<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 mod=1e9+7;

    int countPalindromicSubsequences(string s) {
        int n=s.size();

        long long dp[1002][1002];

        for(int i=n-1;i>=0;i--){
            dp[i][i]=1;
            for(int j=i+1;j<n;j++){

                if(s[i]!=s[j]){
                    dp[i][j]=dp[i+1][j] + dp[i][j-1]-dp[i+1][j-1];
                }
                else{
                    dp[i][j]=2 * dp[i+1][j-1];

                    int l=i+1,r=j-1;

                    while(l<=r && s[i]!=s[l]) l++;
                    while(l<=r && s[i]!=s[r]) r--;

                    if(l > r) dp[i][j]+=2;
                    else if(l==r) dp[i][j]+=1;
                    else{
                        dp[i][j]-=dp[l+1][r-1];
                    }
                }
            dp[i][j] = (dp[i][j] + mod) % mod;
            }
        }
    return (int)dp[0][n-1];
    }
};
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