Count Vowel Substrings of a String
<p>A <strong>substring</strong> is a contiguous (non-empty) sequence of characters within a string.</p>

<p>A <strong>vowel substring</strong> is a substring that <strong>only</strong> consists of vowels (<code>&#39;a&#39;</code>, <code>&#39;e&#39;</code>, <code>&#39;i&#39;</code>, <code>&#39;o&#39;</code>, and <code>&#39;u&#39;</code>) and has <strong>all five</strong> vowels present in it.</p>

<p>Given a string <code>word</code>, return <em>the number of <strong>vowel substrings</strong> in</em> <code>word</code>.</p>

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

<pre>
<strong>Input:</strong> word = &quot;aeiouu&quot;
<strong>Output:</strong> 2
<strong>Explanation:</strong> The vowel substrings of word are as follows (underlined):
- &quot;<strong><u>aeiou</u></strong>u&quot;
- &quot;<strong><u>aeiouu</u></strong>&quot;
</pre>

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

<pre>
<strong>Input:</strong> word = &quot;unicornarihan&quot;
<strong>Output:</strong> 0
<strong>Explanation:</strong> Not all 5 vowels are present, so there are no vowel substrings.
</pre>

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

<pre>
<strong>Input:</strong> word = &quot;cuaieuouac&quot;
<strong>Output:</strong> 7
<strong>Explanation:</strong> The vowel substrings of word are as follows (underlined):
- &quot;c<strong><u>uaieuo</u></strong>uac&quot;
- &quot;c<strong><u>uaieuou</u></strong>ac&quot;
- &quot;c<strong><u>uaieuoua</u></strong>c&quot;
- &quot;cu<strong><u>aieuo</u></strong>uac&quot;
- &quot;cu<strong><u>aieuou</u></strong>ac&quot;
- &quot;cu<strong><u>aieuoua</u></strong>c&quot;
- &quot;cua<strong><u>ieuoua</u></strong>c&quot;
</pre>

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

<ul>
	<li><code>1 &lt;= word.length &lt;= 100</code></li>
	<li><code>word</code> consists of lowercase English letters only.</li>
</ul>
Hash Table; String
func countVowelSubstrings(word string) int {
	ans, n := 0, len(word)
	for i := range word {
		t := map[byte]bool{}
		for j := i; j < n; j++ {
			c := word[j]
			if !(c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u') {
				break
			}
			t[c] = true
			if len(t) == 5 {
				ans++
			}
		}
	}
	return ans
}