Count All Palindromic Subsequence in a given String

Find how many palindromic subsequence (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 Subprolems. 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<iostream>
#include<cstring>
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; 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 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+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; i++)
{
int k = L + i - 1;
if (k < N){
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 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
```

Output:

`Total palindromic subsequence are : 6`

Time Complexity : O(N2)

This article is contributed by Nishant_sing(pintu). If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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