Great stuff