<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)
        memo = {}
        return self.lps(s,0,len(s) - 1,memo)
    def lps(self,s,left,right,memo):
        if (left,right) in memo:
            return memo[(left,right)]
        if left > right:
            return 0
        if left == right:
            return 1
        if s[left] == s[right]:
            memo[(left,right)] = self.lps(s,left + 1,right - 1,memo) + 2
        else:
            memo[(left,right)] = max(self.lps(s,left,right - 1,memo),self.lps(s,left + 1 ,right,memo))
        
        return memo[(left,right)]
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