How can dynamic programming be applied to solve the K-Palindrome problem? Implement the solution.?

The K-Palindrome problem is a variant of the classic palindrome problem, where instead of finding whether a string is a palindrome or not, we need to find if a string can be converted into a palindrome by changing at most k characters.

Dynamic programming is an excellent approach to solve problems like this because it allows us to break down the larger problem into smaller subproblems and reuse the precomputed results multiple times.

Here’s an approach to solve the K-Palindrome problem using dynamic programming:

1. We first define a two-dimensional array ‘dp[][]‘, where ‘dp[i][j]‘ represents whether the substring from index ‘i‘ to index ‘j‘ of the given string is a palindrome or not.

2. We then initialize the diagonal elements of the array to 1, as every substring of length 1 is a palindrome.

3. Next, we iterate through the array to fill in the remaining elements, starting with the substrings of length 2 and increasing the length for each iteration.

4. While iterating, we compare the characters at the start and end of the substring. If they are the same, we check whether the substring ‘dp[i+1][j-1]‘ is a palindrome or not, and set ‘dp[i][j]‘ accordingly.

5. If the characters at the start and end of the substring are different, we take the minimum of ‘dp[i+1][j]‘ and ‘dp[i][j-1]‘ and add 1. This is because, since we can only modify at most k characters to make it a palindrome, we need to keep track of the number of modifications we make as we go along.

6. Finally, if ‘dp[0][n-1] <= 2*k‘, where ‘n‘ is the length of the input string, we can say that the string can be converted into a palindrome by changing at most k characters.

Here’s the Java code that implements this algorithm:

public static boolean isKPalindrome(String s, int k) {
    int n = s.length();
    int[][] dp = new int[n][n];

    for (int i = 0; i < n; i++) {
        dp[i][i] = 1;
    }

    for (int len = 2; len <= n; len++) {
        for (int i = 0; i <= n - len; i++) {
            int j = i + len - 1;
            if (s.charAt(i) == s.charAt(j)) {
                if (len == 2) {
                    dp[i][j] = 2;
                } else {
                    dp[i][j] = dp[i+1][j-1] + 2;
                }
            } else {
                dp[i][j] = Math.max(dp[i+1][j], dp[i][j-1]);
            }
        }
    }

    return n - dp[0][n-1] <= k * 2;
}

Let’s test this function with an example:

String s = "abcdba";
int k = 1;
System.out.println(isKPalindrome(s, k));  // Output: true

In this example, the input string ‘abcdba‘ can be converted into a palindrome by changing just one character. We get the expected output of ‘true‘ from our function.