Count All Palindromic Subsequence in a given String

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 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

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)

Recursive Approach

Here is the recursive code to the above approach

Java

// Java code to Count Palindromic Subsequence
// in a given String
public class GFG {
    // Function return the total palindromic
    // subsequence
    static int solve(String str, int i, int j)
    {
        if (i == j) //  base case when index is same
            return 1;

        if (i > j)
            return 0;

        if (str.charAt(i) == str.charAt(j)) {

            return 1 + solve(str, i + 1, j)
                + solve(str, i, j - 1);
        }

        else
            return solve(str, i + 1, j)
                + solve(str, i, j - 1)
                - solve(str, i + 1, j - 1);
    }

    // Driver program
    public static void main(String args[])
    {
        String str = "abcb";
        System.out.println(
            "Total palindromic "
            + "subsequence are : "
            + solve(str, 0, str.length() - 1));
    }
}
// This code is contributed by Ankit Jha

Output

Total palindromic subsequence are : 6

If we draw recursion tree of the 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 <cstring>
#include <iostream>
using namespace std;

// Function return the total palindromic subsequence
int 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[0][N - 1];
}

// Driver program
int main()
{
    string str = "abcb";
    cout << "Total palindromic subsequence are : "
         << countPS(str) << endl;
    return 0;
}

Java

// Java code to Count Palindromic Subsequence
// in a given String
public 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[0][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[0][N - 1]


# Driver program
str = "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 String
using 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.

PHP

<?php
// Counts Palindromic Subsequence in
// a given String

// Function return the total
// palindromic subsequence
function countPS($str)
{
    $N = strlen($str);

    // create a 2D array to store the
    // count of palindromic subsequence
    $cps = array_fill(0, $N + 1,
           array_fill(0, $N + 1, NULL));

    // palindromic subsequence of length 1
    for ($i = 0; $i < $N; $i++)
        $cps[$i][$i] = 1;

    // check subsequence of length L
    // is palindrome or not
    for ($L = 2; $L <= $N; $L++)
    {
        for ($i = 0; $i <= $N-$L; $i++)
        {
            $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 $cps[0][$N - 1];
}

// Driver Code
$str = "abcb";
echo "Total palindromic subsequence are : " .
                        countPS($str) . "\n";

// This code is contributed by ita_c
?>

Javascript

<script>

// Javascript code to Count Palindromic Subsequence
// in a given String

    // Function return the total palindromic
    // subsequence
    function countPS(str)
    {
        let N = str.length;

        // create a 2D array to store the count
        // of palindromic subsequence
        let cps = new Array(N);
        for(let i=0;i<N;i++)
        {
            cps[i]=new Array(N);
            for(let j=0;j<N;j++)
            {
                cps[i][j]=0;
            }
        }

        // palindromic subsequence of length 1
        for (let i = 0; i < N; i++)
            cps[i][i] = 1;

        // check subsequence of length L is
        // palindrome or not
        for (let L = 2; L <= N; L++) {
            for (let i = 0; i <= N-L; i++) {
                let 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[0][N - 1];
    }

    // Driver program
    let str = "abcb";
    document.write("Total palindromic "
                           + "subsequence are : "
                           + countPS(str));

    // This code is contributed by avanitrachhadiya2155

</script>

Output

Total palindromic subsequence are : 6

Time Complexity: O(N²)
Auxiliary Space: O(N²)

Another approach: (Using memoization)

C++

// C++ program to counts Palindromic Subsequence
// in a given String using recursion
#include <bits/stdc++.h>
using namespace std;

int n, dp[1000][1000];
string str = "abcb";

// Function return the total
// palindromic subsequence
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
int 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[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.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]

    else if (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 code
if __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 recursion
using 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

<?php
// PHP program to counts Palindromic Subsequence
// in a given String using recursion
$dp = array_fill(0, 100,
      array_fill(0, 1000, -1));

$str = "abcb";
$n = strlen($str);

// Function return the total
// palindromic subsequence
function countPS($i, $j)
{
    global $str, $dp, $n;

    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
echo "Total palindromic subsequence are : " .
                          countPS(0, $n - 1);

// This code is contributed by mits
?>

Javascript

<script>
// Javascript program to counts Palindromic Subsequence
// in a given String using recursion

    let n;
    let dp=new Array(1000);
    for(let i=0;i<1000;i++)
    {
        dp[i]=new Array(1000);
        for(let j=0;j<1000;j++)
        {
            dp[i][j]=-1;
        }
    }


    let str = "abcb";

    // Function return the total
    // palindromic subsequence
    function countPS(i,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
     n = str.length;
     document.write("Total palindromic subsequence"
                           + "are : " + countPS(0, n - 1));

    // This code is contributed by rag2127

</script>

Output

Total palindromic subsequence are : 6

Time Complexity : O(N²),
Auxiliary Space: O(N²)

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