<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:
        n = len(s)
        memo = [[-1] * n for _ in range(n)]

        def dfs(start: int, end: int)->int:
            if start > end:
                return 0
            
            if start == end:
                return 1
            
            if -1 != memo[start][end]:
                return memo[start][end]

            cnt = 0
            for ch in "abcd":
                first, last = start, end
                while first <= last and s[first] != ch:
                    first += 1
                while last >= first and s[last] != ch:
                    last -= 1
                if first <= last:
                    cnt += 1

                if first < last:
                    cnt += 1
                    cnt += dfs(first + 1, last - 1)

            memo[start][end] = cnt % (10**9 + 7)
            return memo[start][end]
        
        # return dfs(0, n - 1)
        dp = [[0] * n for _ in range(n + 1)]
        for start in range(n - 1, -1, -1):
            dp[start][start] = 1
            for end in range(n - 1, start, -1):
                cnt = 0
                for ch in "abcd":
                    first, last = start, end
                    while first <= last and s[first] != ch:
                        first += 1
                    while last >= first and s[last] != ch:
                        last -= 1

                    if first <= last:
                        cnt += 1
                    if first < last:
                        cnt += 1
                        cnt += dp[first + 1][last - 1]
                
                dp[start][end] = cnt % (10**9 + 7)
        
        return dp[0][n - 1]


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