Interview Question: Sub String Palindrome using java.

Recent day interviews you may faced this question in programing interview. Find the sub string palindrome of given string input.

We have many way to solve it. Here I added coding snippet which I tried to solve it.

First one I used two for loop to solve it. Its very simple and easy.

private static List<String> subStringPalindrome(String value) {
    List<String> list = new ArrayList<>();
    for(int i =0 ; i < value.length(); i++){
        for(int j = i+1 ; j < value.length(); j++){
            StringBuffer s1 = new StringBuffer(value.substring(i,j));
            String s2 = s1.reverse().toString();
            if(value.substring(i,j).contentEquals(s2) && s1.length()> 1){
                list.add(s1.toString());
            }

        }
    }
    return list;
}

List<String> list = subStringPalindrome("eetestseeammadam");
Output: [ee, ete, estse, sts, ee, amma, mm, ada]
Its very simple and easy. This code look simple and easy  and I tried another way using recursion.

private static List<String> printPalindromeSubString(String value;) {
    List<String> list = new ArrayList<>();
    if(value.equals(new StringBuilder(value).reverse()))
        list.add(value);
    for(int i=0; i <= value.length(); i++){
        if(i > 1 && i < value.length() -1){
            StringBuffer reverse = new StringBuffer(value.substring(i-2, i+1));
            if(value.substring(i-2, i+1).contentEquals(reverse.reverse())){
                System.out.println("Called2");
                list.add(value.substring(i-2, i+1));
                if(i>2){
                    if(i+2 <= value.length()){
                        new Test().callRecursion(value,value.substring(i-2, i+1), i-1, i, list );
                    }
                }
            }
        }
        if(i > 0 && i < value.length() ){
            StringBuffer stringBuffer = new StringBuffer(value.substring(i-1, i+1));
            if(value.substring(i-1, i+1).contentEquals(stringBuffer.reverse())){
                list.add(value.substring(i-1, i+1));
                System.out.println("Called");
                if(i > 1){
                    new Test().callRecursion(value,value.substring(i-1, i+1), i, i, list );
                }
            }
        }
    }
    return list;
}

private List<String> callRecursion(String value, String reversed, int i, int j, List<String> list){
    if(j+1 >= value.length())
        return list;
    else if(value.charAt(i-2) == value.charAt(j+1)){
        String s = value.charAt(i-2) +reversed+ value.charAt(j+1);
        list.add(s);

        callRecursion(value, s, i-1, j+1, list);
    }
    System.out.println(list);
    return list;
}
List<String> list = printPalindromeSubString("eetestseeammadam");
Here i used recursion to find palindrome on given substring. And the output is 
[ee, ete, sts, estse, ee, mm, amma, ada, madam]

And the output is  If you check the output from two logic, there is a mismatch.. First one print only 8 palindrome and recursion logic print 9 palindrome. 'madam' missing in first logic. 
Lets make this blog more interactive with techies, Please find the missing condition or part of code on comment area.✌✌ 

Thanks in advance.