<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, string: str) -> int:
        # Define the modulus to ensure the return value is within the expected range.
        MOD = 10**9 + 7
      
        # Get the length of the string to iterate through it.
        length = len(string)
      
        # Initialize a 3D array to store the count of palindromic subsequences.
        # dp[i][j][k] will store the count for the substring string[i:j+1] with character 'abcd'[k].
        dp = [[[0] * 4 for _ in range(length)] for _ in range(length)]
      
        # Base case initialization for substrings of length 1.
        for index, character in enumerate(string):
            dp[index][index][ord(character) - ord('a')] = 1
      
        # Starting from substrings of length 2, build up solutions for longer substrings.
        for substring_length in range(2, length + 1):
            for start in range(length - substring_length + 1):
                end = start + substring_length - 1
                for char in 'abcd':
                    char_index = ord(char) - ord('a')
                    # If the current characters at start and end indices match and match 'char',
                    # count the inner subsequences twice (for including both start and end characters,
                    # and excluding both), and add the counts of all subsequences in between.
                    if string[start] == string[end] == char:
                        dp[start][end][char_index] = (2 + sum(dp[start + 1][end - 1])) % MOD
                    # If only the start character matches 'char', carry over the count from the previous.
                    elif string[start] == char:
                        dp[start][end][char_index] = dp[start][end - 1][char_index]
                    # If only the end character matches 'char', carry over the count from the previous.
                    elif string[end] == char:
                        dp[start][end][char_index] = dp[start + 1][end][char_index]
                    # If neither character matches 'char', the count remains the same as the inner substring.
                    else:
                        dp[start][end][char_index] = dp[start + 1][end - 1][char_index]
      
        # Finally, sum up all the counts for the entire string and each character 'abcd',
        # and apply the modulus for the result.
        return sum(dp[0][length - 1]) % MOD
109093