<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>

r'''
# Requirements

n = len(s)
1 <= n <= 1000
s[i] is a-z

# Edge cases

1. n == 1

2. Only palindromes of length 1 present in s

# Solutions

1. Brute force, backtracking
    - O(2^n * n) time, O(n) auxiliary space
2. DP / memo
    - O(n^2) time, O(n^2) auxiliary space
'''
class Solution:
    def longestPalindromeSubseq(self, s: str) -> int:
        n, max_length = len(s), 1
        memo = [[-1] * n for _ in range(n)]

        def expand(l: int, r: int) -> int:
            if l < 0 or r >= n:
                return 0
            if memo[l][r] != -1:
                return memo[l][r]

            if s[l] == s[r]:
                memo[l][r] = (2 if l != r else 1) + expand(l - 1, r + 1)
            else:
                memo[l][r] = max(expand(l - 1, r), expand(l, r + 1))

            return memo[l][r]

        for l in range(n):
            max_length = max(max_length, expand(l, l))
            max_length = max(max_length, expand(l, l + 1))

        return max_length
135543