<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:
        
        '''
        We check for each indec where the cnt of each vowel is even and where it is odd
        False -> Even and True -> Odd
        Then where ever we find same vector again, we got even length there
        Example: At ind = 1, we have (False, True, False, False, False) => e is odd at index 1
        Again at ind = 5 we have (False, True, False, False, False) => here also no. of e is odd
        which means if we take string from ind = 2 to 5, we will get even occurance of all vowels
        as say if at index = 1, cnt[e] = 1 and at ind = 5, cnt[e] = 3, form index 2 to 5, we have
        3-1 = 2 occurance of e i.e. even
        '''
        
        integrals = [(False, False, False, False, False)]
        mapping = {"a": 0, "i": 1, "u": 2, "e": 3, "o": 4}

        for v in s:
            vector = list(integrals[-1])
            if v in mapping:
                vector[mapping[v]] = not vector[mapping[v]]
            integrals.append(tuple(vector))

        seen = {}
        res = 0

        for i, v in enumerate(integrals):
            if v in seen:
                res = max(res, i - seen[v]) 
            else:
                seen[v] = i

        return res
570