<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:
        if len(s) == 0:
            return 0
        
        vowels = ['a', 'e', 'i', 'o', 'u']
        counts = [(0, 0, 0, 0, 0)] # counts[i] contains vowel counts until index i (without index i)
        for i in range(len(s)):
            if s[i] == 'a':
                counts.append(((counts[-1][0]+1)%2, counts[-1][1], counts[-1][2], counts[-1][3], counts[-1][4]))
            elif s[i] == 'e':
                counts.append((counts[-1][0], (counts[-1][1]+1)%2, counts[-1][2], counts[-1][3], counts[-1][4]))
            elif s[i] == 'i':
                counts.append((counts[-1][0], counts[-1][1], (counts[-1][2]+1)%2, counts[-1][3], counts[-1][4]))
            elif s[i] == 'o':
                counts.append((counts[-1][0], counts[-1][1], counts[-1][2], (counts[-1][3]+1)%2, counts[-1][4]))
            elif s[i] == 'u':
                counts.append((counts[-1][0], counts[-1][1], counts[-1][2], counts[-1][3], (counts[-1][4]+1)%2))
            else:
                counts.append(counts[-1])
        
        observed = {}
        longest = 0
        for i in range(len(counts)):
            if counts[i] in observed:
                longest = max(longest, i-observed[counts[i]])
            else:
                observed[counts[i]] = i
        return longest
532