<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 int MOD = 1e9 + 7;
        int n = s.size();
        vector<vector<int> > dp(n, vector<int> (n, 0));
        for (int i = 0; i < n; i++) dp[i][i] = 1;
        for (int len = 2; len <= n; len++) {
            for (int i = 0; i <= n - len; i++) {
                int j = i + len - 1;
                if (s[i] == s[j]) {
                    int low = i + 1, high = j - 1;
                    while (low <= high && s[i] != s[low]) low++;
                    while (low <= high && s[j] != s[high]) high--;
                    if (low > high) dp[i][j] = (dp[i+1][j-1] * 2 + 2) % MOD;
                    else if (low == high) dp[i][j] = (dp[i+1][j-1] * 2 + 1) % MOD;
                    else dp[i][j] = (((dp[i+1][j-1] * 2 - dp[low+1][high-1]) % MOD) + MOD) % MOD;
                } else {
                    dp[i][j] = (((dp[i+1][j] + dp[i][j-1] - dp[i+1][j-1]) % MOD) + MOD) % MOD;
                }
                dp[i][j] = dp[i][j] % MOD;
            }
        }
        return dp[0][n-1];
    }
};
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