<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:
        last_even_odd_pos = [-1] * 32
        char_bits = [0] * 26
        char_shift = [0] * 26
        idx = 0
        a_code = ord("a")
        a_idx = ord("a") - a_code
        e_idx = ord("e") - a_code
        i_idx = ord("i") - a_code
        o_idx = ord("o") - a_code
        u_idx = ord("u") - a_code
        char_shift[e_idx] = 1
        char_shift[i_idx] = 2
        char_shift[o_idx] = 3
        char_shift[u_idx] = 4
        even_odd_idx = 0
        for c in s:
            char_idx = ord(c) - a_code
            char_bits[char_idx] ^= 1 << char_shift[char_idx]
            even_odd_idx = (
                char_bits[a_idx]
                + char_bits[e_idx]
                + char_bits[i_idx]
                + char_bits[o_idx]
                + char_bits[u_idx]
            )
            last_even_odd_pos[even_odd_idx] = idx
            idx += 1

        for i in range(len(char_bits)):
            char_bits[i] = 0

        idx = 0
        max_dist = 0
        even_odd_idx = 0
        for c in s:
            dist = last_even_odd_pos[even_odd_idx] - idx + 1
            max_dist = max(dist, max_dist)
            char_idx = ord(c) - a_code
            char_bits[char_idx] ^= 1 << char_shift[char_idx]
            even_odd_idx = (
                char_bits[a_idx]
                + char_bits[e_idx]
                + char_bits[i_idx]
                + char_bits[o_idx]
                + char_bits[u_idx]
            )
            idx += 1

        return max_dist

636