<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>

#define ll long long
#define pb push_back
int mod = 1e9+7;
class Solution {
public:
    int countPalindromicSubsequences(string s) {
        int n = s.size();
        vector<vector<ll>>dp(n,vector<ll>(n,0));
        for(int i =0;i<n;i++){
          dp[i][i] = 1LL;
          if(i+1<n)
          dp[i][i+1] = 2LL;
        }

       // gap strategy : filling dp with gap
       for(int gap = 2;gap<n;gap++){
        for(int i =0;i<n-gap;i++){
          int j = i+ gap;
          if(s[i] == s[j]){
            vector<int>pos;
            for(int k = i+1;k<j;k++){
              if(pos.size()<2 && s[k] == s[i]){
                pos.pb(k);
              }
              else if(pos.size()==2 && s[k] == s[i])pos[1] = k;
            }
            if(!pos.size()){
              dp[i][j] = (2LL*dp[i+1][j-1] + 2LL)%mod;
            }
            else if(pos.size() == 1){
              dp[i][j] = (2LL*dp[i+1][j-1] + 1LL)%mod;
            }
            else
            if(pos[0] != pos[1]-1){
              dp[i][j] = (2LL*dp[i+1][j-1]  - dp[pos[0]+1][pos[1]-1]+mod)%mod;
            }
            else {
              dp[i][j] = 2LL*dp[i+1][j-1];
            }
          }
          else{
            dp[i][j] = (dp[i+1][j] + dp[i][j-1] - dp[i+1][j-1] + mod)%mod;
          }
          dp[i][j] = (dp[i][j] + mod)%mod;
        }
       }
       return dp[0][n-1]%mod;
    }
};
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