<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:
        # 初始化元音 a, e, i, o, u 的奇偶計數
        count = [0] * 5
        
        # 字典存儲每個位元掩碼狀態的首次出現位置
        Map = {0: -1}
        
        ret = 0
        
        # 遍歷字串
        for i, char in enumerate(s):
            if char == 'a':
                count[0] = (count[0] + 1) % 2  # 切換 'a' 的奇偶次數
            if char == 'e':
                count[1] = (count[1] + 1) % 2  # 切換 'e' 的奇偶次數
            if char == 'i':
                count[2] = (count[2] + 1) % 2  # 切換 'i' 的奇偶次數
            if char == 'o':
                count[3] = (count[3] + 1) % 2  # 切換 'o' 的奇偶次數
            if char == 'u':
                count[4] = (count[4] + 1) % 2  # 切換 'u' 的奇偶次數
            
            # 計算當前位元掩碼
            code = 0
            for j in range(5):
                code = code * 2 + count[j]
            
            # 檢查這個位元掩碼是否已經出現過
            if code in Map:
                ret = max(ret, i - Map[code])
            else:
                Map[code] = i        
        return ret

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