<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 {
public:
    int countPalindromicSubsequences(string s) {
        const long MOD = 1e9 + 7; // For avoiding overflow due to large numbers
        int n = s.length();
        
        // dp[i][j] will store the number of different palindromic subsequences in s[i:j]
        vector<vector<long>> dp(n, vector<long>(n, 0));
        
        // Each single character is a palindromic subsequence
        for (int i = 0; i < n; ++i) {
            dp[i][i] = 1;
        }
        
        // Fill the dp array for substrings of length 2 to n
        for (int len = 2; len <= n; ++len) {
            for (int i = 0; i < n - len + 1; ++i) {
                int j = i + len - 1;
                if (s[i] == s[j]) {
                    // If s[i] == s[j], calculate how many unique subsequences can be formed
                    // between i+1 and j-1
                    int l = i + 1;
                    int r = j - 1;
                    
                    // Find the first occurrence of s[i] inside (i, j)
                    while (l <= r && s[l] != s[i]) ++l;
                    // Find the last occurrence of s[j] inside (i, j)
                    while (l <= r && s[r] != s[j]) --r;
                    
                    if (l > r) {
                        // If there are no s[i] between (i+1, j-1), then the palindromes are
                        // "s[i] + (palindromes in s[i+1:j-1]) + s[j]" + s[i], s[j]
                        dp[i][j] = (2 * dp[i + 1][j - 1] + 2) % MOD;
                    } else if (l == r) {
                        // If there is exactly one s[i] inside (i+1, j-1),
                        // "s[i] + palindromes between i+1 and j-1" + s[j]
                        dp[i][j] = (2 * dp[i + 1][j - 1] + 1) % MOD;
                    } else {
                        // If there are at least two occurrences of s[i] inside (i+1, j-1),
                        // We can count the palindromes between i+1 and j-1 only once
                        dp[i][j] = (2 * dp[i + 1][j - 1] - dp[l + 1][r - 1] + MOD) % MOD;
                    }
                } else {
                    // If s[i] != s[j], the palindromes in the current substring are the union of:
                    // - palindromes in s[i+1:j] and palindromes in s[i:j-1] minus double counted ones
                    dp[i][j] = (dp[i + 1][j] + dp[i][j - 1] - dp[i + 1][j - 1] + MOD) % MOD;
                }
            }
        }
        
        // The answer for the full string is in dp[0][n-1]
        return dp[0][n - 1];
    }
};
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