<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:
        # Mapping vowels to bit positions
        vowel_to_bit = {'a': 0, 'e': 1, 'i': 2, 'o': 3, 'u': 4}
        
        # Prefix mask to first occurrence mapping
        mask_to_index = {0: -1}  # Mask 0 is initially at index -1
        prefix_mask = 0
        max_length = 0
        
        for i, char in enumerate(s):
            if char in vowel_to_bit:
                # Toggle the corresponding bit for the vowel
                prefix_mask ^= (1 << vowel_to_bit[char])
            
            # Check if this mask has been seen before
            if prefix_mask in mask_to_index:
                # Calculate the length of the substring
                start_index = mask_to_index[prefix_mask]
                max_length = max(max_length, i - start_index)
            else:
                # Record the first occurrence of this prefix mask
                mask_to_index[prefix_mask] = i
        
        return max_length

# Example usage
if __name__ == "__main__":
    # Create an instance of the Solution class
    sol = Solution()
    
    # Test the method with example inputs
    print(sol.findTheLongestSubstring("eleetminicoworoep"))  # Output: 13
    print(sol.findTheLongestSubstring("leetcodeisgreat"))   # Output: 5
    print(sol.findTheLongestSubstring("bcbcbc"))            # Output: 6

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