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

• Last Updated : 11 May, 2021

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;` `    ``// 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`

## Python3

 `# Python3 program to implement``# the above approach` `# Function to return the count of``# possible palindromic strings``def` `findNewString(s):` `    ``ans ``=` `0` `    ``# Stores the frequencies``    ``# of each character``    ``freq ``=` `[``0``] ``*` `26` `    ``# Stores the length of``    ``# the string``    ``n ``=` `len``(s)` `    ``for` `i ``in` `range``(n):` `        ``# Increase the number of swaps,``        ``# the current character make with``        ``# its previous occurrences``        ``ans ``+``=` `freq[``ord``(s[i]) ``-` `ord``(``'a'``)]` `        ``# Increase frequency``        ``freq[``ord``(s[i]) ``-` `ord``(``'a'``)] ``+``=` `1``    ` `    ``return` `ans` `# Driver Code``s ``=` `"aaabaaa"` `print``(findNewString(s))` `# This code is contributed by code_hunt`

## 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``;` `    ``// 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`

## Javascript

 ``
Output:

`15`

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

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