<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:
        # all vowels start at count = 0 and are all considered even
        bitmask = 0b11111
        prefixes = {}
        longest = 0

        for i, letter in enumerate(s):
            # flip the corresponding bit for each vowel
            if letter == 'a':
                bitmask ^= 0b00001
            if letter == 'e':
                bitmask ^= 0b00010
            if letter == 'i':
                bitmask ^= 0b00100
            if letter == 'o':
                bitmask ^= 0b01000
            if letter == 'u':
                bitmask ^= 0b10000

            # if all vowel counts are even
            if bitmask == 0b11111:
                longest = i + 1
            
            else:
                # needs a bitmask that flips all 0s in the current bitmask
                # this mask is the current bitmask with all bits flipped
                goal = bitmask ^ 0b11111
                for prefix, j in prefixes.items():
                    if prefix == goal:
                        longest = max(longest, i - j)
                        break
            
            # add the list of negated bitmask into the prefix dictionary
            # only really need to keep the smallest prefix  index for a given negated bitmask
            negated_bitmask = bitmask ^ 0b11111
            if negated_bitmask not in prefixes:
                prefixes[negated_bitmask] = i

        return longest
        
731