<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:
        n = len(s)
        #st1 = set('aeiou')
        #st2 = set(s)
        #if len(st1.intersection(st2)) == 0 : return n
        tmp = dict()
        tmp['a']=0; tmp['e']=1; tmp['i']=2
        tmp['o']=3; tmp['u']=4
        d = dict()
        cur = [0,0,0,0,0]
        if s[0] in tmp:
            curi = s[0]
            indx = tmp[curi]
            cur[indx]+=1
        d[tuple(cur)] = [0]
        d[tuple([0,0,0,0,0])] = [-1]            ###
       
        res = 0
        for i in range(1,n):
            curi = s[i]
            if curi in tmp:
                indx = tmp[curi]
                cur[indx]+=1
                cur[indx] %=2
            if tuple(cur) in d:
                d[tuple(cur)] += [i]
                #curL = len(s[d[tuple(cur)][0]:i+1])
                #if cur != [0,0,0,0,0]: curL -=1
                #res = max(res,curL)
                #print(d)                                #
                #print(s[d[tuple(cur)]:i+1],cur)         #
                #print('\t!',res,'\n\n')                 #
            else: d[tuple(cur)] = [i]
        #print(d)
        for k in d:
            cur = d[k]
            ln = cur[-1] - cur[0]
            if ln > res : res = ln
        return res

        
522