<p>Given the string <code>s</code>, return the size of the longest substring containing each vowel an even number of times. That is, &#39;a&#39;, &#39;e&#39;, &#39;i&#39;, &#39;o&#39;, and &#39;u&#39; must appear an even number of times.</p>

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

<pre>
<strong>Input:</strong> s = &quot;eleetminicoworoep&quot;
<strong>Output:</strong> 13
<strong>Explanation: </strong>The longest substring is &quot;leetminicowor&quot; which contains two each of the vowels: <strong>e</strong>, <strong>i</strong> and <strong>o</strong> and zero of the vowels: <strong>a</strong> and <strong>u</strong>.
</pre>

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

<pre>
<strong>Input:</strong> s = &quot;leetcodeisgreat&quot;
<strong>Output:</strong> 5
<strong>Explanation:</strong> The longest substring is &quot;leetc&quot; which contains two e&#39;s.
</pre>

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

<pre>
<strong>Input:</strong> s = &quot;bcbcbc&quot;
<strong>Output:</strong> 6
<strong>Explanation:</strong> In this case, the given string &quot;bcbcbc&quot; is the longest because all vowels: <strong>a</strong>, <strong>e</strong>, <strong>i</strong>, <strong>o</strong> and <strong>u</strong> appear zero times.
</pre>

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

<ul>
	<li><code>1 &lt;= s.length &lt;= 5 x 10^5</code></li>
	<li><code>s</code>&nbsp;contains only lowercase English letters.</li>
</ul>

class Solution {
public:
	static size_t findTheLongestSubstring(const std::string& str)
	{
		const auto n = str.length();
		const std::unordered_set<char> sVowels = { 'a', 'e', 'i', 'o', 'u' };
		std::unordered_map<int32_t, size_t> m = { { 0, std::numeric_limits<size_t>::max() } };
		size_t szMaxLength = 0;
		int32_t iMask = 0;
		for (size_t i = 0; i != n; ++i)
		{
			const auto c = str[i];
			if (sVowels.find(c) != sVowels.cend())
			{
				iMask ^= 1 << (c - 'a');
				if (m.find(iMask) == m.cend())
				{
					m.emplace(std::make_pair(iMask, i));
					continue;
				}
			}
			const auto szCurrentLength = i - m.at(iMask);
			if (szMaxLength < szCurrentLength)
				szMaxLength = szCurrentLength;
		}
		return szMaxLength;
	}
};
16000