Implement the solution to the Longest Common Substring with K Mismatches problem using dynamic programming.?

The Longest Common Substring with K Mismatches problem is defined as finding the longest common substring between two given strings with at most k mismatched characters.

One way to solve this problem is by using dynamic programming. Let’s approach this problem by defining a 2D array to represent the lengths of the common substrings between the two strings.

Let’s assume "a" and "b" are the two input strings.

1. Create a 2D array dp with dimensions (length of string "a" + 1) x (length of string "b" + 1).

2. Initialize all values in the first row and column to 0, as there can’t be any common substrings with an empty string.

3. Create two variables, "max_len" and "end_pos", to keep track of the longest common substring length found so far and its ending position in string "a".

4. Loop through each element (i, j) in the 2D array dp, where i represents the index in string "a" and j represents the index in string "b".

5. If the characters at indices i-1 in string "a" and j-1 in string "b" are the same, set dp[i][j] to dp[i-1][j-1] + 1, which means a new common substring of length dp[i-1][j-1] + 1 is found.

6. If the characters are not the same, set dp[i][j] to 0, because there is no common substring at these indices.

7. At each iteration, check if dp[i][j] is greater than max_len, if so, update max_len and end_pos to i and j, respectively.

8. Once the loop is finished, the longest common substring with at most k mismatches can be found by iterating through the length of max_len to 0, and checking each substring of that length to see if it has at most k mismatches.

Here is the Java implementation of the solution:

public static String longestCommonSubstring(String a, String b, int k) {
    // Initialize the 2D array
    int[][] dp = new int[a.length() + 1][b.length() + 1];
    int max_len = 0, end_pos = 0;
    
    // Loop through each element of the 2D array
    for (int i = 1; i <= a.length(); i++) {
        for (int j = 1; j <= b.length(); j++) {
            // If the characters are the same, update dp array value
            if (a.charAt(i-1) == b.charAt(j-1)) {
                dp[i][j] = dp[i-1][j-1] + 1;
            }
            // If the characters are different, set dp array value to 0
            else {
                dp[i][j] = 0;
            }

            // Check for the longest common substring length found so far
            if (dp[i][j] > max_len) {
                max_len = dp[i][j];
                end_pos = i;
            }
        }
    }

    // Find the longest common substring with at most k mismatches
    for (int len = max_len; len > 0; len--) {
        for (int i = 0; i <= end_pos - len; i++) {
            int mismatches = 0;
            for (int j = i; j < i + len; j++) {
                if (a.charAt(j) != b.charAt(j-i)) {
                    mismatches++;
                }
            }
            if (mismatches <= k) {
                return a.substring(i, i + len);
            }
        }
    }

    // No common substring with at most k mismatches found
    return "";
}

In this implementation, we’re using nested loops to iterate through each element of the 2D array. We also have another nested loop to find the longest common substring with at most k mismatches.