/*
--- Day 14: Extended Polymerization ---

The incredible pressures at this depth are starting to put a strain on your submarine. The submarine
has polymerization equipment that would produce suitable materials to reinforce the submarine, and
the nearby volcanically-active caves should even have the necessary input elements in sufficient
quantities.

The submarine manual contains instructions for finding the optimal polymer formula; specifically, it
offers a polymer template and a list of pair insertion rules (your puzzle input). You just need to
work out what polymer would result after repeating the pair insertion process a few times.

For example:

```
NNCB

CH -> B
HH -> N
CB -> H
NH -> C
HB -> C
HC -> B
HN -> C
NN -> C
BH -> H
NC -> B
NB -> B
BN -> B
BB -> N
BC -> B
CC -> N
CN -> C
```

The first line is the polymer template - this is the starting point of the process.

The following section defines the pair insertion rules. A rule like `AB -> C` means that when
elements `A` and `B` are immediately adjacent, element `C` should be inserted between them. These
insertions all happen simultaneously.

So, starting with the polymer template `NNCB`, the first step simultaneously considers all three
pairs:

- The first pair (`NN`) matches the rule `NN -> C`, so element `C` is inserted between the first `N`
  and the second `N`.
- The second pair (`NC`) matches the rule `NC -> B`, so element `B` is inserted between the `N` and
  the `C`.
- The third pair (`CB`) matches the rule `CB -> H`, so element `H` is inserted between the `C` and
  the `B`.

Note that these pairs overlap: the second element of one pair is the first element of the next pair.
Also, because all pairs are considered simultaneously, inserted elements are not considered to be
part of a pair until the next step.

After the first step of this process, the polymer becomes `NCNBCHB`.

Here are the results of a few steps using the above rules:

```
Template:     NNCB
After step 1: NCNBCHB
After step 2: NBCCNBBBCBHCB
After step 3: NBBBCNCCNBBNBNBBCHBHHBCHB
After step 4: NBBNBNBBCCNBCNCCNBBNBBNBBBNBBNBBCBHCBHHNHCBBCBHCB
```

This polymer grows quickly. After step 5, it has length 97; After step 10, it has length 3073. After
step 10, `B` occurs 1749 times, `C` occurs 298 times, `H` occurs 161 times, and `N` occurs 865
times; taking the quantity of the most common element (`B`, 1749) and subtracting the quantity of
the least common element (`H`, 161) produces 1749 - 161 = 1588.

Apply 10 steps of pair insertion to the polymer template and find the most and least common elements
in the result. What do you get if you take the quantity of the most common element and subtract the
quantity of the least common element?
*/

package com.curtislb.adventofcode.year2021.day14.part1

import com.curtislb.adventofcode.year2021.day14.polymer.PolymerizationProcess
import java.nio.file.Path
import java.nio.file.Paths

/**
 * Returns the solution to the puzzle for 2021, day 14, part 1.
 *
 * @param inputPath The path to the input file for this puzzle.
 * @param stepCount The number of pair insertion steps to apply to the polymer template.
 */
fun solve(inputPath: Path = Paths.get("..", "input", "input.txt"), stepCount: Int = 10): Long {
    val polymerization = PolymerizationProcess.fromFile(inputPath.toFile())
    polymerization.applyPairInsertions(stepCount)
    return polymerization.maxElementCountDifference()
}

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

/*
--- Day 14: Extended Polymerization ---

The incredible pressures at this depth are starting to put a strain on your submarine. The submarine
has polymerization equipment that would produce suitable materials to reinforce the submarine, and
the nearby volcanically-active caves should even have the necessary input elements in sufficient
quantities.

The submarine manual contains instructions for finding the optimal polymer formula; specifically, it
offers a polymer template and a list of pair insertion rules (your puzzle input). You just need to
work out what polymer would result after repeating the pair insertion process a few times.

For example:

```
NNCB

CH -> B
HH -> N
CB -> H
NH -> C
HB -> C
HC -> B
HN -> C
NN -> C
BH -> H
NC -> B
NB -> B
BN -> B
BB -> N
BC -> B
CC -> N
CN -> C
```

The first line is the polymer template - this is the starting point of the process.

The following section defines the pair insertion rules. A rule like `AB -> C` means that when
elements `A` and `B` are immediately adjacent, element `C` should be inserted between them. These
insertions all happen simultaneously.

So, starting with the polymer template `NNCB`, the first step simultaneously considers all three
pairs:

- The first pair (`NN`) matches the rule `NN -> C`, so element `C` is inserted between the first `N`
  and the second `N`.
- The second pair (`NC`) matches the rule `NC -> B`, so element `B` is inserted between the `N` and
  the `C`.
- The third pair (`CB`) matches the rule `CB -> H`, so element `H` is inserted between the `C` and
  the `B`.

Note that these pairs overlap: the second element of one pair is the first element of the next pair.
Also, because all pairs are considered simultaneously, inserted elements are not considered to be
part of a pair until the next step.

After the first step of this process, the polymer becomes `NCNBCHB`.

Here are the results of a few steps using the above rules:

```
Template:     NNCB
After step 1: NCNBCHB
After step 2: NBCCNBBBCBHCB
After step 3: NBBBCNCCNBBNBNBBCHBHHBCHB
After step 4: NBBNBNBBCCNBCNCCNBBNBBNBBBNBBNBBCBHCBHHNHCBBCBHCB
```

This polymer grows quickly. After step 5, it has length 97; After step 10, it has length 3073. After
step 10, `B` occurs 1749 times, `C` occurs 298 times, `H` occurs 161 times, and `N` occurs 865
times; taking the quantity of the most common element (`B`, 1749) and subtracting the quantity of
the least common element (`H`, 161) produces 1749 - 161 = 1588.

Apply 10 steps of pair insertion to the polymer template and find the most and least common elements
in the result. What do you get if you take the quantity of the most common element and subtract the
quantity of the least common element?
*/

package com.curtislb.adventofcode.year2021.day14.part1

import com.curtislb.adventofcode.year2021.day14.polymer.PolymerizationProcess
import java.nio.file.Path
import java.nio.file.Paths

/**
 * Returns the solution to the puzzle for 2021, day 14, part 1.
 *
 * @param inputPath The path to the input file for this puzzle.
 * @param stepCount The number of pair insertion steps to apply to the polymer template.
 */
fun solve(inputPath: Path = Paths.get("..", "input", "input.txt"), stepCount: Int = 10): Long {
    val polymerization = PolymerizationProcess.fromFile(inputPath.toFile())
    polymerization.applyPairInsertions(stepCount)
    return polymerization.maxElementCountDifference()
}

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