<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:
        # Optimisations:
        # - cache of dps (i, j)
        # - initial dict store of all letter idxs
        # - need to store which letters already used to avoid double counting
        cache = [[0] * len(s) for _ in range(len(s))]
        
        def dp(i, j) -> int:
            if cache[i][j]:
                return cache[i][j]
            count = 1  # for no letters picked

            used = set()
            for i_idx in range(i, j + 1):
                if s[i_idx] in used:
                    continue
                used.add(s[i_idx])
                count += 1 # for single letter picked
                for j_idx in range(j, i_idx, -1):
                    if s[j_idx] == s[i_idx]:
                        count += dp(i_idx + 1, j_idx - 1)
                        break

            cache[i][j] = count
            return count

        return (dp(0, len(s) - 1) - 1) % (10 ** 9 + 7)  # - 1 to account for blank string
74566