<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>

# Problem:
#   - Given string 's'
#   - Return size of longest substring containing each vowel an even number of times

# Logic:
#   - Use a bitmask to determine if all vowels occur an even amt of times
#   - Keep track of the first occurrence of each bitmask pattern (with matching criteria)
#   - If a bitmask pattern has already occurred, then this the current substring is an extension of the first occurrence

# Approach:
#   1. Create bitmask
#   2. Create first occurrence dictionary
#   3. Build dictionary, bitmask, and max length
class Solution:
    def findTheLongestSubstring(self, s: str) -> int:
        mask = 0
        first_occurrence = {0 : -1}
        vowel_to_bit = { letter : position for position, letter in enumerate('aeiou') }
        max_length = 0

        for idx, letter in enumerate(s):
            if letter in vowel_to_bit:
                # Toggle the bit position in the bitmask
                bit_position = vowel_to_bit[letter]
                mask ^= (1 << bit_position)
            
            # Check if this pattern has occurred previously
            if mask in first_occurrence:
                max_length = max(max_length, idx - first_occurrence[mask])
            else:
                first_occurrence[mask] = idx
        
        return max_length
22000