/*
--- Part Two ---

The crabs don't seem interested in your proposed solution. Perhaps you misunderstand crab
engineering?

As it turns out, crab submarine engines don't burn fuel at a constant rate. Instead, each change of
1 step in horizontal position costs 1 more unit of fuel than the last: the first step costs 1, the
second step costs 2, the third step costs 3, and so on.

As each crab moves, moving further becomes more expensive. This changes the best horizontal position
to align them all on; in the example above, this becomes 5:

- Move from 16 to 5: 66 fuel
- Move from 1 to 5: 10 fuel
- Move from 2 to 5: 6 fuel
- Move from 0 to 5: 15 fuel
- Move from 4 to 5: 1 fuel
- Move from 2 to 5: 6 fuel
- Move from 7 to 5: 3 fuel
- Move from 1 to 5: 10 fuel
- Move from 2 to 5: 6 fuel
- Move from 14 to 5: 45 fuel

This costs a total of 168 fuel. This is the new cheapest possible outcome; the old alignment
position (2) now costs 206 fuel instead.

Determine the horizontal position that the crabs can align to using the least fuel possible so they
can make you an escape route! How much fuel must they spend to align to that position?
*/

package com.curtislb.adventofcode.year2021.day07.part2

import com.curtislb.adventofcode.common.comparison.MinMax
import com.curtislb.adventofcode.common.comparison.minMaxOrNull
import com.curtislb.adventofcode.common.number.triangleNumber
import com.curtislb.adventofcode.common.parse.parseInts
import java.nio.file.Path
import java.nio.file.Paths
import kotlin.math.abs

/**
 * Returns the solution to the puzzle for 2021, day 7, part 2.
 *
 * @param inputPath The path to the input file for this puzzle.
 */
fun solve(inputPath: Path = Paths.get("..", "input", "input.txt")): Int? {
    val positions = inputPath.toFile().readText().parseInts()

    // The optimal alignment position must be in the range of starting positions
    val minMaxPositions = positions.minMaxOrNull() ?: MinMax(0, 0)

    // Fuel usage over a distance of n units is the nth triangular number
    return (minMaxPositions.min..minMaxPositions.max).minOfOrNull { position ->
        positions.sumOf { triangleNumber(abs(it - position)) }
    }
}

fun main() = when (val solution = solve()) {
    null -> println("No solution found.")
    else -> println(solution)
}

/*
--- Part Two ---

The crabs don't seem interested in your proposed solution. Perhaps you misunderstand crab
engineering?

As it turns out, crab submarine engines don't burn fuel at a constant rate. Instead, each change of
1 step in horizontal position costs 1 more unit of fuel than the last: the first step costs 1, the
second step costs 2, the third step costs 3, and so on.

As each crab moves, moving further becomes more expensive. This changes the best horizontal position
to align them all on; in the example above, this becomes 5:

- Move from 16 to 5: 66 fuel
- Move from 1 to 5: 10 fuel
- Move from 2 to 5: 6 fuel
- Move from 0 to 5: 15 fuel
- Move from 4 to 5: 1 fuel
- Move from 2 to 5: 6 fuel
- Move from 7 to 5: 3 fuel
- Move from 1 to 5: 10 fuel
- Move from 2 to 5: 6 fuel
- Move from 14 to 5: 45 fuel

This costs a total of 168 fuel. This is the new cheapest possible outcome; the old alignment
position (2) now costs 206 fuel instead.

Determine the horizontal position that the crabs can align to using the least fuel possible so they
can make you an escape route! How much fuel must they spend to align to that position?
*/

package com.curtislb.adventofcode.year2021.day07.part2

import com.curtislb.adventofcode.common.comparison.MinMax
import com.curtislb.adventofcode.common.comparison.minMaxOrNull
import com.curtislb.adventofcode.common.number.triangleNumber
import com.curtislb.adventofcode.common.parse.parseInts
import java.nio.file.Path
import java.nio.file.Paths
import kotlin.math.abs

/**
 * Returns the solution to the puzzle for 2021, day 7, part 2.
 *
 * @param inputPath The path to the input file for this puzzle.
 */
fun solve(inputPath: Path = Paths.get("..", "input", "input.txt")): Int? {
    val positions = inputPath.toFile().readText().parseInts()

    // The optimal alignment position must be in the range of starting positions
    val minMaxPositions = positions.minMaxOrNull() ?: MinMax(0, 0)

    // Fuel usage over a distance of n units is the nth triangular number
    return (minMaxPositions.min..minMaxPositions.max).minOfOrNull { position ->
        positions.sumOf { triangleNumber(abs(it - position)) }
    }
}

fun main() = when (val solution = solve()) {
    null -> println("No solution found.")
    else -> println(solution)
}
