<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 {
    int dp[1001][1001]; 

private:
    int LCS(string& s, string& rev_s, int n, int m) {
        if (n == 0 || m == 0) {
            return 0;
        }
        if (dp[n][m] != -1) {
            return dp[n][m]; 
        }
        if (s[n - 1] == rev_s[m - 1]) {
            return dp[n][m] = 1 + LCS(s, rev_s, n - 1, m - 1);
        } else {
            return dp[n][m] = max(LCS(s, rev_s, n, m - 1),
                                  LCS(s, rev_s, n - 1, m));
        }
    }

public:
    int longestPalindromeSubseq(string s) {
        memset(dp, -1, sizeof(dp));
        string rev_s = s; 
        reverse(rev_s.begin(), rev_s.end()); 
        int n = s.size();
        return LCS(s, rev_s, n, n); 
    }
};

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