<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 int findTheLongestSubstring(string s) {
            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;
        const int n=s.size();
        vector<char> bmask(n+1, 0);// 1-indexed prefix sum
        int first_seen[32]; // index for different bmask first seen
        memset(first_seen, -1, sizeof(first_seen));
        int len=0;
        first_seen[0]=0; // no vowels for bmask[0]
        for(int i=0; i<n; i++){
            const char x=vow[s[i]-'a'];
            const char curr=bmask[i+1]=bmask[i]^(x==-1?0:(1<<x));
            if (first_seen[curr]==-1) first_seen[curr]=i+1;
            len=max(len, i+1-first_seen[curr]);
        }
        return len;
    
    }
};
19800