<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:
        vowel_count = {'a': 0, 'e': 0, 'i': 0, 'o': 0, 'u': 0}
        first_occurrence = {tuple(vowel_count.values()): -1}  # Tuple tracks the counts of each vowel
        max_length = 0
        start = 0
        
        for end in range(len(s)):
            ch = s[end]
            
            # If the character is a vowel, toggle its count between 0 (even) and 1 (odd)
            if ch in vowel_count:
                vowel_count[ch] = 1 - vowel_count[ch]
                
            # Check the current state of vowel counts
            state = tuple(vowel_count.values())
            
            # If we've seen this state before, calculate the possible max length
            if state in first_occurrence:
                max_length = max(max_length, end - first_occurrence[state])
            else:
                # Store the first occurrence of this state
                first_occurrence[state] = end

        return max_length

645