<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:
        
        # def f(i,j):

        #     if i == j:
        #         return 1
        #     if i>j:
        #         return 0
        #     if (i,j) in dp:
        #         return dp[(i,j)]
        #     if s[i] == s[j]:
        #         dp[(i,j)] = 1 + f(i+1,j) + f(i,j-1)
        #     else:
        #         dp[(i,j)] = f(i+1,j) + f(i,j-1) - f(i+1,j-1)
        #     return dp[(i,j)]
        # dp = {}
        # n = len(s)
        # mod = 10**9 + 7
        # return f(0,n-1) % mod


        # n = len(s)
        # dp = [[-1] * 1000 for _ in range(1000)]

        # def helper(s, i, j):
        #     if dp[i][j] != -1:
        #         return dp[i][j]

        #     mod = pow(10, 9) + 7

        #     if i > j:
        #         return 0
        #     elif i == j:
        #         dp[i][j] = 1
        #         return dp[i][j]
        #     elif s[i] == s[j]:
        #         dp[i][j] = (1 + helper(s, i + 1, j) + helper(s, i, j - 1)) % mod
        #         return dp[i][j]
        #     else:
        #         dp[i][j] = (helper(s, i + 1, j) + helper(s, i, j - 1) - helper(s, i + 1, j - 1)) % mod
        #         return dp[i][j]

        # return helper(s, 0, n - 1)



class Solution:
    def countPalindromicSubsequences(self, s: str) -> int:
        n = len(s)
        dp = [[-1] * n for _ in range(n)]

        def helper(i, j):
            if i > j:
                return 0
            if i == j:
                dp[i][j] = 1
                return dp[i][j]
            if dp[i][j] != -1:
                return dp[i][j]

            mod = 10**9 + 7

            if s[i] == s[j]:
                low, high = i + 1, j - 1
                while low <= high and s[low] != s[i]:
                    low += 1
                while low <= high and s[high] != s[j]:
                    high -= 1

                if low > high:
                    dp[i][j] = 2 * helper(i + 1, j - 1) + 2
                elif low == high:
                    dp[i][j] = 2 * helper(i + 1, j - 1) + 1
                else:
                    dp[i][j] = 2 * helper(i + 1, j - 1) - helper(low + 1, high - 1)
            else:
                dp[i][j] = helper(i + 1, j) + helper(i, j - 1) - helper(i + 1, j - 1)

            return dp[i][j] % mod

        return helper(0, n - 1)

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