Count All Palindromic Subsequence in a given String

• Difficulty Level : Hard
• Last Updated : 14 Apr, 2021

Find how many palindromic subsequences (need not necessarily be distinct) can be formed in a given string. Note that the empty string is not considered as a palindrome.
Examples:

Input : str = "abcd"
Output : 4
Explanation :- palindromic  subsequence are : "a" ,"b", "c" ,"d"

Input : str = "aab"
Output : 4
Explanation :- palindromic subsequence are :"a", "a", "b", "aa"

Input : str = "aaaa"
Output : 15

Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

The above problem can be recursively defined.

Initial Values : i= 0, j= n-1;

CountPS(i,j)
// Every single character of a string is a palindrome
// subsequence
if i == j
return 1 // palindrome of length 1

// If first and last characters are same, then we
// consider it as palindrome subsequence and check
// for the rest subsequence (i+1, j), (i, j-1)
Else if (str[i] == str[j)]
return   countPS(i+1, j) + countPS(i, j-1) + 1;

else
// check for rest sub-sequence and  remove common
// palindromic subsequences as they are counted
// twice when we do countPS(i+1, j) + countPS(i,j-1)
return countPS(i+1, j) + countPS(i, j-1) - countPS(i+1, j-1) If we draw recursion tree of above recursive solution, we can observe overlapping Subproblems. Since the problem has overlapping subproblems, we can solve it efficiently using Dynamic Programming. Below is Dynamic Programming based solution.

C++

 // Counts Palindromic Subsequence in a given String#include #include using namespace std; // Function return the total palindromic subsequenceint countPS(string str){    int N = str.length();     // create a 2D array to store the count of palindromic    // subsequence    int cps[N + 1][N + 1];    memset(cps, 0, sizeof(cps));     // palindromic subsequence of length 1    for (int i = 0; i < N; i++)        cps[i][i] = 1;     // check subsequence of length L is palindrome or not    for (int L = 2; L <= N; L++) {        for (int i = 0; i <= N-L; i++) {            int k = L + i - 1;            if (str[i] == str[k])                cps[i][k]                    = cps[i][k - 1] + cps[i + 1][k] + 1;            else                cps[i][k] = cps[i][k - 1] + cps[i + 1][k]                            - cps[i + 1][k - 1];        }    }     // return total palindromic subsequence    return cps[N - 1];} // Driver programint main(){    string str = "abcb";    cout << "Total palindromic subsequence are : "         << countPS(str) << endl;    return 0;}

Java

 // Java code to Count Palindromic Subsequence// in a given Stringpublic class GFG {    // Function return the total palindromic    // subsequence    static int countPS(String str)    {        int N = str.length();         // create a 2D array to store the count        // of palindromic subsequence        int[][] cps = new int[N][N];         // palindromic subsequence of length 1        for (int i = 0; i < N; i++)            cps[i][i] = 1;         // check subsequence of length L is        // palindrome or not        for (int L = 2; L <= N; L++) {            for (int i = 0; i <= N-L; i++) {                int k = L + i - 1;              if (str.charAt(i) == str.charAt(k)) {                cps[i][k] = cps[i][k - 1]                                    + cps[i + 1][k] + 1;              }else{                cps[i][k] = cps[i][k - 1]                                    + cps[i + 1][k]                                    - cps[i + 1][k - 1];              }            }        }         // return total palindromic subsequence        return cps[N - 1];    }     // Driver program    public static void main(String args[])    {        String str = "abcb";        System.out.println("Total palindromic "                           + "subsequence are : "                           + countPS(str));    }}// This code is contributed by Sumit Ghosh

Python3

 # Python3 code to Count Palindromic# Subsequence in a given String # Function return the total# palindromic subsequence  def countPS(str):     N = len(str)     # Create a 2D array to store the count    # of palindromic subsequence    cps = [[0 for i in range(N + 2)]for j in range(N + 2)]     # palindromic subsequence of length 1    for i in range(N):        cps[i][i] = 1     # check subsequence of length L    # is palindrome or not    for L in range(2, N + 1):         for i in range(N):            k = L + i - 1            if (k < N):                if (str[i] == str[k]):                    cps[i][k] = (cps[i][k - 1] +                                 cps[i + 1][k] + 1)                else:                    cps[i][k] = (cps[i][k - 1] +                                 cps[i + 1][k] -                                 cps[i + 1][k - 1])     # return total palindromic subsequence    return cps[N - 1]  # Driver programstr = "abcb"print("Total palindromic subsequence are : ", countPS(str))  # This code is contributed by Anant Agarwal.

C#

 // C# code to Count Palindromic Subsequence// Subsequence in a given Stringusing System; class GFG {     // Function return the total    // palindromic subsequence    static int countPS(string str)    {        int N = str.Length;         // create a 2D array to store the        // count of palindromic subsequence        int[, ] cps = new int[N + 1, N + 1];         // palindromic subsequence        // of length 1        for (int i = 0; i < N; i++)            cps[i, i] = 1;         // check subsequence of length        // L is palindrome or not        for (int L = 2; L <= N; L++) {            for (int i = 0; i <= N-L; i++) {                int k = L + i - 1;                if (k < N) {                    if (str[i] == str[k])                        cps[i, k] = cps[i, k - 1]                                    + cps[i + 1, k] + 1;                    else                        cps[i, k] = cps[i, k - 1]                                    + cps[i + 1, k]                                    - cps[i + 1, k - 1];                }            }        }         // return total palindromic        // subsequence        return cps[0, N - 1];    }     // Driver Code    public static void Main()    {        string str = "abcb";        Console.Write("Total palindromic "                      + "subsequence are : "                      + countPS(str));    }} // This code is contributed by nitin mittal.



Javascript



Output:

Total palindromic subsequence are : 6

Time Complexity : O(N2)

Another approach: (Using recursion)

C++

 // C++ program to counts Palindromic Subsequence// in a given String using recursion#include using namespace std; int n, dp;string str = "abcb"; // Function return the total// palindromic subsequenceint countPS(int i, int j){     if (i > j)        return 0;     if (dp[i][j] != -1)        return dp[i][j];     if (i == j)        return dp[i][j] = 1;     else if (str[i] == str[j])        return dp[i][j]               = countPS(i + 1, j) +                countPS(i, j - 1) + 1;     else        return dp[i][j] = countPS(i + 1, j) +                          countPS(i, j - 1) -                          countPS(i + 1, j - 1);} // Driver codeint main(){    memset(dp, -1, sizeof(dp));    n = str.size();    cout << "Total palindromic subsequence are : "         << countPS(0, n - 1) << endl;    return 0;}// this code is contributed by Kushdeep Mittal

Java

 // Java program to counts Palindromic Subsequence// in a given String using recursion class GFG {    static int n;    static int[][] dp = new int;     static String str = "abcb";     // Function return the total    // palindromic subsequence    static int countPS(int i, int j)    {         if (i > j)            return 0;         if (dp[i][j] != -1)            return dp[i][j];             if (i == j)            return dp[i][j] = 1;         else if (str.charAt(i) == str.charAt(j))            return dp[i][j]                = countPS(i + 1, j) +                    countPS(i, j - 1) + 1;         else            return dp[i][j] = countPS(i + 1, j) +                              countPS(i, j - 1) -                              countPS(i + 1, j - 1);    }     // Driver code    public static void main(String[] args)    {        for (int i = 0; i < 1000; i++)            for (int j = 0; j < 1000; j++)                dp[i][j] = -1;         n = str.length();        System.out.println("Total palindromic subsequence"                           + "are : " + countPS(0, n - 1));    }} // This code is contributed by Ryuga

Python 3

 # Python 3 program to counts Palindromic# Subsequence in a given String using recursion str = "abcb" # Function return the total# palindromic subsequence  def countPS(i, j):     if(i > j):        return 0     if(dp[i][j] != -1):        return dp[i][j]     if(i == j):        dp[i][j] = 1        return dp[i][j]     elif (str[i] == str[j]):        dp[i][j] = (countPS(i + 1, j) +                    countPS(i, j - 1) + 1)        return dp[i][j]    else:        dp[i][j] = (countPS(i + 1, j) +                    countPS(i, j - 1) -                    countPS(i + 1, j - 1))        return dp[i][j]  # Driver codeif __name__ == "__main__":     dp = [[-1 for x in range(1000)]          for y in range(1000)]     n = len(str)    print("Total palindromic subsequence are :",          countPS(0, n - 1)) # This code is contributed by ita_c

C#

 // C# program to counts Palindromic Subsequence// in a given String using recursionusing System; class GFG {    static int n;    static int[, ] dp = new int[1000, 1000];     static string str = "abcb";     // Function return the total    // palindromic subsequence    static int countPS(int i, int j)    {         if (i > j)            return 0;         if (dp[i, j] != -1)            return dp[i, j];         if (i == j)            return dp[i, j] = 1;         else if (str[i] == str[j])            return dp[i, j]                = countPS(i + 1, j) +                countPS(i, j - 1) + 1;         else            return dp[i, j] = countPS(i + 1, j)                              + countPS(i, j - 1)                              - countPS(i + 1, j - 1);    }     // Driver code    static void Main()    {        for (int i = 0; i < 1000; i++)            for (int j = 0; j < 1000; j++)                dp[i, j] = -1;         n = str.Length;        Console.Write("Total palindromic subsequence"                      + "are : " + countPS(0, n - 1));    }} // This code is contributed by DrRoot_

PHP

 \$j)        return 0;         if(\$dp[\$i][\$j] != -1)        return \$dp[\$i][\$j];         if(\$i == \$j)        return \$dp[\$i][\$j] = 1;         else if (\$str[\$i] == \$str[\$j])        return \$dp[\$i][\$j] = countPS(\$i + 1, \$j) +                             countPS(\$i, \$j - 1) + 1;         else        return \$dp[\$i][\$j] = countPS(\$i + 1, \$j) +                             countPS(\$i, \$j - 1) -                             countPS(\$i + 1, \$j - 1);} // Driver codeecho "Total palindromic subsequence are : " .                          countPS(0, \$n - 1);         // This code is contributed by mits?>

Javascript



Output:

Total palindromic subsequence are : 6

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