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 same the 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 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 0Explanation :"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 0Explanation :"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-pallindromes 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 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 left half to be 2-palindrome and hence, making 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.

Lets 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 :**

// A C++ program which counts different // palindromic characteristics of a string. #include <bits/stdc++.h> using namespace std; const int MAX_STR_LEN = 1000; bool P[MAX_STR_LEN][MAX_STR_LEN]; int Kpal[MAX_STR_LEN]; // A C++ function which checks whether a // substring str[i..j] of a given string // is a palindrome or not. void checkSubStrPal(string str, int n) { // P[i][j] = true if substring str // [i..j] is palindrome, else false memset(P, false, sizeof(P)); // palindrome of single length for (int i = 0; i < n; i++) P[i][i] = true; // palindrome of length 2 for (int i = 0; i < n - 1; i++) if (str[i] == str[i + 1]) P[i][i + 1] = true; // Palindromes of length more then 2. // This loop is similar to Matrix Chain // Multiplication. We start with a gap of // length 2 and fill P table in a way that // gap between starting and ending indexes // increases one by one by outer loop. for (int gap = 2; gap < n; gap++) { // Pick starting point for current gap for (int i = 0; i < n - gap; i++) { // Set ending point int j = gap + i; // If current string is palindrome if (str[i] == str[j] && P[i + 1][j - 1]) P[i][j] = true; } } } // A C++ function which recursively // counts if a string str [i..j] is // a k-palindromic string or not. void countKPalindromes(int i, int j, int k) { // terminating condition for a // string which is a k-palindrome. if (i == j) { Kpal[k]++; return; } // terminating condition for a // string which is not a k-palindrome. if (P[i][j] == false) return; // increases the counter for the // string if it is a k-palindrome. Kpal[k]++; // mid is middle pointer of // the string str [i...j]. int mid = (i + j) / 2; // if length of string which is // (j - i + 1) is odd than we have // to subtract one from mid. // else if even then no change. if ((j - i + 1) % 2 == 1) mid--; // if the string is k-palindrome // then we check if it is a // (k+1) - palindrome or not by // just sending any of one half of // the string to the Count_k_Palindrome // function. countKPalindromes(i, mid, k + 1); } void printKPalindromes(string s) { // Count of k-palindromes is equal // to zero initially. memset(Kpal, 0, sizeof(Kpal)); // Finding all palindromic // substrings of given string int n = s.length(); checkSubStrPal(s, n); // counting k-palindromes for each and // every substring of given string. . for (int i = 0; i < n; i++) for (int j = 0; j < n - i; j++) countKPalindromes(j, j + i, 1); // Output the number of K-palindromic // substrings of a given string. for (int i = 1; i <= n; i++) cout << Kpal[i] << " "; cout << "\n"; } // Driver code int main() { string s = "abacaba"; printKPalindromes(s); return 0; }

**Output:**

12 4 1 0 0 0 0

**Time Complexity : ** O()

**Auxiliary Space : ** O()

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