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How Can I Efficiently Calculate the Damerau-Levenshtein Distance Between Two Strings?

Patricia Arquette
Release: 2025-01-15 09:39:45
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How Can I Efficiently Calculate the Damerau-Levenshtein Distance Between Two Strings?

Use the Damerau-Levenshtein algorithm to calculate the distance similarity of a given string

Determining the similarity between strings is crucial in a variety of applications such as spell checking and text comparison. Damerau-Levenshtein distance is an efficient measure that calculates the minimum number of edits (insertions, deletions, substitutions, or transpositions) required to transform one string into another.

Performance optimization of Damerau-Levenshtein algorithm

For optimal performance when calculating the Damerau-Levenshtein distance, consider the following key points:

  • Convert a string to an array of integers: Comparing arrays of integers is much faster than character arrays.
  • Short-circuit mechanism: If the current distance exceeds the specified threshold, stop calculation.
  • Rotate collection of arrays: Use three arrays instead of large matrices to reduce memory overhead.
  • Optimize array slicing: Make sure arrays are aligned with shorter strings.

Code implementation

The following optimized C# code snippet implements the Damerau-Levenshtein algorithm:

<code class="language-csharp">public static int DamerauLevenshteinDistance(int[] source, int[] target, int threshold) {
    int length1 = source.Length;
    int length2 = target.Length;

    if (Math.Abs(length1 - length2) > threshold) { return int.MaxValue; }

    if (length1 > length2) {
        Swap(ref target, ref source);
        Swap(ref length1, ref length2);
    }

    int maxi = length1;
    int maxj = length2;

    int[] dCurrent = new int[maxi + 1];
    int[] dMinus1 = new int[maxi + 1];
    int[] dMinus2 = new int[maxi + 1];
    int[] dSwap;

    for (int i = 0; i  1 && j > 1 && source[im2] == target[jm1] && source[im1] == target[j - 2])
                min = Math.Min(min, dMinus2[im2] + cost);
            dCurrent[i] = min;
            if (min  threshold) { return int.MaxValue; }
    }

    int result = dCurrent[maxi];
    return (result > threshold) ? int.MaxValue : result;
}</code>
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