If you are not familiar with the problem, you can read it here.

This has been a simple problem, could be solved even with brute forcing, but I went the maths path and implemented a quadratic equation solver instead.

It's easy to compute how far our boat moves if we wait for x ms at the beginning. The solution to this equation tells us the x-es for which we break the record distance.

Part 2 is just Part 1 with bigger numbers.

namespace AdventOfCode.Y2023.Day06;

using System;
using System.Linq;
using System.Text.RegularExpressions;

[ProblemName("Wait For It")]
class Solution : Solver {

    public object PartOne(string input) => Solve(input);
    public object PartTwo(string input) => Solve(input.Replace(" ", ""));

    long Solve(string input) {
        var rows = input.Split("\n");
        var times = Parse(rows[0]);
        var records = Parse(rows[1]);

        var res = 1L;
        for (var i = 0; i < times.Length; i++) {
            res *= WinningMoves(times[i], records[i]);
        }
        return res;
    }

    long WinningMoves(long time, long record) {
        // If we wait x ms, our boat moves `(time - x) * x` millimeters.
        // This breaks the record when `(time - x) * x > record`
        // or `-x^2  + time * x - record > 0`.

        // get the roots first
        var (x1, x2) = SolveEq(-1, time, -record);
        
        // integers in between the roots
        var maxX = (long)Math.Ceiling(x2) - 1;
        var minX = (long)Math.Floor(x1) + 1;
        return maxX - minX + 1; 
    }

    // solves ax^2 + bx + c = 0 (supposing two different roots)
    (double, double) SolveEq(long a, long b, long c) {
        var d = Math.Sqrt(b * b - 4 * a * c);
        var x1 = (-b - d) / (2 * a);
        var x2 = (-b + d) / (2 * a);
        return (Math.Min(x1, x2), Math.Max(x1, x2));
    }

    long[] Parse(string input) => (
        from m in Regex.Matches(input, @"\d+")
        select long.Parse(m.Value)
    ).ToArray();
}

Please ☆ my repo if you like it!