<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:
        '''
        idea: 2-d dp, dp[i][j] = count dif pal subseq in s[i:j+1] 
        consider the case where s[i] == s[j] or s[i] != s[j]
        '''
        n = len(s)
        x = 10 ** (9) + 7
        next_occur, prev_occur = [n] * n, [-1] * n
        occur = {}
        for i in range(n):
            c = s[i]
            if c in occur:
                next_occur[occur[c]] = i 
            occur[c] = i 
        occur = {}
        for j in range(n - 1, -1, -1):
            c = s[j]
            if c in occur:
                prev_occur[occur[c]] = j 
            occur[c] = j
        dp = [[0] * n for _ in range(n)]
        for i in range(n):
            dp[i][i] = 1 
        def driver(i, j):
            if i > j:
                return 0 
            if dp[i][j]:
                return dp[i][j]
            if s[i] == s[j]:
                nexti, prevj = next_occur[i], prev_occur[j]
                if nexti > prevj:
                    dp[i][j] =  2 * driver(i + 1, j - 1) + 2 
                elif nexti == prevj:
                    dp[i][j] = 2 * driver(i + 1, j - 1) + 1
                else:
                    dp[i][j] =  2 * driver(i + 1, j - 1) - driver(nexti + 1, prevj - 1)
            else:
                dp[i][j] =  driver(i, j - 1) + driver(i + 1, j) - driver(i + 1, j - 1)
            dp[i][j] %= x
            return dp[i][j]
        return driver(0, n - 1)

        
96538