Count Different Palindromic Subsequences

This approach involves generating every possible non-empty subsequence of the given string s. For each generated subsequence, we check if it is a palindrome. To count only the unique palindromic subsequences, we store them in a hash set. The final answer is the size of this set.

• Initialize an empty HashSet<String> called palindromes.
• Define a recursive function generate(index, currentSubsequence).
• Base Case: If index reaches the end of the string:
  • If currentSubsequence is not empty and is a palindrome, add it to the palindromes set.
  • Return.
• Recursive Step:
  • Call generate(index + 1, currentSubsequence) (character at index is not included).
  • Call generate(index + 1, currentSubsequence + s.charAt(index)) (character at index is included).
• Start the process by calling generate(0, "").
• The answer is palindromes.size(). Note that this approach is not feasible for the given constraints and does not handle the modulo arithmetic.

We can use a recursive helper function to generate all subsequences. The function would take the current index and the subsequence built so far. At each index i, we have two choices: either include the character s[i] in the current subsequence or not. This leads to 2^N total subsequences. The base case for the recursion is when we have traversed the entire string. At this point, if the generated subsequence is non-empty, we check if it's a palindrome. A palindrome check for a string of length k can be done in O(k) time. A HashSet<String> is used to store the unique palindromic subsequences found. The final result is the size of the set. This method is too slow for the problem's constraints but serves as a conceptual starting point.

import java.util.HashSet;
import java.util.Set;

class Solution {
    Set<String> palindromes = new HashSet<>();
    String s;
    int n;

    // This method is for illustration and will Time Limit Exceed.
    public int countPalindromicSubsequences(String s) {
        this.s = s;
        this.n = s.length();
        generate(0, new StringBuilder());
        return palindromes.size();
    }

    private void generate(int index, StringBuilder current) {
        if (index == n) {
            if (current.length() > 0 && isPalindrome(current.toString())) {
                palindromes.add(current.toString());
            }
            return;
        }

        // Exclude s.charAt(index)
        generate(index + 1, current);

        // Include s.charAt(index)
        current.append(s.charAt(index));
        generate(index + 1, current);
        current.deleteCharAt(current.length() - 1); // backtrack
    }

    private boolean isPalindrome(String str) {
        int left = 0, right = str.length() - 1;
        while (left < right) {
            if (str.charAt(left) != str.charAt(right)) {
                return false;
            }
            left++;
            right--;
        }
        return true;
    }
}