<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:
    # Given a string s

    # Return the longest substring size containing each vowel and even number of times
    #   Since we are looking at substrings we can use a sliding window
    #   We could use some sort of dp to keep track of vowel counts more efficiently 
    def findTheLongestSubstring(self, s: str) -> int:
        constSet = {"a", "e", "i", "o", "u"}
        vowelSet = {"a", "e", "i", "o", "u"}

        for c in s:
            if c in constSet:
                if c in vowelSet:
                    vowelSet.remove(c)
                else:
                    vowelSet.add(c)

        maxSize = 0
        start = 0
        newVowelSet = vowelSet.copy()
        while start < len(s):
            newVowelSet = vowelSet.copy()
            end = len(s) - 1
            while end >= start + maxSize:
                if len(newVowelSet) == 5:
                    maxSize = max(end - start + 1, maxSize)
                    break
                else:
                    if s[end] in constSet:
                        if s[end] in newVowelSet:
                            newVowelSet.remove(s[end])
                        else:
                            newVowelSet.add(s[end])
                    end -= 1
            if s[start] in constSet:
                if s[start] in vowelSet:
                    vowelSet.remove(s[start])
                else:
                    vowelSet.add(s[start])
            start += 1
        return maxSize
            
            
            

        

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