Question

lets say we have a

string[] array = {"telekinesis", "laureate", "Allequalsfive", "Indulgence"};

and we need to find a word within this array

normally we’d do following: (or use any similar method to find a string)

bool result = array.Contains("laureate"); // returns true

In my case, the word that I am searching for, may have errors in it (as the title suggests).

For example, I can’t distinguish a difference between letters “I”(large “i”) and “l”(small “L”) and “1”(number one).

Is there any way how I can find a word such as “Allequalsfive” or “A11equalsfive” or “AIIequalsfive”? (loose search) Normally result will be “false”.

If only I can specify to ignore some letters.. (the sequence is constant, other letters are constants).

Answer 1

With the help of extension methods & Levenshtein Distance algorithm

var array = new string[]{ "telekinesis", "laureate", 
                          "Allequalsfive", "Indulgence" };

bool b = array.LooseContains("A11equalsfive", 2); //returns true

–

public static class UsefulExtensions
{
    public static bool LooseContains(this IEnumerable<string> list, string word,int distance)
    {
        foreach (var s in list)
            if (s.LevenshteinDistance(word) <= distance) return true;
        return false;
    }

    //
    //http://www.merriampark.com/ldcsharp.htm
    //
    public static int LevenshteinDistance(this string s, string t)
    {
        int n = s.Length;
        int m = t.Length;
        int[,] d = new int[n + 1, m + 1];

        // 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 = (char.ToUpperInvariant(t[j - 1]) == char.ToUpperInvariant(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];
    }
}