<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:
    # DP on strings
    def longestPalindromeSubseq(self, s: str) -> int:
        def f(i, j): # Longest Common Subsequence Problem
            if i >= n or j >= n:
                return 0

            if dp[i][j] != -1: return dp[i][j]
            
            if s[i] == revS[j]:
                return 1 + f(i+1, j+1)

            dp[i][j] = max(f(i, j+1), f(i+1, j))
            return dp[i][j]
        
        n = len(s)
        dp = [[-1] * (n) for i in range(n)]
        revS = s[::-1]
        return f(0, 0)

    # Space Optimised
    # Time complexity = O(n**2)
    # Space complexity = O(n) 
    def longestPalindromeSubseq2(self, s: str) -> int:
        def lcs(s1, s2): # Longest Common Subsequence Problem
            n = len(s1)
            prev = [0] * (n+1)
            for i in range(1, n+1):
                curr = [0] * (n+1)
                for j in range(1, n+1):
                    if s1[i-1] == s2[j-1]:
                        curr[j] = 1 + prev[j-1]
                    else:
                        curr[j] = max(prev[j], curr[j-1])
                prev = curr
            return curr[n]

        return lcs(s, s[::-1]) # calculate lcs of s and reverse of s to get longest palindromic subsequence

    
74981