aoc 2023 - #4 part 1+2

.js

{
	const input = document.body.innerText.trim()

	const cardsRaw = input.split("\n")
	
	// Parse
	/** @typedef {{idx: number, winning : number[], scratched : number[]}} Card */
	/** @type {Card[]} */
	const cards = []
	for (let i = 0; i < cardsRaw.length; i++) {
		const cardStr = cardsRaw[i];
		const [first, second] = cardStr.split("|")

		function parseNumbers (str) {
			return str.split(" ")
				.map(s => s.trim())
				.filter(s => s !== "")
				.map(s => Number(s))
		}

		const winning = parseNumbers(first.split(":")[1])
		const scratched = parseNumbers(second)

		cards.push({idx: i, winning, scratched})
	}

	function answerP1() {
		let pointsSum = 0
		for (const card of cards) {
			let value = 0
			for (const number of card.winning) {
				if (card.scratched.includes(number)) {
					value = value === 0 ? 1 : value * 2
				}
			}
			pointsSum += value
		}
		return pointsSum
	}

	function answerP2() {
		/** @type {Card[]} */
		const cardsNew = [...cards]
		for (let i = 0; i < cardsNew.length; i++) {
			const card = cardsNew[i]
			let matches = 0
			for (const number of card.winning) {
				if (card.scratched.includes(number)) {
					matches ++
				}
			}

			for (let j = 1; j < matches + 1; j++) {
				cardsNew.push(cards[card.idx+j])
			}
		}

		return cardsNew.length
	}

	console.log("answer p1", answerP1())
	console.log("answer p2", answerP2())
}

could be made efficiently without copying cards in the P2. Instead cache amount of copies in the original card data and sum up copies that way, pretty much no LOC difference.