Count Palindromic Subsequences
<p>Given a string of digits <code>s</code>, return <em>the number of <strong>palindromic subsequences</strong> of</em> <code>s</code><em> having length </em><code>5</code>. Since the answer may be very large, return it <strong>modulo</strong> <code>10<sup>9</sup> + 7</code>.</p>

<p><strong>Note:</strong></p>

<ul>
	<li>A string is <strong>palindromic</strong> if it reads the same forward and backward.</li>
	<li>A <strong>subsequence</strong> is a string that can be derived from another string by deleting some or no characters without changing the order of the remaining characters.</li>
</ul>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;103301&quot;
<strong>Output:</strong> 2
<strong>Explanation:</strong> 
There are 6 possible subsequences of length 5: &quot;10330&quot;,&quot;10331&quot;,&quot;10301&quot;,&quot;10301&quot;,&quot;13301&quot;,&quot;03301&quot;. 
Two of them (both equal to &quot;10301&quot;) are palindromic.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;0000000&quot;
<strong>Output:</strong> 21
<strong>Explanation:</strong> All 21 subsequences are &quot;00000&quot;, which is palindromic.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;9999900000&quot;
<strong>Output:</strong> 2
<strong>Explanation:</strong> The only two palindromic subsequences are &quot;99999&quot; and &quot;00000&quot;.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= s.length &lt;= 10<sup>4</sup></code></li>
	<li><code>s</code> consists of digits.</li>
</ul>
String; Dynamic Programming
func countPalindromes(s string) int {
	n := len(s)
	pre := [10010][10][10]int{}
	suf := [10010][10][10]int{}
	t := make([]int, n)
	for i, c := range s {
		t[i] = int(c - '0')
	}
	c := [10]int{}
	for i := 1; i <= n; i++ {
		v := t[i-1]
		for j := 0; j < 10; j++ {
			for k := 0; k < 10; k++ {
				pre[i][j][k] = pre[i-1][j][k]
			}
		}
		for j := 0; j < 10; j++ {
			pre[i][j][v] += c[j]
		}
		c[v]++
	}
	c = [10]int{}
	for i := n; i > 0; i-- {
		v := t[i-1]
		for j := 0; j < 10; j++ {
			for k := 0; k < 10; k++ {
				suf[i][j][k] = suf[i+1][j][k]
			}
		}
		for j := 0; j < 10; j++ {
			suf[i][j][v] += c[j]
		}
		c[v]++
	}
	ans := 0
	const mod int = 1e9 + 7
	for i := 1; i <= n; i++ {
		for j := 0; j < 10; j++ {
			for k := 0; k < 10; k++ {
				ans += pre[i-1][j][k] * suf[i+1][j][k]
				ans %= mod
			}
		}
	}
	return ans
}