<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:
        # use mask to store parity of aeiou
        # each vowel occupies a bit starting from `a` in the rightmost bit
        # if a bit it set, it means we saw that vowel an odd number of times
        bitmask = 0
        max_length = 0
        # record the last position we saw a particular value of the bitmask
        # at max bitmask will be 31 if all set, so allocate array of length 32
        mapy = [None] * 32
        # at the beginning, we have all counts = 0 so bitmask = 0
        # set to -1 since the first could be a vowel, then that would break
        mapy[bitmask] = -1
        # process the string
        for i, c in enumerate(s):
            # update mask
            if c == 'a': bitmask ^= 1
            elif c == 'e': bitmask ^= 2
            elif c == 'i': bitmask ^= 4
            elif c == 'o': bitmask ^= 8
            elif c == 'u': bitmask ^= 16

            prev = mapy[bitmask]
            if prev is None: 
                # first time seeing bitmask, set to current position
                # note we only set this once on the first occurrence
                # so if we encounter it multiple times in the future, we will
                # always calculate length relative to the first occurrence.
                mapy[bitmask] = i
            else:
                # we've seen it before at position `prev`
                # it means from prev to i we flipped each bit an even number of times
                # so the substring from prev to i should have even count for each vowel
                print(prev, i)
                # if prev == -1 then it means whole thing up to `i` is good
                # if prev != -1 do we want to include the index of prev?
                # this is actually doing s[prev+1:i+1]
                max_length = max(max_length, i - prev)

        return max_length
23000