<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:
        ans = 0
        n = len(s)
        memo = [[None] * n for _ in range(n)]

        def dp(i, j):
            if i < 0 or j >= len(s):
                return 0

            if memo[i][j] is not None:
                return memo[i][j]

            elif i == j:
                memo[i][j] = 1 + dp(i - 1, j + 1)

            elif s[i] == s[j]:
                memo[i][j] = 2 + dp(i - 1, j + 1)

            elif s[i] != s[j]:
                memo[i][j] = max(dp(i - 1, j), dp(i, j + 1))

            return memo[i][j]

        for i in range(len(s)):
            # odd
            ans = max(ans, dp(i, i))

            # even
            ans = max(ans, dp(i, i + 1))

        return ans
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