<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:
        dp = {}
        def lps(i, j):
            # pass
            if (i, j) in dp:
                return dp[(i, j)] 
            if i > j:
                return 0
            if i == j:
                return 1
            if s[i] == s[j]:
                dp[(i, j)]= 2 + lps(i+1, j -1)
                
            else:
                dp[(i, j)] = max(lps(i, j-1), lps(i+1, j))
            return dp[(i, j)]  
        

        n = len(s)
        dp = [[0 for i in range(n)] for j in range(n)]
        for i in range(n):
            dp[i][i] = 1
        for i in range(n-2, -1, -1):
            for j in range(i+1, n):
                if s[i] == s[j]:
                    dp[i][j] = 2+ dp[i+1][j-1]
                else:
                    dp[i][j] = max(dp[i][j-1], dp[i+1][j])
        return dp[0][n-1]
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