/*
 * This is the classic Levenshtein Distance algorithm. It compares
 * two strings and returns the difference in an integer, where the
 * higher the integer is, the greater the distance.
 *
 * The code is in a static class, and thus can be called like this:
 * LevenshteinDistance.ComputeDistance(x, y);
 * 
 * Samuel Allen (Dot Net Perls) worked on this particular implementation.
 * It is in the public domain, as I cannot hold copyrights to such a
 * popular algorithm! I might as well patent the IF statement.
 * 
 * Samuel Allen, http://dotnetperls.com, April 2008
 * Posted on Google Code, dotnetperls-controls.
 * 
 * */

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;

/// <summary>
/// Contains the Levenshtein Distance algorithm for computing the
/// distance between two strings.
/// </summary>
public static class LevenshteinDistance
{
    /// <summary>
    /// Compute the distance between two strings (the parameters).
    /// </summary>
    /// <param name="s">The first of the two strings you want to compare.</param>
    /// <param name="t">The second of the two strings you want to compare.</param>
    /// <returns>The Levenshtein Distance (higher is a bigger difference).</returns>
    public static int ComputeDistance(string s, string t)
    {
        int n = s.Length;
        int m = t.Length;
        int[,] d = new int[n + 1, m + 1]; // matrix

        // Step 1
        if (n == 0)
        {
            return m;
        }

        if (m == 0)
        {
            return n;
        }

        // Step 2
        for (int i = 0; i <= n; d[i, 0] = i++)
        {
        }

        for (int j = 0; j <= m; d[0, j] = j++)
        {
        }

        // Step 3
        for (int i = 1; i <= n; i++)
        {
            //Step 4
            for (int j = 1; j <= m; j++)
            {
                // Step 5
                int cost = (t[j - 1] == s[i - 1]) ? 0 : 1;

                // Step 6
                d[i, j] = Math.Min(
                    Math.Min(d[i - 1, j] + 1, d[i, j - 1] + 1),
                    d[i - 1, j - 1] + cost);
            }
        }
        // Step 7
        return d[n, m];
    }
}