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# Count of unique palindromic strings of length X from given string

• Last Updated : 07 Jun, 2021

Given a string s and an integer X, our task is to find the number of distinct palindromic strings of length X from the given string.

Examples:

Input: s = “aaa”, X = 2
Output:
Explanation:
Here all the characters of the string is the same so we can only make a one different string {aa} of length 2.

Input: s = “aabaabbc”, X = 3
Output:
Explanation:
To make a palindromic string of length 3 we have 6 possible strings aaa, aba, aca, bab, bcb, bbb.

Naive Approach: The naive method is to generate all the possible subsequence of length X, then check if that subsequence forms a palindrome or not.

Time Complexity: O(2N)
Auxiliary Space: O(X)

Efficient Approach: The idea is to find the frequency of all the characters of the string. Count the different numbers of the characters we can put at the particular position and decrease the number of count by 2 because for palindromic string the position (i) and (X – i) will be the same. If the length of X is odd then we have to put the only single char at that position X/2.

Below is the implementation of the above approach:

## C++

 // C++ implementation to count different// palindromic string of length X// from the given string S #include using namespace std; // Function to count different// palindromic string of length X// from the given string Slong long findways(string s, int x){    // Base case    if (x > (int)s.length())        return 0;     long long int n = (int)s.length();     // Create the frequency array    int freq[26];     // Intitalise frequency array with 0    memset(freq, 0, sizeof freq);     // Count the frequency in the string    for (int i = 0; i < n; ++i)        freq[s[i] - 'a']++;     multiset se;     for (int i = 0; i < 26; ++i)        if (freq[i] > 0)            // Store frequency of the char            se.insert(freq[i]);     long long ans = 1;     for (int i = 0; i < x / 2; ++i) {        long long int count = 0;        for (auto u : se) {            // check the frequency which            // is greater than zero            if (u >= 2)                 // No. of different char we can                // put at the position of                // the i and x - i                count++;        }         if (count == 0)            return 0;         else            ans = ans * count;         // Iterator pointing to the        // last element of the set        auto p = se.end();        p--;        int val = *p;        se.erase(p);        if (val > 2)            // decrease the value of the char            // we put on the position i and n - i            se.insert(val - 2);    }     if (x % 2 != 0) {        long long int count = 0;        for (auto u : se)            // different no of char we can            // put at the position x/2            if (u > 0)                count++;         ans = ans * count;    }     // Return total no of    // different string    return ans;} // Driver codeint main(){    string s = "aaa";    int x = 2;    cout << findways(s, x);     return 0;}

## Java

 // Java implementation to count different// palindromic string of length X from the// given string Simport java.util.Arrays;import java.util.HashSet;import java.util.Set; class GFG{ // Function to count different// palindromic string of length X// from the given string Sstatic int findways(String s, int x){         // Base case    if (x > s.length())        return 0;     int n = s.length();     // Create the frequency array    int[] freq = new int[26];     // Intitalise frequency array with 0    Arrays.fill(freq, 0);     // Count the frequency in the string    for(int i = 0; i < n; ++i)        freq[s.charAt(i) - 'a']++;     // multiset se;    Set se = new HashSet<>();     for(int i = 0; i < 26; ++i)        if (freq[i] > 0)                     // Store frequency of the char            se.add(freq[i]);     int ans = 1;     for(int i = 0; i < x / 2; ++i)    {        int count = 0;        for(int u : se)        {                         // Check the frequency which            // is greater than zero            if (u >= 2)                 // No. of different char we can                // put at the position of                // the i and x - i                count++;        }         if (count == 0)            return 0;        else            ans = ans * count;         // Iterator pointing to the        // last element of the set        int p = (int)se.toArray()[se.size() - 1];        int val = p;        se.remove(p);                 if (val > 2)                     // Decrease the value of the char            // we put on the position i and n - i            se.add(val - 2);    }     if (x % 2 != 0)    {        int count = 0;                 for(int u : se)                     // Different no of char we can            // put at the position x/2            if (u > 0)                count++;         ans = ans * count;    }     // Return total no of    // different string    return ans;} // Driver codepublic static void main(String[] args){    String s = "aaa";    int x = 2;         System.out.println(findways(s, x));}} // This code is contributed by sanjeev2552

## Python3

 # Python3 implementation to count# different palindromic string of# length X from the given string S # Function to count different# palindromic string of length X# from the given string Sdef findways(s, x):     # Base case    if(x > len(s)):        return 0     n = len(s)     # Create the frequency array    # Intitalise frequency array with 0    freq = [0] * 26     # Count the frequency in the string    for i in range(n):        freq[ord(s[i]) - ord('a')] += 1     se = set()     for i in range(26):        if(freq[i] > 0):                         # Store frequency of the char            se.add(freq[i])     ans = 1    for i in range(x // 2):        count = 0        for u in se:                         # Check the frequency which            # is greater than zero            if(u >= 2):                 # No. of different char we can                # put at the position of                # the i and x - i                count += 1         if(count == 0):            return 0        else:            ans *= count         # Iterator pointing to the        # last element of the set        p = list(se)        val = p[-1]        p.pop(-1)        se = set(p)         if(val > 2):                         # Decrease the value of the char            # we put on the position i and n - i            se.add(val - 2)     if(x % 2 != 0):        count = 0        for u in se:                         # Different no of char we can            # put at the position x/2            if(u > 0):                count += 1         ans = ans * count     # Return total no of    # different string    return ans # Driver codeif __name__ == '__main__':     s = "aaa"    x = 2         print(findways(s, x)) # This code is contributed by Shivam Singh

## C#

 // C# implementation to count different// palindromic string of length X from the// given string Susing System;using System.Collections.Generic; class GFG {     // Function to count different    // palindromic string of length X    // from the given string S    static int findways(string s, int x)    {         // Base case        if (x > s.Length)            return 0;        int n = s.Length;         // Create the frequency array        int[] freq = new int[26];         // Count the frequency in the string        for (int i = 0; i < n; ++i)            freq[s[i] - 'a']++;         // multiset se;        HashSet se = new HashSet();        for (int i = 0; i < 26; ++i)            if (freq[i] > 0)                 // Store frequency of the char                se.Add(freq[i]);         int ans = 1;        for (int i = 0; i < x / 2; ++i)        {            int count = 0;            foreach(int u in se)            {                 // Check the frequency which                // is greater than zero                if (u >= 2)                     // No. of different char we can                    // put at the position of                    // the i and x - i                    count++;            }             if (count == 0)                return 0;            else                ans = ans * count;             // Iterator pointing to the            // last element of the set            int[] arr = new int[se.Count];            se.CopyTo(arr);            int p = arr[se.Count - 1];            int val = p;            se.Remove(p);            if (val > 2)                 // Decrease the value of the char                // we put on the position i and n - i                se.Add(val - 2);        }         if (x % 2 != 0)        {            int count = 0;             foreach(int u in se)                 // Different no of char we can                // put at the position x/2                if (u > 0) count++;            ans = ans * count;        }         // Return total no of        // different string        return ans;    }   // Driver code    static public void Main()    {        string s = "aaa";        int x = 2;        Console.WriteLine(findways(s, x));    }} // This code is contributed by dharanendralv23

## Javascript


Output:
1

Time Complexity: O(N + 26 * X)

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