<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[1005][1005];
    vector<int>places[26];
    string s;
    int searchindex(int num, int end){
        int l = 0, r = places[num].size(), g = (l+r)/2;
        while(l + 1 < r){
            g = (l+r)/2;
            if(places[num][g] > end) r = g;
            else l = g;
        }
        return places[num][l];
    }

    int findans( int start, int end){
        if(start > end)return 0;
        if(dp[start][end] != -1)return dp[start][end];
        int ans = 1;
        int num = s[start]-'a';
        int l = 0, r = places[num].size(), g;
        while(l + 1 < r){
            g = (l+r)/2;
            if(places[num][g] > end) r = g;
            else l = g;
        }
        int last = places[num][l];

        if(last > start){
            ans = max(ans, 2 + findans(start+1, last -1));
        }

        ans = max(ans, findans(start+1, end));
        dp[start][end] = ans;
        return ans;
    }
public:
    int longestPalindromeSubseq(string s1) {
        s = s1;
        for(int i=0;i<1005;i++){
            for(int j=0;j<1005;j++)dp[i][j] = -1;
        }
        for(int i=0;i<s.size();i++){
            places[s[i]-'a'].push_back(i);
        }
        return findans(0, s.size()-1);
    }
};
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