get length of palindromes, within k steps code example

Example 1: k palindrome

public static int longestPalindrome(String s){
        int len = s.length();
        int[][] cal = new int[len][len];
        for(int i=0;i<len;i++){
            cal[i][i] = 1; //considering strings of length = 1
        }
        for(int i=0;i<len-1;i++){
            //considering strings of length = 2
            if (s.charAt(i) == s.charAt(i+1)){
                cal[i][i+1] = 2;
            }else{
                cal[i][i+1] = 0;
            }
        }

        for(int p = len-1; p>=0; p--){
            for(int q=p+2; q<len; q++){
                if (s.charAt(p)==s.charAt(q)){
                    cal[p][q] = 2 + cal[p+1][q-1];
                }else{
                    cal[p][q] = max(cal[p+1][q], cal[p][q-1]);
                }
            }
        }
        return cal[0][len-1];
    }


OR++++++++++++++++++++++++++++++++++++++++++++++++++


public class KPalindrome {
    public static void main(String[] args) {
        KPalindrome kPalindrome = new KPalindrome();
        String s = args[0];
        int k = Integer.parseInt(args[1]);
        if (kPalindrome.testIt(s, k)) {
            System.out.println("YES");
        }
        else {
            System.out.println("NO");
        }
    }

    boolean testIt(String s, int k) {
        if (s.length() <= 1) {
            return true;
        }

        while (s.charAt(0) == s.charAt(s.length()-1)) {
            s = s.substring(1, s.length()-1);

            if (s.length() <= 1) {
                return true;
            }
        }

        if (k == 0) {
            return false;
        }

        // Try to remove the first or last character
        return testIt(s.substring(0, s.length() - 1), k - 1) || testIt(s.substring(1, s.length()), k - 1);
    }
}

Example 2: formula for finding a mathematical palindrome

how to check if a number is a palindrome or not