Find the Longest Substring Containing Vowels in Even Counts
<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>
Bit Manipulation; Hash Table; String; Prefix Sum
class Solution {
    public int findTheLongestSubstring(String s) {
        String vowels = "aeiou";
        int[] d = new int[32];
        Arrays.fill(d, 1 << 29);
        d[0] = 0;
        int ans = 0, mask = 0;
        for (int i = 1; i <= s.length(); ++i) {
            char c = s.charAt(i - 1);
            for (int j = 0; j < 5; ++j) {
                if (c == vowels.charAt(j)) {
                    mask ^= 1 << j;
                    break;
                }
            }
            ans = Math.max(ans, i - d[mask]);
            d[mask] = Math.min(d[mask], i);
        }
        return ans;
    }
}
