<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>

#pragma GCC optimize("O3,unroll-loops")
#pragma GCC target("avx2,bmi,bmi2,lzcnt,popcnt")
#pragma GCC ivdep

const bool f = []()
    {
        ios_base::sync_with_stdio (false);
        cin.tie (nullptr);
        cout.tie (nullptr);
        return false;
    }();

class Solution 
{
public:
    int findTheLongestSubstring (string_view s) 
    {
        const auto toMask = [](char ch) -> uint8_t
            {
                switch (ch)
                {
                    case 'a':
                        return 1 << 0;
                    case 'e':
                        return 1 << 1;
                    case 'i':
                        return 1 << 2;
                    case 'o':
                        return 1 << 3;
                    case 'u':
                        return 1 << 4;
                }

                return 0;
            };

        uint8_t msk[s.size() + 1];
        msk[0] = 0;
        uint8_t cur = 0;
        for (size_t ii = 0; ii < s.size(); ++ii)
        {
            cur ^= toMask (s[ii]);
            msk[ii + 1] = cur;
        }

        size_t res = 0;
        size_t hsh[256] = {};
        for (size_t ii = 0; ii < s.size() + 1; ++ii)
        {
            if (hsh[msk[ii]] == 0)
                hsh[msk[ii]] = ii + 1;
            else
                res = max (res, ii + 1 - hsh[msk[ii]]);
        }

        return res;
    }
};
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