Understanding the Algorithm

The dynamic programming algorithm for calculating edit distance involves the following steps:

1. Create a matrix to store the subproblem results. The dimensions of the matrix are based on the lengths of the two input strings.

2. Fill the first row and first column of the matrix with incremental values representing the number of deletions or insertions required.

3. Iterate over each cell of the matrix, comparing the characters of the substrings. If the characters match, the value in the current cell is set to the diagonal cell's value. If the characters don't match, the value in the current cell is set to the minimum value among the adjacent cells plus 1 (representing the cost of substitution).

4. The bottom-right cell of the matrix represents the edit distance between the two input strings.

Here's an implementation of the dynamic programming algorithm in C#:

1{code}

In this implementation, we calculate the edit distance between two strings, str1 and str2. We create a matrix dp with dimensions based on the lengths of the two strings and fill the first row and first column with incremental values. Then, we use nested loops to iterate over each cell of the matrix and fill it based on the comparison of characters. Finally, we return the value in the bottom-right cell, which represents the edit distance.

By understanding the steps involved in the dynamic programming algorithm, we can efficiently calculate the edit distance between two strings.

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}

public static int EditDistance(string str1, string str2)

{

    int m = str1.Length;

    int n = str2.Length;

​

    // Create a matrix to store the subproblem results

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

​

    // Fill the first row and first column with incremental values

    for (int i = 0; i <= m; i++)

    {

        dp[i, 0] = i;

    }

​

    for (int j = 0; j <= n; j++)

    {

        dp[0, j] = j;

    }

​

    // Fill the matrix using the dynamic programming approach

    for (int i = 1; i <= m; i++)

    {

        for (int j = 1; j <= n; j++)

        {

            if (str1[i - 1] == str2[j - 1])

            {

                dp[i, j] = dp[i - 1, j - 1];

            }

            else