# Solving Leetcode Interviews in Seconds with AI: Maximum Product of the Length of Two Palindromic Subsequences


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "2002" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

	# Problem Statement
	> Given a string s, find two disjoint palindromic subsequences of s such that the product of their lengths is maximized. The two subsequences are disjoint if they do not both pick a character at the same index. Return the maximum possible product of the lengths of the two palindromic subsequences. A subsequence is a string that can be derived from another string by deleting some or no characters without changing the order of the remaining characters. A string is palindromic if it reads the same forward and backward.   Example 1:   Input: s = "leetcodecom" Output: 9 Explanation: An optimal solution is to choose "ete" for the 1st subsequence and "cdc" for the 2nd subsequence. The product of their lengths is: 3 * 3 = 9.  Example 2:  Input: s = "bb" Output: 1 Explanation: An optimal solution is to choose "b" (the first character) for the 1st subsequence and "b" (the second character) for the 2nd subsequence. The product of their lengths is: 1 * 1 = 1.  Example 3:  Input: s = "accbcaxxcxx" Output: 25 Explanation: An optimal solution is to choose "accca" for the 1st subsequence and "xxcxx" for the 2nd subsequence. The product of their lengths is: 5 * 5 = 25.    Constraints:  2 <= s.length <= 12 s consists of lowercase English letters only.  

	# Explanation
	Here's the breakdown of the approach, complexities, and the Python code:

*   **High-Level Approach:**
    *   **Bitmasking for Subsets:** Generate all possible subsets of indices of the given string `s` using bitmasking. Each bit in the mask represents whether an index is included in the subset or not.
    *   **Palindrome Check and Length Calculation:** For each subset and its complement (the indices not in the subset), check if the corresponding subsequences are palindromes. If they are, calculate their lengths.
    *   **Maximize Product:** Keep track of the maximum product of the lengths of two disjoint palindromic subsequences found so far.

*   **Complexity:**
    *   **Runtime Complexity:** O(2<sup>N</sup> * N), where N is the length of the string `s`. This comes from iterating through all possible subsets (2<sup>N</sup>) and checking if each subset forms a palindrome (O(N)).
    *   **Storage Complexity:** O(N), primarily due to storing subsequences.

	
	# Code
	```python
	def maxProduct(s: str) -> int:
    n = len(s)
    max_product = 0

    def is_palindrome(subsequence):
        return subsequence == subsequence[::-1]

    for i in range(1 << n):
        subset1_indices = []
        subset2_indices = []

        for j in range(n):
            if (i >> j) & 1:
                subset1_indices.append(j)
            else:
                subset2_indices.append(j)

        if not subset1_indices or not subset2_indices:
            continue

        subset1 = "".join(s[k] for k in subset1_indices)
        subset2 = "".join(s[k] for k in subset2_indices)

        if is_palindrome(subset1) and is_palindrome(subset2):
            max_product = max(max_product, len(subset1) * len(subset2))

    return max_product
	```