Home » Interview coding problems/challenges

Count the number of palindromic subsequences in a given string

Here, we are going to learn how to count the number of palindromic subsequences in a given string using Dynamic programming? Submitted by Souvik Saha, on March 31, 2020

Problem statement

Given a string you have to count the total number of palindromic subsequences in the giving string and print the value.

    Input:
    T Test case
    T no of input string will be given to you.

    E.g.
    3
    
    abc
    aa
    abcc
    
    Constrain 
    1≤ length (string) ≤100
    
    Output:
    Print the count of the palindromic sub sequences from the given string.

Example

    T=3

    Input:
    abc
    
    Output:
    3 ("a", "b", "c")
    
    Input:
    aa
    
    Output:
    3 ("a", "a", "aa")
    
    Input:
    abcc
    
    Output:
    5 ("a", "b", "c", "c", "cc")

Explanation with example:

Let there is a string str.

Now possible arrangements are:

1. Single middle characters like aba
2. Paired middle characters like bb

Let, f(a,b) = count of number of palindromic subsequences from index a to index b.

Considering the above two facts:

1. If there is only one character then f(a,a) = 1
2. If there is a substring starting from index a to index b then,
   1. If str[a] and str[b] both are same then
      f(a,b)=f(a+1,b)+f(a,b-1)+1
   2. If str[a] and str[b] both are not same then
      f(a,b)=f(a+1,b)+f(a,b-1)-f(a+1,b-1)

For, str = abbaa

From the above, it is understandable that one function is called repeatedly so for the large input the repetition will be very high. Because of this problem we are using a Dynamic programming approach to avoid repetition. We are using the memorization process to solve the problem.

Problem solution

Recursive algorithm:

int Function(str,pos,l):
	if str.length()== l
		return
	for a=pos to str.length()-l 
		if str[a]==str[a+l-1]
			return Function(str,a,l-1)+Function(str,a+1,l-1)+1
		else
			return Function(str,a,l-1)+Function(str,a+1,l-1)-Function(str,a+1,l-2)

DP conversion:

int Function(str,n):
	for len=1 to str.length()
		for a=0 to str.length()-len
			b=pos+len-1
			if str[a]==str[b]
				arr[a][b]=arr[a+1][b]+ arr[a][b-1]+1
			else
				arr[a][b]=arr[a+1][b]+arr[a][b-1]-arr[a+1][b-1]
	return arr[0][len-1]

C++ Implementation

#include <bits/stdc++.h>
using namespace std;

int count_palindrome(string str)
{
    int len = str.length();
    int arr[len][len] = { 0 };
    for (int i = 0; i < len; i++) {
        arr[i][i] = 1;
    }
    for (int l = 2; l <= len; l++) {
        for (int i = 0; i <= len - l; i++) {
            int j = i + l - 1;
            if (str[i] == str[j]) {
                arr[i][j] = arr[i + 1][j] + arr[i][j - 1] + 1;
            }
            else {
                arr[i][j] = arr[i + 1][j] + arr[i][j - 1] - arr[i + 1][j - 1];
            }
        }
    }
    return arr[0][len - 1];
}

int main()
{
    int t;
    cout << "Test Case : ";
    cin >> t;
    while (t--) {
        cout << "Enter the string : ";
        string str;
        cin >> str;
        cout << "Number of palindromic subsequences are : " << count_palindrome(str) << endl;
    }

    return 0;
}

Output

Test Case : 3
Enter the string : abc
Number of palindromic subsequences are : 3
Enter the string : aaa
Number of palindromic subsequences are : 7
Enter the string : abcc
Number of palindromic subsequences are : 5