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AoC &#Day 红鼻子报告(C# 和 Python)

Mary-Kate Olsen
Mary-Kate Olsen原创
2024-12-07 21:02:15976浏览

AoC

今天的解决方案比前一天稍微复杂一点,需要更多的代码,但仍然非常简单。

解决方案的基本概念:

第 1 部分必须满足以下条件才有效:

  • 我们需要决定数字是否都在增加,或者逐个减少。

  • 数字之间的差距不能超过3

第 1 部分

var reports = File.ReadAllLines("./input1.txt")
    .Select(x => x.Split(" "))
    .ToList();

Part1(reports);

void Part1(List<string[]> input)
{
    var validReports = input
        .Select(report => report.Select(int.Parse).ToList())
        .Count(IsValidReport);

    Console.WriteLine(validReports);
}

bool IsValidReport(List<int> levels)
{
    // We don't care about the numbers, just wether going up or down, not both
    var isIncreasing = IsIncreasing(levels);
    var isDecreasing = IsDecreasing(levels);

    if (!isIncreasing && !isDecreasing) return false;

    // Check that all adjacent levels differ by at least 1 and at most 3
    for (var i = 0; i < levels.Count - 1; i++)
    {
        var diff = Math.Abs(levels[i + 1] - levels[i]);
        if (diff is < 1 or > 3)
        {
            return false;
        }
    }

    return true;
}

bool IsIncreasing(List<int> numbers)
{
    for (var i = 1; i < numbers.Count; i++)
    {
        if (numbers[i] < numbers[i - 1]) return false;
    }

    return true;
}

bool IsDecreasing(List<int> numbers)
{
    for (var i = 1; i < numbers.Count; i++)
    {
        if (numbers[i] > numbers[i - 1]) return false;
    }

    return true;
}

一个简单的解释是,在将报告传递给我们的助手 IsValidReport() 方法之前,我们使用 LINQ 将字符串数字解析为整数。

此方法检查所有数字是否在增加或减少。这是通过将 id 与有序列表进行比较来完成的,这将证明所有数字都朝一个方向发展。

一旦我们知道它们都在一个方向(而不是带),我们就可以检查每个数字是否彼此相差 3 个以内。使用 Math.Abs​​() 方法返回绝对数,帮助我们计算负数。

注意:绝对数,是数字和零之间的数字,即 -5 和 5 都会返回 5;

第2部分

对于第 2 部分,我们有稍微复杂的说明,但听起来比实际更难。

第二部分介绍了人为错误的警告,基本概念是您需要知道是否可以从报告中删除任何 1 项,并且该报告仍将被视为有效报告。

void Part2(List<string[]> input)
{
    var validReports = 0;

    foreach (var ints in input.Select(report => report.Select(int.Parse).ToList()))
    {
        // If valid as-is, count it
        if (IsValidReport(ints))
        {
            validReports++;
            continue;
        }

        // Check if removing any single level makes it valid
        if (ints.Select((t, i) => ints.Where((_, index) => index != i).ToList())
            .Any(IsValidReport))
        {
            validReports++;
        }
    }

    Console.WriteLine(validReports);
}

在这里,我们最大限度地发挥 LINQ 的功能,解析字符串报告,并使用它来修改 ints 数组。我们通过循环遍历数组并删除索引处的当前项来修改,从而修改了整数数组。

修改后,我们通过验证器运行修改后的报告。这可以使用 LINQ Select() 方法一次性完成,并结合 Any(),一旦满足条件(即修改后的报告有效),Any() 就会短路。

为什么只检查单个有效的修改报告?

这是因为它规定如果可以删除任何一项,则报告可以被视为有效。它没有说明特定的项目,因此一旦我们找到匹配的项目(删除后意味着报告有效),我们就可以停止查看所有其他项目。

然后我们增加 validReports 计数器,以跟踪有效的报告。

这是我将其转换为 Python 的尝试:

def main():
    # Read the input file and split each line into a list of strings
    with open("./input1.txt") as file:
        reports = [line.split() for line in file]

    part1(reports)
    part2(reports)


def part1(reports):
    # Count valid reports using a list comprehension and the IsValidReport function

    valid_reports = 0
    for report in reports:
        levels = list(map(int, report))

        if is_valid_report(levels):
            valid_reports += 1

    print(valid_reports)


def part2(reports):
    valid_reports = 0

    for report in reports:
        ints = list(map(int, report))

        # If valid as-is, count it
        if is_valid_report(ints):
            valid_reports += 1
            continue

        # Check if removing any single level makes it valid
        valid_with_dampener = False

        # Iterate over each element in ints
        for i in range(len(ints)):
            # Create a new list without the element at index i
            modified_report = [x for j, x in enumerate(ints) if j != i]

            # Check if the modified report is valid
            if is_valid_report(modified_report):
                valid_with_dampener = True
                break

        if valid_with_dampener:
            valid_reports += 1

    print(valid_reports)


def is_valid_report(levels):
    # Check if the sequence is either all increasing or all decreasing
    is_increasing = is_increasing_sequence(levels)
    is_decreasing = is_decreasing_sequence(levels)

    if not is_increasing and not is_decreasing:
        return False

    # Check that all adjacent levels differ by at least 1 and at most 3
    for i in range(len(levels) - 1):
        diff = abs(levels[i + 1] - levels[i])
        if diff < 1 or diff > 3:
            return False

    return True


def is_increasing_sequence(numbers):
    for i in range(1, len(numbers)):
        if numbers[i] < numbers[i - 1]:
            return False
    return True


def is_decreasing_sequence(numbers):
    for i in range(1, len(numbers)):
        if numbers[i] > numbers[i - 1]:
            return False
    return True


if __name__ == "__main__":
    main()

对于不太冗长的解决方案,我们可以用修改和检查替换第 2 部分内部 for 循环,就像我们在 C# 解决方案中使用 LINQ 所做的那样,如下所示:

var reports = File.ReadAllLines("./input1.txt")
    .Select(x => x.Split(" "))
    .ToList();

Part1(reports);

void Part1(List<string[]> input)
{
    var validReports = input
        .Select(report => report.Select(int.Parse).ToList())
        .Count(IsValidReport);

    Console.WriteLine(validReports);
}

bool IsValidReport(List<int> levels)
{
    // We don't care about the numbers, just wether going up or down, not both
    var isIncreasing = IsIncreasing(levels);
    var isDecreasing = IsDecreasing(levels);

    if (!isIncreasing && !isDecreasing) return false;

    // Check that all adjacent levels differ by at least 1 and at most 3
    for (var i = 0; i < levels.Count - 1; i++)
    {
        var diff = Math.Abs(levels[i + 1] - levels[i]);
        if (diff is < 1 or > 3)
        {
            return false;
        }
    }

    return true;
}

bool IsIncreasing(List<int> numbers)
{
    for (var i = 1; i < numbers.Count; i++)
    {
        if (numbers[i] < numbers[i - 1]) return false;
    }

    return true;
}

bool IsDecreasing(List<int> numbers)
{
    for (var i = 1; i < numbers.Count; i++)
    {
        if (numbers[i] > numbers[i - 1]) return false;
    }

    return true;
}

但我认为你可以同意,带有更好命名变量的详细方法更具可读性。

感谢您的阅读,明天再次加入我,并一如既往地在 Twitter 上关注我

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