<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>

from collections import defaultdict

class Solution:
    def findTheLongestSubstring(self, s: str) -> int:
        seen = defaultdict(tuple)
        seen[(0,0,0,0,0)] = -1
        ans = 0
        curr_a = curr_e = curr_i = curr_o = curr_u = 0
        for i in range(len(s)):
            if s[i] == 'a':
                curr_a += 1
                curr_a %= 2
            elif s[i] == 'e':
                curr_e += 1
                curr_e %= 2
            elif s[i] == 'i':
                curr_i += 1
                curr_i %= 2
            elif s[i] == 'o':
                curr_o += 1
                curr_o %= 2
            elif s[i] == 'u':
                curr_u += 1
                curr_u %= 2
            
            if (curr_a, curr_e, curr_i, curr_o, curr_u) in seen:
                ans = max(ans, i - seen[(curr_a, curr_e, curr_i, curr_o, curr_u)])
            
            else:
                seen[(curr_a, curr_e, curr_i, curr_o, curr_u)] = i
        
        return ans





        
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