<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:
        reverseString = s[::-1]
        indexPos = len(s) - 1
        dp = {}
        for i in range(len(s) + 1):
            for j in range(len(s) + 1):
                if i == 0:
                    dp[(i, j)] = 0
                if j == 0:
                    dp[(i, j)] = 0
        
        for k in range(1, len(s) + 1):
            for l in range(1, len(s) + 1):
                if s[k - 1] == reverseString[l - 1]:
                    dp[(k, l)] = 1 + dp[(k - 1, l - 1)]
                else:
                    dp[(k, l)] = max(dp[(k - 1, l)], dp[(k, l - 1)])

        return dp[(len(s), len(s))]

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