<p>Given a string s, return <em>the number of different non-empty palindromic subsequences in</em> <code>s</code>. Since the answer may be very large, return it <strong>modulo</strong> <code>10<sup>9</sup> + 7</code>.</p>

<p>A subsequence of a string is obtained by deleting zero or more characters from the string.</p>

<p>A sequence is palindromic if it is equal to the sequence reversed.</p>

<p>Two sequences <code>a<sub>1</sub>, a<sub>2</sub>, ...</code> and <code>b<sub>1</sub>, b<sub>2</sub>, ...</code> are different if there is some <code>i</code> for which <code>a<sub>i</sub> != b<sub>i</sub></code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;bccb&quot;
<strong>Output:</strong> 6
<strong>Explanation:</strong> The 6 different non-empty palindromic subsequences are &#39;b&#39;, &#39;c&#39;, &#39;bb&#39;, &#39;cc&#39;, &#39;bcb&#39;, &#39;bccb&#39;.
Note that &#39;bcb&#39; is counted only once, even though it occurs twice.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;abcdabcdabcdabcdabcdabcdabcdabcddcbadcbadcbadcbadcbadcbadcbadcba&quot;
<strong>Output:</strong> 104860361
<strong>Explanation:</strong> There are 3104860382 different non-empty palindromic subsequences, which is 104860361 modulo 10<sup>9</sup> + 7.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= s.length &lt;= 1000</code></li>
	<li><code>s[i]</code> is either <code>&#39;a&#39;</code>, <code>&#39;b&#39;</code>, <code>&#39;c&#39;</code>, or <code>&#39;d&#39;</code>.</li>
</ul>

const int M = 1e9+7 ;
class Solution {
public:
    int countPalindromicSubsequences(string s) {
        int n = s.size() ;
        int dp[n+1][n+1] ;
        memset(dp,0,sizeof(dp)) ;
        for(int i = 1;i<=n;i++)dp[i][i] = 1 ;
        vector<int>nextInd(26,-1) ;

        for(int i = n;i>=1;i--){
            int next = nextInd[s[i-1]-'a'] ;
            int prev = i ;
            for(int j = i+1;j<=n;j++){
                if(s[i-1]!=s[j-1]){
                    dp[i][j] = dp[i][j-1]+dp[i+1][j] - dp[i+1][j-1] ;
                    if(dp[i][j]<0)dp[i][j] += M ;
                    dp[i][j] %= M ; 
                }
                else{
                    if(next==j){
                        dp[i][j] = 2*dp[i+1][j-1]%M +2 ;
                        dp[i][j] %= M ;
                    }
                    else if(next==prev){
                        dp[i][j] = 2*dp[i+1][j-1]%M +1 ;
                        dp[i][j] %= M ;
                    }
                    else{
                        dp[i][j] = 2*dp[i+1][j-1]%M - dp[next+1][prev-1]+M ;
                        dp[i][j] %= M ;
                    }
                }
                if(s[j-1]==s[i-1])prev = j ;
            }
            nextInd[s[i-1]-'a'] = i ;

        }
        return dp[1][n] ;
    }
};
13593