<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:
    def findTheLongestSubstring(self, s: str) -> int:
        view = 0
        yindex = {0:-1}
        best = 0
        for i, c in enumerate(s):
            view ^= 1 << ("aeiou".find(c) + 1) >> 1
            if view not in yindex:
                yindex[view] = i
            best = max(best, i - yindex[view])
        return best
        def s2(s):
            def bit(c):
                return 1 << ("aeiou".find(c) + 1) >> 1
            view = 0
            yindex = {0:-1}
            best = 0
            for i, c in enumerate(s):
                view ^= 1 << ("aeiou".find(c) + 1) >> 1
                if view not in yindex:
                    yindex[view] = i
                best = max(best, i - yindex[view])
            return best
        return s2(s)
            
             
        
        
        
        
        
        
        
        
        
        
        
        
        
        
        
        
        
        
        
        
        n = len(s)
        mem = [Counter() for _ in range(n)]
        best = 0
        for k in range(1, n+1):
            for i in range(0, n-k+1):
                j = i + k - 1
                if s[j] in 'aeiou':
                    mem[i][s[j]] += 1
                if not any(v&1 for v in mem[i].values()):
                    best = j-i+1
        return best
            
466