<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:
    int findTheLongestSubstring(string s) {
        const int n = s.length();
        char vow[26];
        memset(vow, -1, 26);
        vow[0] = 0;
        vow['e' - 'a'] = 1;
        vow['i' - 'a'] = 2;
        vow['o' - 'a'] = 3;
        vow['u' - 'a'] = 4;
        
        vector<char> prefix(n + 1, 0);
        
        int first_seen[32];
        memset(first_seen, -1, 32 * 4);

        int len = 0;
        first_seen[0] = 0;

        for(int i = 0; i < n; i++) {
            const char x = vow[s[i] - 'a'];
            const char cur = prefix[i + 1] = prefix[i] ^ (x == -1? 0: (1 << x));

            if(first_seen[cur] == -1) first_seen[cur] = i + 1;
            len = max(len, i + 1 - first_seen[cur]);
        }

        return len;
    }
};
19500