<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:

    typedef long long int ll;
    int mod = (1e9+7);

    ll dp[1005][1005][5];

    ll find(string &s,int l,int r, char ch){

        if(l>r) return 0;

        if(l==r){
            if(s[l] == ch) return 1;
            return 0;
        }

        if(dp[l][r][ch-'a'] != -1) return dp[l][r][ch-'a'] ;

        long long int ways = 0;

        if(s[l] == s[r] && s[l] == ch){
            ways = 2;
            for(int k = 0 ;k<4 ;k++){
                ways = (ways + find(s,l+1,r-1,k + 'a'))%mod;
            }
        }else{

            ways = (ways + find(s,l+1,r,ch)) % mod;
            ways = (ways + find(s,l,r-1,ch)) % mod;
            ways = (ways - find(s,l+1,r-1,ch)) % mod;
            if(ways <0) ways += mod;
        }

        return dp[l][r][ch-'a']  = ways % mod;
    } 
    
    int countPalindromicSubsequences(string s) {

        int n = s.size();
    

        memset(dp,-1,sizeof(dp));

        ll ans = 0;
        for(int k = 0; k < 4 ; k ++){
            ans = (ans %mod + (find(s , 0,n-1, 'a' + k)) % mod)%mod;
        }

        return ( ans % mod );
        
    }
};
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