<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:

        # helper function
        def helper(l: int) -> bool: 
            
            vowels = {"a", "e", "i", "o", "u"}
            counter = defaultdict(int)
            p1, p2 = 0, l-1

            # initial loop to form first subarray with length l
            for c in range(l): 
                if s[c] in vowels: 
                    counter[s[c]] = 1 if not counter[s[c]] else 0

            while True: 
                # check condition
                state = True
                for vowel in counter: 
                    if counter[vowel] == 1: 
                        state = False
                if state: 
                    return True
                elif p2 == LEN - 1: 
                    return False
                else:
                    p2 += 1
                    if s[p2] in vowels: 
                        counter[s[p2]] = 1 if not counter[s[p2]] else 0
                    if s[p1] in vowels: 
                        counter[s[p1]] = 1 if not counter[s[p1]] else 0
                    p1 += 1

        LEN = len(s)
        for i in range(LEN, -1, -1): 
            if helper(i): 
                return i
645