Longest Common Subsequence in C#

Longest Common

Subsequence in C#

Welcome to the Longest Common Subsequence in C# page! Here, you'll find the source code for this program as well as a description of how the program works.

Current Solution

using System.Collections.Generic;

if (
    args is not [var raw1, var raw2]
    || !TryParseList(raw1.AsSpan(), out var a)
    || !TryParseList(raw2.AsSpan(), out var b)
)
{
    Console.WriteLine(
        """
Usage: please provide two lists in the format "1, 2, 3, 4, 5"
"""
    );
    return;
}

Console.WriteLine(string.Join(", ", LCS(a, b)));

static bool TryParseList(ReadOnlySpan<char> span, out List<int> numbers)
{
    numbers = new(span.Count(',') + 1);

    while (!span.IsEmpty)
    {
        int comma = span.IndexOf(',');
        var token = comma >= 0 ? span[..comma] : span;

        span = comma >= 0 ? span[(comma + 1)..] : [];

        if (!int.TryParse(token, out int n))
            return false;

        numbers.Add(n);
    }

    return true;
}

static List<int> LCS(List<int> a, List<int> b)
{
    int n = a.Count, m = b.Count;
    if (n == 0 || m == 0) return [];

    int[,] dp = new int[n + 1, m + 1];

    for (int i = 1; i <= n; i++)
    for (int j = 1; j <= m; j++)
        dp[i, j] = a[i - 1] == b[j - 1]
            ? dp[i - 1, j - 1] + 1
            : Math.Max(dp[i - 1, j], dp[i, j - 1]);

    var result = new List<int>(dp[n, m]);

    int i2 = n, j2 = m;

    while (i2 > 0 && j2 > 0)
    {
        if (a[i2 - 1] == b[j2 - 1])
        {
            result.Add(a[i2 - 1]);
            i2--;
            j2--;
        }
        else if (dp[i2 - 1, j2] >= dp[i2, j2 - 1])
        {
            i2--;
        }
        else
        {
            j2--;
        }
    }

    result.Reverse();
    return result;
}

Longest Common Subsequence in C# was written by:

• Parker Johansen
• Ștefan-Iulian Alecu

If you see anything you'd like to change or update, please consider contributing.

How to Implement the Solution

No 'How to Implement the Solution' section available. Please consider contributing.

How to Run the Solution

No 'How to Run the Solution' section available. Please consider contributing.