# Find the count of palindromic sub-string of a string in its sorted form

Given string str consisting of lowercase English alphabets, the task is to find the total number of palindromic sub-strings present in the sorted form of str.

Examples:

Input: str = “acbbd”
Output:
All palindromic sub-string in it’s sorted form (“abbcd”) are “a”, “b”, “b”, “bb”, “c” and “d”.

Input: str = “abbabdbd”
Output: 16

Naive approach: One way is to sort the given string and then count the total number of sub-strings present which are palindromes. For finding a number of palindromic sub-strings this approach can be used which has a time complexity of O(n^2).

Optimized approach: An efficient way is to count the frequency of each character and then for each frequency total number of palindromes will (n*(n+1))/2 as all the palindromic sub-strings of a sorted string will consist of the same character.

For example, palindromic sub-string for the string “aabbbcd” will be “a”, “aa”, …, “bbb”, “c”, … etc. Time complexity for this approach will be O(n).

• Create a hash table for storing the frequencies of each character of the string str.
• Traverse the hash table and for each non-zero frequency add (hash[i] * (hash[i]+1)) / 2 to the sum.
• Print the sum in the end.

Below is the implementation of the above approach:

## C++

 `// CPP program to find the count of palindromic sub-string` `// of a string in it's ascending form` `#include ` `using` `namespace` `std;`   `const` `int` `MAX_CHAR = 26;`   `// function to return count of palindromic sub-string` `int` `countPalindrome(string str)` `{` `    ``int` `n = str.size();` `    ``int` `sum = 0;`   `    ``// calculate frequency` `    ``int` `hashTable[MAX_CHAR];` `    ``for` `(``int` `i = 0; i < n; i++)` `        ``hashTable[str[i] - ``'a'``]++;`   `    ``// calculate count of palindromic sub-string` `    ``for` `(``int` `i = 0; i < 26; i++) {` `        ``if` `(hashTable[i])` `            ``sum += (hashTable[i] * (hashTable[i] + 1) / 2);` `    ``}`   `    ``// return result` `    ``return` `sum;` `}`   `// driver program` `int` `main()` `{` `    ``string str = ``"ananananddd"``;`   `    ``cout << countPalindrome(str);` `    ``return` `0;` `}`

## Java

 `// Java program to find the count of palindromic sub-string ` `// of a string in it's ascending form `   `class` `GFG {`   `    ``final` `static` `int` `MAX_CHAR = ``26``;`   `// function to return count of palindromic sub-string ` `    ``static` `int` `countPalindrome(String str) {` `        ``int` `n = str.length();` `        ``int` `sum = ``0``;`   `        ``// calculate frequency ` `        ``int` `hashTable[] = ``new` `int``[MAX_CHAR];` `        ``for` `(``int` `i = ``0``; i < n; i++) {` `            ``hashTable[str.charAt(i) - ``'a'``]++;` `        ``}`   `        ``// calculate count of palindromic sub-string ` `        ``for` `(``int` `i = ``0``; i < ``26``; i++) {` `            ``if` `(hashTable[i] != ``0``) {` `                ``sum += (hashTable[i] * (hashTable[i] + ``1``) / ``2``);` `            ``}` `        ``}`   `        ``// return result ` `        ``return` `sum;` `    ``}`   `// driver program ` `    ``public` `static` `void` `main(String[] args) {` `        ``String str = ``"ananananddd"``;`   `        ``System.out.println(countPalindrome(str));`   `    ``}` `}`

## Python3

 `# Python3 program to find the count of ` `# palindromic sub-string of a string ` `# in it's ascending form` `MAX_CHAR ``=` `26`   `# function to return count of ` `# palindromic sub-string` `def` `countPalindrome(``str``):`   `    ``n ``=` `len` `(``str``)` `    ``sum` `=` `0`   `    ``# calculate frequency` `    ``hashTable ``=` `[``0``] ``*` `MAX_CHAR` `    ``for` `i ``in` `range``(n):` `        ``hashTable[``ord``(``str``[i]) ``-` `                  ``ord``(``'a'``)] ``+``=` `1`   `    ``# calculate count of palindromic` `    ``# sub-string` `    ``for` `i ``in` `range``(``26``) :` `        ``if` `(hashTable[i]):` `            ``sum` `+``=` `(hashTable[i] ``*` `                   ``(hashTable[i] ``+` `1``) ``/``/` `2``)`   `    ``# return result` `    ``return` `sum`   `# Driver Code` `if` `__name__ ``=``=` `"__main__"``:`   `    ``str` `=` `"ananananddd"`   `    ``print` `(countPalindrome(``str``))`   `# This code is contributed by ita_c`

## C#

 `// C# program to find the count of palindromic sub-string ` `// of a string in it's ascending form ` `using` `System;` `                    `  `public` `class` `GFG{` ` `  `    ``readonly` `static` `int` `MAX_CHAR = 26;` ` `  `// function to return count of palindromic sub-string ` `    ``static` `int` `countPalindrome(String str) {` `        ``int` `n = str.Length;` `        ``int` `sum = 0;` ` `  `        ``// calculate frequency ` `        ``int` `[]hashTable = ``new` `int``[MAX_CHAR];` `        ``for` `(``int` `i = 0; i < n; i++) {` `            ``hashTable[str[i] - ``'a'``]++;` `        ``}` ` `  `        ``// calculate count of palindromic sub-string ` `        ``for` `(``int` `i = 0; i < 26; i++) {` `            ``if` `(hashTable[i] != 0) {` `                ``sum += (hashTable[i] * (hashTable[i] + 1) / 2);` `            ``}` `        ``}` ` `  `        ``// return result ` `        ``return` `sum;` `    ``}` ` `  `// driver program ` `    ``public` `static` `void` `Main() {` `        ``String str = ``"ananananddd"``;` ` `  `        ``Console.Write(countPalindrome(str));` ` `  `    ``}` `}`   `// This code is contributed by Rajput-Ji`

## PHP

 ``

## Javascript

 ``

Output

`32626385`

Complexity Analysis:

• Time Complexity: O(N), as we are using a loop to traverse N times. Where N is the length of the string.
• Auxiliary Space: O(26), as we are using extra space for the hash table.

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