<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:
        n = len(s)
        dparray = [[0 for _ in range(i + 1)] for i in range(n)] # only upper half used # dparray[end][start]

        def solver(start, end):
            if start > end:
                return 0
            if dparray[end][start] > 0:
                return dparray[end][start]
            if start == end:
                dparray[end][start] = 1
            else:
                if s[start] == s[end]:
                    dparray[end][start] = solver(start + 1, end - 1) + 2
                else:
                    dparray[end][start] = max( solver(start + 1, end) , solver(start, end - 1))
            return dparray[end][start]
        
        return solver(0, n-1)

                
                

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