<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 isvowel(self, s: str) -> bool:
        return s[0] == "a" or s[0] == "e" or s[0] == "i" or s[0] == "o" or s[0] == "u"

    def valide_len(self, s: str) -> bool:
        dic = {}
        n = len(s)
        val = True
        i, mx = 0, 0
        while i < n:
            if self.isvowel(s[i]):
                if s[i] not in dic:
                    dic[s[i]] = 1
                else:
                    dic[s[i]] += 1
                val = True
                for v in dic:
                    if dic[v] % 2 == 1:
                        val = False
            if val:
                mx = i + 1
            i+=1
        return mx

    def findTheLongestSubstring(self, s: str) -> int:
        n = len(s)
        val, i, mx = 0, 0, 0
        dic = {}
        while i < n:
            if self.isvowel(s[i]):
                dic[i] = i
            i+=1
        i = 0
        cur_mx = 0
        while i < n:
            cur_mx = 0
            val = 0
            while i < n and self.isvowel(s[i]) == 0:
                i+=1
                cur_mx+=1
            if i < n:
                val = self.valide_len(s[i:])
            mx = max(mx ,val + cur_mx)
            if mx > n - i:
                return mx
            i+=1
        return mx
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