<p>Given a string <code>s</code>, find <em>the longest palindromic <strong>subsequence</strong>&#39;s length in</em> <code>s</code>.</p>

<p>A <strong>subsequence</strong> is a sequence that can be derived from another sequence by deleting some or no elements without changing the order of the remaining elements.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;bbbab&quot;
<strong>Output:</strong> 4
<strong>Explanation:</strong> One possible longest palindromic subsequence is &quot;bbbb&quot;.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;cbbd&quot;
<strong>Output:</strong> 2
<strong>Explanation:</strong> One possible longest palindromic subsequence is &quot;bb&quot;.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= s.length &lt;= 1000</code></li>
	<li><code>s</code> consists only of lowercase English letters.</li>
</ul>

class Solution:
    def longestPalindromeSubseq(self, s: str) -> int:
        rev_s = s[::-1]
        n = len(s)
        dp = [[-1] * (n + 1) for _ in range(n + 1)]

        def rec(i: int, j: int) -> int:
            # Base case: if either string is exhausted
            if i == n or j == n:
                return 0
            
            # If the value has already been computed, return it
            if dp[i][j] != -1:
                return dp[i][j]
            
            # Initialize answer
            answer = 0
            
            # If characters match, increment the length of LCS and recurse
            if s[i] == rev_s[j]:
                answer = 1 + rec(i + 1, j + 1)
            else:
                # Recur by either skipping the current character of s or rev_s
                answer = max(rec(i + 1, j), rec(i, j + 1))
            
            # Store the computed value in dp table
            dp[i][j] = answer
            
            return answer
        
        # Compute LCS length starting from (0, 0)
        return rec(0, 0)

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