<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 {
public:

    int longestPalindromeSubseq(string s) {
        int n = s.size();
        int dp[1001][1001];
        memset(dp, -1,sizeof(dp));
        function<int(int,int)> find = [&](int i, int j)->int {
            if (i>j) return 0;
            int& res = dp[i][j];
            if (res != -1) return res;
            int ans= 0;
            if (s[i] == s[j]) {
                if (i==j) {
                    ans += 1;
                } else {
                    ans = max(ans, 2 + find(i+1, j-1));
                }
            }
            // ko chon vi tri i, chon vi tri j
            ans = max(ans, find(i+1, j));
            // chon i, ko chon j
            ans = max(ans, find(i, j-1));
            return res = ans;
        };
        return find(0, n-1);
    }
};
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