<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>

#pragma GCC optimize("O3", "unroll-loops")

auto init = []() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    return 'c';
}();

class Solution {
public:
    int findTheLongestSubstring(string s) {
        int n = s.size();
        int dp[32];
        fill_n(dp, 32, -2);
        dp[0] = -1;
        int maxInterval = 0;
        int bitFlag = 0;
        for(int i = 0; i < n; i++){
            char c = s[i];
            bitFlag = newBitFlag(c, bitFlag);
            if(dp[bitFlag] == -2) dp[bitFlag] = i;
            if(dp[bitFlag] != -2) maxInterval = max(maxInterval, i - dp[bitFlag]);
        }
        return maxInterval;
    }

    inline int newBitFlag(char c, int curr){
        int bit = 0;
        if(c == 'a') bit = 1;
        if(c == 'e') bit = 1<<1;
        if(c == 'i') bit = 1<<2;
        if(c == 'o') bit = 1<<3;
        if(c == 'u') bit = 1<<4;
        return curr ^ bit;
    }
};
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