<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:
        if len(s) == 1:
            return 1

        dp = [[0]*1000 for i in range(1000)]
        #solve(len(s), len(s))
        l=0
        r = len(s)-1
        def solve(l, r):
            if l== r:
                return 1
            if l == r+1:
                return 0
            if dp[l][r] >0:
                return dp[l][r]
            if s[l] == s[r]:
                dp[l][r] = solve(l+1, r-1) + 2

            dp[l][r] = max(dp[l][r], solve(l+1, r))
            dp[l][r] = max(dp[l][r], solve(l, r-1))
            return dp[l][r]
        return solve(l, r)
        
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