<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:

        d = defaultdict(list)
        ans = 0
        mask = 0
        
        for i in range(0, len(s)):
            if s[i] in ["a", "e", "i", "o", "u"]:
                index = ["a", "e", "i", "o", "u"].index(s[i])
                # print(i)
                # print(bin(mask)[2:], bin(1 << index)[2:], bin((~(1 << index) & ((1 << index)-1)))[2:])
                
                if (1 << index) & mask:
                    # print("mask_and")
                    # print(bin(mask)[2:], bin(1<<index)[2:])
                    # mask &= (~(1 << index) & ((1 << (index+1))-1))
                    mask &= ~(1 << index)
                    # print(mask, ~(1 << index))
                else:
                    # print("mask_or")
                    mask |= (1 << index)
                
                # print(mask, index, bin(mask)[2:], i, s[i])

            if mask == 0:
                # print(count)
                ans = i + 1
            d[mask].append(i)

        # print(d) 
        for k, v in d.items():
            if len(v) > 1:
                ans = max(ans, v[-1]-v[0])
        return ans

            
        

         
238