<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:
        d = dict()

        ptr = -1
        maxlen = 0
        self.vowels = {
            'a': 0,
            'e': 1,
            'i': 2,
            'o': 3,
            'u': 4
        }
        self.index = ['T', 'T', 'T', 'T', 'T']
        d[''.join(self.index)] = -1

        while (ptr+1 < len(s)):
            ptr += 1
            if s[ptr] in self.vowels:
                if self.index[self.vowels[s[ptr]]] == 'F':
                    self.index[self.vowels[s[ptr]]] = 'T'
                else:
                    self.index[self.vowels[s[ptr]]] = 'F'

            index_str = ''.join(self.index)

            if index_str in d:
                maxlen = max(maxlen, ptr-d[index_str])
            else:
                d[index_str] = ptr

        return min(maxlen, len(s))
485