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

Given a 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: 6
All palindromic sub-string in it’s sorted form (“abbcd”) are “a”, “b”, “b”, “bb”, “c” and “d”.

Input: str = “abbabdbd”
Output: 16

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 number of palindromic sub-strings this approach can be used which has 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

 ` `

Output:

```26
```

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