<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 checkIfEven(self, amount: List[int]) -> bool:
        for x in amount:
            if (x % 2):
                return False
        return True

    def findTheLongestSubstring(self, s: str) -> int:
        amount = [0,0,0,0,0] # e, i, o, a, u
        vowels = []

        for i in range(len(s)):
            x = s[i]

            if (x == 'e'):
                amount[0] += 1
                vowels.append((i, amount.copy()))
            elif (x == 'i'):
                amount[1] += 1
                vowels.append((i, amount.copy()))
            elif (x == 'o'):
                amount[2] += 1
                vowels.append((i, amount.copy()))
            elif (x == 'u'):
                amount[3] += 1
                vowels.append((i, amount.copy()))   
            elif (x == 'a'):
                amount[4] += 1
                vowels.append((i, amount.copy()))

        high = len(s)

        if (self.checkIfEven(amount)):
            return high
        
        if (len(vowels) == 1):
            return max(vowels[0][0], high - (vowels[0][0] + 1))

        longest = 0
        
        for i in range(len(vowels) - 1, -1, -1):
            current = vowels[i]
            if self.checkIfEven(current[1]):
                return vowels[i+1][0]

            low = 0

            for vowel in vowels:
                foo = list(map(lambda a, b: a - b, current[1], vowel[1]))
                
                temp = vowel[0] + 1
                if (temp != high):
                    low = vowel[0] + 1
                else:
                    high = high - 1

                length = (high - low)

                if (length <= longest):
                    break
                elif (sum(foo) == 1):
                    longest = max(longest, current[0] - low, high - (current[0] + 1))
                elif (self.checkIfEven(foo)):
                    longest = length
                    break
            
            high = current[0]

            if (high <= longest):
                break
        return longest
172