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Advent of code - Day 11

qmenoret profile image Quentin Ménoret ・2 min read

Are you participating in the Advent of code this year?

If you don't know what the advent of code is, it's a website where you'll find a daily challenge (every day it gets harder). It's a really fun event, you should participate!

I try to solve the exercises using either JavaScript or TypeScript and will share my solutions daily (with one day delay so no one can cheat!). I only share the solution for the second part.

Part 1 of day 11 was a simple game of life exercise. I liked the part 2 more, it was a bit different for once.

Here is my solution for day #11:

function isOccupied(rows, x, y, direction, log = false) {
  if (log) console.log(x, y, direction)
  if (!(rows[x] || [])[y]) return false
  if (rows[x][y] === 'L') return false
  if (rows[x][y] === '#') return true
  return isOccupied(rows, x + direction[0], y + direction[1], direction, log)
}

function getNextTurn(rows) {
  return rows.map((row, rowIndex) =>
    row.map((seat, seatIndex) => {
      if (seat === '.') return '.'
      const numberOfSeatTaken = [
        [-1, -1],
        [-1, 0],
        [-1, 1],
        [1, -1],
        [1, 0],
        [1, 1],
        [0, -1],
        [0, 1],
      ].filter(([x, y]) => {
        return isOccupied(rows, rowIndex + x, seatIndex + y, [x, y])
      }).length
      if (seat === '#') return numberOfSeatTaken >= 5 ? 'L' : '#'
      return numberOfSeatTaken > 0 ? 'L' : '#'
    }),
  )
}

function areEquals(rows1, rows2) {
  return rows1.map((x) => x.join('')).join('\n') === rows2.map((x) => x.join('')).join('\n')
}

let previous = input
let current = getNextTurn(previous)

while (!areEquals(previous, current)) {
  previous = current
  current = getNextTurn(previous)
}

console.log(current.reduce((acc, row) => row.reduce((acc, seat) => (seat === '#' ? acc + 1 : acc), acc), 0))
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Feel free to share your solution in the comments!


Photo by Markus Spiske on Unsplash

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