<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:
        
        def _lps_r(t1, t2, i, j):
            if i == len(t1) or j == len(t2):
                return 0

            ans = 0
            if t1[i] == t2[j]:
                ans = 1 + _lcs_r(t1, t2, i + 1, j + 1)
            else:
                ans = 0 + max(_lcs_r(t1, t2, i, j + 1), _lcs_r(
                    t1, t2, i + 1, j))

            return ans

        def _lps_m(t1, t2, i, j, dp):
            if i == len(t1) or j == len(t2):
                return 0

            if dp[i][j] != -1:
                return dp[i][j]

            ans = 0
            if t1[i] == t2[j]:
                ans = 1 + _lps_m(t1, t2, i + 1, j + 1, dp)
            else:
                ans = 0 + max(_lps_m(t1, t2, i, j + 1, dp),
                              _lps_m(t1, t2, i + 1, j, dp))

            dp[i][j] = ans

            return dp[i][j]
        
        str1 = s
        str2 = s[::-1]
        i = 0
        j = 0

        # # Recursion
        # ans = _lps_r(str1, str2)

        # DP: Memoziation
        dp = [[-1 for _ in range(len(str2))] for _ in range(len(str1))]
        ans = _lps_m(str1, str2, i, j, dp)

        return ans
35793