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 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;       // 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 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;       // 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 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 =  * 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;           // 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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