<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>

def vowel_count_even(vcount):
    if not vcount:
        return True
    for k,v in vcount.items():
        if v%2 != 0:
            print(v%2)
            return False
    return True

def update_count(rem, add, vcount):
    vowels = ['a', 'e', 'i', 'o', 'u']

    if rem in vowels:
        remcount = vcount.get(rem)
        remcount -= 1
        if remcount >= 0:
            vcount[rem] = remcount
        else:
            vcount.remove(rem)
    
    if add in vowels:
        addcount = vcount.get(add,0)
        addcount += 1
        vcount[add] = addcount
    



class Solution:
    def findTheLongestSubstring(self, s: str) -> int:
        vowels = ['a', 'e', 'i', 'o', 'u']
        # vcount = {}
        # for c in s:
        #     if c in vowels:
        #         count = vcount.get(c, 0)
        #         vcount[c] = count+1
        # if vowel_count_even(vcount):
        #     return len(s)

        ws = len(s)

        while ws > 0:
            vcount = {}
            ss = s[:ws]
            for c in ss:
                if c in vowels:
                    count = vcount.get(c, 0)
                    vcount[c] = count+1
            # print(vcount)
            if vowel_count_even(vcount):
                # print('even')
                return len(ss)

            # print(s[:ws])
            # print(vcount)
            num_steps = len(s)-ws

            # print('num steps',num_steps)
            for i in range(num_steps):
                update_count(s[i], s[i+ws], vcount)
                if vowel_count_even(vcount):
                    return ws

            ws -= 1


        return 0
23100