<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:
    def countPalindromicSubsequences(self, s: str) -> int:
        n = len(s)
        l = [0]*4
        def nm(x):
            return ord(x)-ord("a")
        def od(x):
            return chr(ord("a")+x)
        x = []
        for i in s:
            l[nm(i)] += 1
            x.append(l.copy())
        
        
        ans = 0
        mod = 10**9+7

        dp = {}
        def f(i,j,next):

            if i > j:
                return 0
            if i == j:
                return nm(s[i]) == next
            
            if (i,j,next) in dp:
                return dp[(i,j,next)]

            total_chars = x[j][next]
            if i:
                total_chars -= x[i-1][next]

            if total_chars == 0:
                return 0
            if total_chars == 1:
                return 1
            
            ans = 2
            iii,jjj = i,j
            while i<n and s[i] != od(next):
                i += 1
            i += 1
            while j>=0 and s[j] != od(next):
                j -= 1
            j -= 1
            for ii in range(4):
                ans += f(i,j,ii)
                ans %= mod
            dp[(iii,jjj,next)] = ans
            return ans
        ans = 0
        for i in range(4):
            xx = f(0,n-1,i)
            ans += xx
            ans %= mod
        return ans



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