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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 <bits/stdc++.h>
using namespace std;
 
// Function to count different
// palindromic string of length X
// from the given string S
long 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<int> 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 code
int 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 S
import 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 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];
 
    // 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<int> se;
    Set<Integer> 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 code
public 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 S
def 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 code
if __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 S
using 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<int> se;
        HashSet<int> se = new HashSet<int>();
        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




<script>
 
// Javascript 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 S
function findways(s, x)
{
    // Base case
    if (x > s.length)
        return 0;
 
    var n = s.length;
 
    // Create the frequency array
    var freq = Array(26).fill(0);
 
    // Count the frequency in the string
    for (var i = 0; i < n; ++i)
        freq[s[i].charCodeAt(0) - 'a'.charCodeAt(0)]++;
 
    var se = [];
 
    for (var i = 0; i < 26; ++i)
        if (freq[i] > 0)
            // Store frequency of the char
            se.push(freq[i]);
 
    var ans = 1;
 
    for (var i = 0; i < x / 2; ++i) {
        var count = 0;
 
        se.forEach(u => {
             
            // 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;
 
        se.sort((a,b)=>a-b)
        var val = se[se.length-1]
        // Iterator pointing to the
        // last element of the set
        se.pop();
 
        if (val > 2)
            // decrease the value of the char
            // we put on the position i and n - i
            se.push(val - 2);
    }
 
    if (x % 2 != 0) {
        var count = 0;
 
        se.forEach(u => {
             
            // 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
var s = "aaa";
var x = 2;
document.write( findways(s, x));
 
// This code is contributed by noob2000.
</script>
Output: 
1

 

Time Complexity: O(N + 26 * X)
 

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