Count palindromic characteristics of a String

Given a string s of length n, count the number of substrings having different types of palindromic characteristics. 
Palindromic Characteristic is the number of k-palindromes in a string where k lies in range [0, n). 

A string is 1-palindrome(or simply palindrome) if and only if it reads the same backward as it reads forward. 
A string is k-palindrome (k > 1) if and only if : 

1. Its left half is equal to its right half. 
2. Its left half and right half should be non-empty (k - 1)-palindrome. The left half of a string of length 't' is its prefix of length ?t/2?, and the right half is the suffix of the same length. 

Note: Each substring is counted as many times as it appears in the string. For example, in the string "aaa" the substring "a" appears 3 times.

Examples :

Input : abba
Output : 6 1 0 0
Explanation : 
"6" 1-palindromes = "a", "b", "b", "a", "bb", "abba".
"1" 2-palindrome = "bb". Because "b" is 1-palindrome 
and "bb" has both left and right parts equal. 
"0" 3-palindrome and 4-palindrome.

Input : abacaba
Output : 12 4 1 0 0 0 0
Explanation : 
"12" 1-palindromes = "a", "b", "a", "c", "a", "b", 
"a", "aba", "aca", "aba", "bacab", "abacaba".
"4" 2-palindromes = "aba", "aca", "aba", "abacaba". 
Because "a" and "aba" are 1-palindromes.

NOTE :  "bacab" is not 2-palindrome as "ba" 
is not 1-palindrome. "1" 3-palindrome = "abacaba". 
Because is "aba" is 2-palindrome. "0" other 
k-palindromes where 4 < = k < = 7.

Approach: 

Take a string s and say it is a 1-palindrome and checking its left half also turns out to be a 1-palindrome then, obviously its right part will always be equal to the left part (as the string is also a 1-palindrome) making the original string to be 2-palindrome. 

Now, similarly, checking the left half of the string, it also turns out to be a 1-palindrome then it will make the left half to be 2-palindrome and hence, making the original string to be 3-palindrome and so on checking it till only 1 alphabet is left or the part is not a 1-palindrome. 

Let's take string = "abacaba". As known "abacaba" is 1-palindrome. So, when checking for its left half "aba", which is also a 1-palindrome, it makes "abacaba" a 2-palindrome. Then, check for "aba"'s left half "a" which is also a 1-palindrome. So it makes "aba" a 2-palindrome and "aba" makes "abacaba" a 3-palindrome. So in other words, if a string is a k-palindrome then it is also a (k-1)-palindrome, (k-2)-palindrome, and so on till 1-palindrome. Code for the same is given below which firsts check each and every substring if it is a palindrome or not and then counts the number of k-palindromic substrings recursively in logarithmic time. 

Below is the implementation of above approach :  

12 4 1 0 0 0 0 

Complexity Analysis:

• Time Complexity : O() 
• Auxiliary Space : O()