<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>

static const unordered_map<char, char> um = []() {
    unordered_map<char, char> um{};
    for (char i = 'a'; i <= 'z'; ++i) {
        um[i];
    }
    um['a'] = 1;
    um['e'] = 2;
    um['i'] = 4;
    um['o'] = 8;
    um['u'] = 16;
    return um;
}();

class Solution {
public:

    int findTheLongestSubstring(const string_view s) {
        vector<int> vv(32, -2);
        vv[0] = -1;

        int res = 0;
        char m = 0;

        for (int i = 0; i < s.size(); ++i) {
            const char c = s[i];

            m ^= um.at(c);

            const int pi = vv[m];
            if (pi == -2) {
                vv[m] = i;
            } else {
                res = max(res, i-pi);
            }
        }

        return res;
    }
};
15800