<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 stateString(self, index, s):
        ans = ""
        for i in range(5):
            if i == index:
                if s[i] == "0":
                    ans += "1"
                else:
                    ans += "0"
            else:
                ans += s[i]
        return ans

    def findTheLongestSubstring(self, s: str) -> int:
        # 00000 => even #11111 => odd
        # aeiou
        v = "aeiou"
        indic = {"a": 0, "e": 1, "i": 2, "o": 3, "u": 4}
        ans = 0
        states = {}
        startState = "00000"
        temp = 0
        if s[0] in v:
            states[self.stateString(indic[s[0]], startState)] = 0
        for i in range(len(s)):
            c = s[i]
            if c in v:
                startState = self.stateString(indic[c], startState)

            if startState == "00000":
                temp = i + 1
            elif startState not in states :      
                states[startState] = i
            else  : 
                temp = i - states[startState]
                
            if temp > ans : 
                ans = temp
        return ans

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