/*
--- Part Two ---

Now that you have the structure of your transmission decoded, you can calculate the value of the
expression it represents.

Literal values (type ID 4) represent a single number as described above. The remaining type IDs are
more interesting:

- Packets with type ID 0 are `sum` packets - their value is the sum of the values of their
  sub-packets. If they only have a single sub-packet, their value is the value of the sub-packet.
- Packets with type ID 1 are `product` packets - their value is the result of multiplying together
  the values of their sub-packets. If they only have a single sub-packet, their value is the value
  of the sub-packet.
- Packets with type ID 2 are `minimum` packets - their value is the minimum of the values of their
  sub-packets.
- Packets with type ID 3 are `maximum` packets - their value is the maximum of the values of their
  sub-packets.
- Packets with type ID 5 are `greater than` packets - their value is 1 if the value of the first
  sub-packet is greater than the value of the second sub-packet; otherwise, their value is 0. These
  packets always have exactly two sub-packets.
- Packets with type ID 6 are `less than` packets - their value is 1 if the value of the first
  sub-packet is less than the value of the second sub-packet; otherwise, their value is 0. These
  packets always have exactly two sub-packets.
- Packets with type ID 7 are `equal to` packets - their value is 1 if the value of the first
  sub-packet is equal to the value of the second sub-packet; otherwise, their value is 0. These
  packets always have exactly two sub-packets.

Using these rules, you can now work out the value of the outermost packet in your BITS transmission.

For example:

- `C200B40A82` finds the sum of 1 and 2, resulting in the value 3.
- `04005AC33890` finds the product of 6 and 9, resulting in the value 54.
- `880086C3E88112` finds the minimum of 7, 8, and 9, resulting in the value 7.
- `CE00C43D881120` finds the maximum of 7, 8, and 9, resulting in the value 9.
- `D8005AC2A8F0` produces 1, because 5 is less than 15.
- `F600BC2D8F` produces 0, because 5 is not greater than 15.
- `9C005AC2F8F0` produces 0, because 5 is not equal to 15.
- `9C0141080250320F1802104A08` produces 1, because 1 + 3 = 2 * 2.

What do you get if you evaluate the expression represented by your hexadecimal-encoded BITS
transmission?
*/

package com.curtislb.adventofcode.year2021.day16.part2

import com.curtislb.adventofcode.year2021.day16.bits.BitsReader
import java.nio.file.Path
import java.nio.file.Paths

/**
 * Returns the solution to the puzzle for 2021, day 16, part 2.
 *
 * @param inputPath The path to the input file for this puzzle.
 */
fun solve(inputPath: Path = Paths.get("..", "input", "input.txt")): Long {
    val bitsReader = BitsReader(inputPath.toFile().readText().trim())
    return bitsReader.evaluatePacket()
}

fun main() {
    println(solve())
}
