<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:
        MOD = 10**9 + 7
        n = len(s)
        dp = [[0] * n for _ in range(n)]
        
        for i in range(n):
            dp[i][i] = 1  # Each single character is a palindrome
        
        for length in range(2, n+1):  # length of the substring
            for i in range(n-length+1):
                j = i + length - 1
                if s[i] == s[j]:
                    low, high = i+1, j-1
                    # Find next occurrence of s[i] inside s[i+1:j]
                    while low <= high and s[low] != s[i]:
                        low += 1
                    while low <= high and s[high] != s[i]:
                        high -= 1
                    
                    if low > high:  # no same characters between i+1 and j-1
                        dp[i][j] = dp[i+1][j-1] * 2 + 2
                    elif low == high:  # one character between i+1 and j-1
                        dp[i][j] = dp[i+1][j-1] * 2 + 1
                    else:  # more than one character between i+1 and j-1
                        dp[i][j] = dp[i+1][j-1] * 2 - dp[low+1][high-1]
                else:
                    dp[i][j] = dp[i][j-1] + dp[i+1][j] - dp[i+1][j-1]
                
                dp[i][j] = (dp[i][j] + MOD) % MOD  # Ensure non-negative modulo
        
        return dp[0][n-1]

# Example usage
sol = Solution()
s = "bccb"
print(sol.countPalindromicSubsequences(s))  # Output the result

30624