<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)
        m = {}

        def solve(l, r):
            if l > r:
                return 0
            if l == r:
                return 1  # Single character is a palindrome
            
            if (l, r) in m:
                return m[(l, r)]
            
            res = 0
            if s[l] == s[r]:
                # Find the next occurrence of s[l] from l+1 and the previous occurrence of s[r] from r-1
                left, right = l + 1, r - 1
                while left <= right and s[left] != s[l]:
                    left += 1
                while left <= right and s[right] != s[r]:
                    right -= 1
                
                if left > right:
                    # No duplicate characters between l and r
                    res = 2 * solve(l + 1, r - 1) + 2
                elif left == right:
                    # One duplicate character between l and r
                    res = 2 * solve(l + 1, r - 1) + 1
                else:
                    # More than one duplicate character between l and r
                    res = 2 * solve(l + 1, r - 1) - solve(left + 1, right - 1)
            else:
                # When characters are different, count subsequences excluding either l or r
                res = solve(l + 1, r) + solve(l, r - 1) - solve(l + 1, r - 1)

            # Store result in memoization dictionary, apply MOD to avoid overflow and negative results
            m[(l, r)] = (res + MOD) % MOD
            return m[(l, r)]

        return solve(0, n - 1)

181284