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++

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// C++ Program to implement
// the above approach
#include <bits/stdc++.h>
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;
}

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Java

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// 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

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Python3

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# 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

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C#

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// 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

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Output: 

15

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

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Improved By : sapnasingh4991, code_hunt