<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(object):
    def __init__(self):
        #lookup table mapping vowels to bit locations for the XOR operation
        self.vowel_XOR_dict = {'a':1,'e':2,'i':4,'o':8,'u':16}

        
    def findTheLongestSubstring(self, s):        
        #min_index_list[vowel_parity = p] is the index of the min length prefix with vowel parity p
        #-1 corresponds to the empty prefix (which has parity 0), thus min_index_list[0]=-1
        min_index_list = {0:-1}
        
        #maximum length of substrings with even # of vowels
        max_length = 0
        
        #current vowel parity
        vowel_parity = 0
        
        #iterate through each character in s
        for i in range(len(s)):          
            #if we see a vowel...
            if s[i] in self.vowel_XOR_dict:
                #since the parity is changing at index i, 
                #then index i-1 is the largest index seen so far at the previous parity
                #so, update the maximum length substring value
                max_length = max(max_length,i-1-min_index_list[vowel_parity])
                
                #update the current parity
                vowel_parity^=self.vowel_XOR_dict[s[i]]
                
                #if we have not seen this parity before, put it in the minimum index list
                if vowel_parity not in min_index_list:
                    min_index_list[vowel_parity] = i
                    
        #update the maximum length substring value for the parity of the last prefix (whole string)      
        max_length = max(max_length,len(s)-1-min_index_list[vowel_parity])
        return max_length
125