<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)
        dp = [[0 for _ in range(n)] for _ in range(n)]

        for i in range(n):
            dp[i][i] = 1 # case "a", "b" "c"...

        for distance in range(1, n):
            for i in range(0, n - distance):
                j = i + distance;
                if s[i] == s[j]:
                    low = i + 1
                    high = j - 1

                    # Get rid of duplicates
                    # find if s[i] exists in s[i+1:j], if it does i.e. when low <= high,
                    # we need to take care of duplicates
                    # In low == high case, there is only one duplicate i.e. s[i] itself
                    # otherwise duplicates will be all palindromic subsequence in
                    # s[low+1:high] range
                    while low <= high and s[low] != s[j]:
                        low += 1
                    while low <= high and s[high] != s[j]:
                        high -= 1

                    if low > high: # s[i] has not been previously encountered
                        # s=abba, both a & b has not been encountered in the inner substring
                        # print(f"l>h {i=}, {j=}, {low=}, {high=}")
                        dp[i][j] = dp[i + 1][j - 1] * 2 + 2
                    elif low == high:
                        # s=ababa, s[i] (=a) has been encountered before, but length is only
                        # 1 char long, so we need just not count s[i] twice
                        # print(f"l==h {i=}, {j=}, {low=}, {high=}")
                        dp[i][j] = dp[i + 1][j - 1] * 2 + 1
                    else:
                        # s=abcadacba, s[i] (=outer a) has been encountered before, so we
                        # need to remove duplicates
                        # print(f"l<h {i=}, {j=}, {low=}, {high=}")
                        dp[i][j] = dp[i + 1][j - 1] * 2 - dp[low + 1][high - 1]
                else:
                    # s[i] != s[j]
                    dp[i][j] = dp[i][j - 1] + dp[i + 1][j] - dp[i + 1][j - 1]

                dp[i][j] = dp[i][j] % (10 ** 9 + 7)

        return dp[0][n - 1]
40040