# Count of Palindromic Strings possible by swapping of a pair of Characters

Given a palindromic string S, the task is to find the count of palindromic strings possible by swapping a pair of character at a time.
Examples:

Input: s = “abba”
Output:
Explanation:
1st Swap: abba -> abba
2nd Swap: abba -> abb
All other swaps will lead to a non-palindromic string.
Therefore, the count of possible strings is 2.
Input: s = “aaabaaa”
Output: 15

Naive Approach:
The simplest approach to solve the problem is to generate all possible pair of characters from the given string and for each pair if swapping them generates a palindromic string or not. If found to be true, increase count. Finally, print the value of count
Time Complexity: O(N3
Auxiliary Space: O(1)
Efficient Approach:
To optimize the above-mentioned approach, calculate the frequencies of each character in the string. For the string to remain a palindrome, only the same character can be swapped in the string.
Follow the steps below to solve the problem:

• Traverse the string.
• For every ith character, increase count with the current frequency of the character. This increases the number of swaps the current character can make with its previous occurrences.
• Increase the frequency of the ith character.
• Finally, after complete traversal of the string, print count.

Below is the implementation of above approach:

## C++

 `// C++ Program to implement` `// the above approach` `#include ` `using` `namespace` `std;`   `// Function to return the count of` `// possible palindromic strings` `long` `long` `findNewString(string s)` `{`   `    ``long` `long` `ans = 0;`   `    ``// Stores the frequencies` `    ``// of each character` `    ``int` `freq[26];`   `    ``// Stores the length of` `    ``// the string` `    ``int` `n = s.length();`   `    ``// Initialize frequencies` `    ``memset``(freq, 0, ``sizeof` `freq);`   `    ``for` `(``int` `i = 0; i < (``int``)s.length(); ++i) {`   `        ``// Increase the number of swaps,` `        ``// the current character make with` `        ``// its previous occurrences` `        ``ans += freq[s[i] - ``'a'``];`   `        ``// Increase frequency` `        ``freq[s[i] - ``'a'``]++;` `    ``}`   `    ``return` `ans;` `}`   `// Driver Code` `int` `main()` `{` `    ``string s = ``"aaabaaa"``;` `    ``cout << findNewString(s) << ``'\n'``;`   `    ``return` `0;` `}`

## Java

 `// Java Program to implement` `// the above approach` `import` `java.util.*;` `class` `GFG{`   `// Function to return the count of` `// possible palindromic Strings` `static` `long` `findNewString(String s)` `{` `    ``long` `ans = ``0``;`   `    ``// Stores the frequencies` `    ``// of each character` `    ``int` `[]freq = ``new` `int``[``26``];`   `    ``// Stores the length of` `    ``// the String` `    ``int` `n = s.length();`   `    ``// Initialize frequencies` `    ``Arrays.fill(freq, ``0``);`   `    ``for` `(``int` `i = ``0``; i < (``int``)s.length(); ++i)` `    ``{`   `        ``// Increase the number of swaps,` `        ``// the current character make with` `        ``// its previous occurrences` `        ``ans += freq[s.charAt(i) - ``'a'``];`   `        ``// Increase frequency` `        ``freq[s.charAt(i) - ``'a'``]++;` `    ``}` `    ``return` `ans;` `}`   `// Driver Code` `public` `static` `void` `main(String[] args)` `{` `    ``String s = ``"aaabaaa"``;` `    ``System.out.print(findNewString(s));` `}` `}`   `// This code is contributed by sapnasingh4991`

## C#

 `// C# Program to implement` `// the above approach` `using` `System;` `class` `GFG{`   `// Function to return the count of` `// possible palindromic Strings` `static` `long` `findNewString(String s)` `{` `    ``long` `ans = 0;`   `    ``// Stores the frequencies` `    ``// of each character` `    ``int` `[]freq = ``new` `int``[26];`   `    ``// Stores the length of` `    ``// the String` `    ``int` `n = s.Length;`   `    ``for` `(``int` `i = 0; i < (``int``)s.Length; ++i)` `    ``{`   `        ``// Increase the number of swaps,` `        ``// the current character make with` `        ``// its previous occurrences` `        ``ans += freq[s[i] - ``'a'``];`   `        ``// Increase frequency` `        ``freq[s[i] - ``'a'``]++;` `    ``}` `    ``return` `ans;` `}`   `// Driver Code` `public` `static` `void` `Main(String[] args)` `{` `    ``String s = ``"aaabaaa"``;` `    ``Console.Write(findNewString(s));` `}` `}`   `// This code is contributed by sapnasingh4991`

Output:

```15

```

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

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Improved By : sapnasingh4991