Count Different Palindromic Subsequences
<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>
String; Dynamic Programming
class Solution {
    private final int MOD = (int) 1e9 + 7;

    public int countPalindromicSubsequences(String s) {
        int n = s.length();
        long[][][] dp = new long[n][n][4];
        for (int i = 0; i < n; ++i) {
            dp[i][i][s.charAt(i) - 'a'] = 1;
        }
        for (int l = 2; l <= n; ++l) {
            for (int i = 0; i + l <= n; ++i) {
                int j = i + l - 1;
                for (char c = 'a'; c <= 'd'; ++c) {
                    int k = c - 'a';
                    if (s.charAt(i) == c && s.charAt(j) == c) {
                        dp[i][j][k] = 2 + dp[i + 1][j - 1][0] + dp[i + 1][j - 1][1]
                            + dp[i + 1][j - 1][2] + dp[i + 1][j - 1][3];
                        dp[i][j][k] %= MOD;
                    } else if (s.charAt(i) == c) {
                        dp[i][j][k] = dp[i][j - 1][k];
                    } else if (s.charAt(j) == c) {
                        dp[i][j][k] = dp[i + 1][j][k];
                    } else {
                        dp[i][j][k] = dp[i + 1][j - 1][k];
                    }
                }
            }
        }
        long ans = 0;
        for (int k = 0; k < 4; ++k) {
            ans += dp[0][n - 1][k];
        }
        return (int) (ans % MOD);
    }
}