<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:
        n = len(s)
        vowels = ['a', 'e', 'i', 'o', 'u']


        # prefix_bitmask[i] & 2**b  = whether there is an odd number of occurences of vowels[b] in s[:i]
        prefix_bitmask = [0 for _ in range(n+1)]

        for i in range(1, n+1):
            c = s[i-1]
            if c in vowels:
                idx = vowels.index(c)
                prefix_bitmask[i] = prefix_bitmask[i-1] ^ 2**idx
            else:
                prefix_bitmask[i] = prefix_bitmask[i-1]
        
        longest = 0
        for i in range(n+1):
            for j in range(i + longest, n+1):
                if prefix_bitmask[i] ^ prefix_bitmask[j] == 0:
                    longest = j - i
        return longest


        # Does it start with s[0]?
        # If no, F(s[1:])
        # If yes, jump out to end counting all counts, and go backwards until you get a valid even
        
        # then if we remove s[0] and it was not a vowel, 

        # for end in range(len(s)):
        #     print(s[start:end])
        #     c = s[end]
        #     if c in hasOddNumber:
        #         hasOddNumber[c] = not hasOddNumber[c]
        #     while True in hasOddNumber.values() and start < end:
        #         start += 1
        #         c = s[start]
        #         if c in hasOddNumber:
        #             hasOddNumber[c] = not hasOddNumber[c]
        #     longest = max(longest, end-start)
        # return longest
                

        

        return 0

664