<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>

class Solution {
public:
    int countPalindromicSubsequences(string s) {
        int n=s.length();
        int md = 1e9+7;
         vector<vector<long long int>> dp(n,vector<long long int>(n,0));
         for(int i=0;i<n;i++)
         {
            dp[i][i] =1;
         }

         vector<int> pre(n),post(n); 
          int a =-1,b=-1,c=-1,d=-1;
          for(int i=0;i<n;i++)
          {
            if(s[i]=='a')
            {
              pre[i] = a;
              a = i;
            }
            else if(s[i]=='b')
            {
              pre[i] = b;
              b = i;
            }
            else if(s[i]=='c')
            {
              pre[i] = c;
              c = i;
            }
            else
            {
              pre[i] =d; 
              d = i;
            }     
          }

           a =-1;b=-1;c=-1;d=-1;
          for(int i=n-1;i>=0;i--)
          {
            if(s[i]=='a')
            {
              post[i] = a;
              a = i;
            }
            else if(s[i]=='b')
            {
              post[i] = b;
              b = i;
            }
            else if(s[i]=='c')
            {
              post[i] = c;
              c = i;
            }
            else
            {
              post[i] =d; 
              d = i;
            }     
          }

        
         for(int i= n-1;i>=0;i--)
         {
            for(int j=i+1;j<n;j++)
            {
                if(s[i]==s[j])
                {
                    if(pre[j] <= i)
                    {
                       dp[i][j] = (2*dp[i+1][j-1])%md;
                       dp[i][j] += 2;
                       dp[i][j] %= md; 
                    }
                    else
                    {
                        if(pre[j]==post[i])
                        {
                           dp[i][j] = (2*dp[i+1][j-1])%md;
                            dp[i][j] += 1;
                           dp[i][j] %= md; 
                        }
                        else
                        {
                            dp[i][j] = (2*dp[i+1][j-1]- dp[post[i]+1][pre[j]-1]+md)%md;
                            dp[i][j] %= md; 
                        }
                    }
                }
                else
                   dp[i][j] = (dp[i+1][j]+dp[i][j-1]-dp[i+1][j-1]+md)%md;

                 
            }
         }
         return dp[0][n-1];
    }
};
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