<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:
        return self.longestCommonSubseq(s, s[::-1])

    def longestCommonSubseq(self, s1: str, s2: str) -> int:
        dp = [[0] * (len(s2)+1) for _ in range(len(s1)+1)]
        # dp[i][j] is max len LCS of s1[:i] and s2[:j]

        for i in range(len(s1)):
            for j in range(len(s2)):
                dp[i+1][j+1] = max(
                    dp[i+1][j],
                    dp[i][j+1],
                    dp[i][j] + int(s1[i]==s2[j])
                )
        return dp[len(s1)][len(s2)]

    def longestPalindromeSubseq1(self, s: str) -> int:
        dp = [[0] * (len(s)+1) for _ in range(len(s)+1)]
        # dp[i][j] is optimal solution of s[i:j] (j exclusive), and
        # dp[i][i] has s at length 0

        for i in range(len(s)):
            dp[i][i] = 1
            if i < len(s)-1:
                dp[i][i+1] = 1
        
        for d in range(2, len(s)+1):
            for i in range(len(s)-d):
                j = i + d
                dp[i][j] = max(
                    dp[i+1][j-1] + int(s[i]==s[j]), 
                    dp[i+1][j],
                    dp[i][j-1])
        print(dp)
        return dp[len(s)-1][len(s)-1]
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