<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:
        t = s[::-1]
        n = len(s)
        dp = [[-1 for i in range(n)] for j in range(n)]
        def dfs(i, j):
            if i == n or j == n:
                return 0
            if dp[i][j] != -1:
                return dp[i][j]
            if s[i] == t[j]:
                return 1 + dfs(i+1, j+1)
            dp[i][j] = max(dfs(i+1, j), dfs(i, j+1))
            return dp[i][j]
        dp = [[0 for i in range(n+1)] for j in range(n+1)]
        for i in range(n-1, -1, -1):
            for j in range(n-1, -1, -1):
                if s[i] == t[j]:
                    dp[i][j] = 1 + dp[i+1][j+1]
                else:
                    dp[i][j] = max(dp[i+1][j], dp[i][j+1])
        return dp[0][0]
        

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