Count Vowels Permutation
<p>Given an integer <code>n</code>, your task is to count how many strings of length <code>n</code> can be formed under the following rules:</p>

<ul>
	<li>Each character is a lower case vowel&nbsp;(<code>&#39;a&#39;</code>, <code>&#39;e&#39;</code>, <code>&#39;i&#39;</code>, <code>&#39;o&#39;</code>, <code>&#39;u&#39;</code>)</li>
	<li>Each vowel&nbsp;<code>&#39;a&#39;</code> may only be followed by an <code>&#39;e&#39;</code>.</li>
	<li>Each vowel&nbsp;<code>&#39;e&#39;</code> may only be followed by an <code>&#39;a&#39;</code>&nbsp;or an <code>&#39;i&#39;</code>.</li>
	<li>Each vowel&nbsp;<code>&#39;i&#39;</code> <strong>may not</strong> be followed by another <code>&#39;i&#39;</code>.</li>
	<li>Each vowel&nbsp;<code>&#39;o&#39;</code> may only be followed by an <code>&#39;i&#39;</code> or a&nbsp;<code>&#39;u&#39;</code>.</li>
	<li>Each vowel&nbsp;<code>&#39;u&#39;</code> may only be followed by an <code>&#39;a&#39;</code>.</li>
</ul>

<p>Since the answer&nbsp;may be too large,&nbsp;return it modulo&nbsp;<code>10^9 + 7</code>.</p>

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

<pre>
<strong>Input:</strong> n = 1
<strong>Output:</strong> 5
<strong>Explanation:</strong> All possible strings are: &quot;a&quot;, &quot;e&quot;, &quot;i&quot; , &quot;o&quot; and &quot;u&quot;.
</pre>

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

<pre>
<strong>Input:</strong> n = 2
<strong>Output:</strong> 10
<strong>Explanation:</strong> All possible strings are: &quot;ae&quot;, &quot;ea&quot;, &quot;ei&quot;, &quot;ia&quot;, &quot;ie&quot;, &quot;io&quot;, &quot;iu&quot;, &quot;oi&quot;, &quot;ou&quot; and &quot;ua&quot;.
</pre>

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

<pre>
<strong>Input:</strong> n = 5
<strong>Output:</strong> 68</pre>

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

<ul>
	<li><code>1 &lt;= n &lt;= 2 * 10^4</code></li>
</ul>
Dynamic Programming
func countVowelPermutation(n int) (ans int) {
	const mod int = 1e9 + 7
	f := make([]int, 5)
	for i := range f {
		f[i] = 1
	}
	for i := 1; i < n; i++ {
		g := make([]int, 5)
		g[0] = (f[1] + f[2] + f[4]) % mod
		g[1] = (f[0] + f[2]) % mod
		g[2] = (f[1] + f[3]) % mod
		g[3] = f[2] % mod
		g[4] = (f[2] + f[3]) % mod
		f = g
	}
	for _, x := range f {
		ans = (ans + x) % mod
	}
	return
}