Count Different Palindromic Subsequences
<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>
String; Dynamic Programming
class Solution:
    def countPalindromicSubsequences(self, s: str) -> int:
        mod = 10**9 + 7
        n = len(s)
        dp = [[[0] * 4 for _ in range(n)] for _ in range(n)]
        for i, c in enumerate(s):
            dp[i][i][ord(c) - ord('a')] = 1
        for l in range(2, n + 1):
            for i in range(n - l + 1):
                j = i + l - 1
                for c in 'abcd':
                    k = ord(c) - ord('a')
                    if s[i] == s[j] == c:
                        dp[i][j][k] = 2 + sum(dp[i + 1][j - 1])
                    elif s[i] == c:
                        dp[i][j][k] = dp[i][j - 1][k]
                    elif s[j] == c:
                        dp[i][j][k] = dp[i + 1][j][k]
                    else:
                        dp[i][j][k] = dp[i + 1][j - 1][k]
        return sum(dp[0][-1]) % mod
