<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 {
public:
    Solution(){
        std::ios::sync_with_stdio(false);
        std::cin.tie(nullptr);
        std::cout.tie(nullptr);
    }
    bool flip(char& c, int& f) {
        switch (c) {
        case 'a': f ^= 1;  break;
        case 'e': f ^= 2;  break;
        case 'i': f ^= 4;  break;
        case 'o': f ^= 8;  break;
        case 'u': f ^= 16; break;
        default: return false;
        }
        return true;
    }
    int findTheLongestSubstring(string s) {
        int p = 0, q = 0, l = s.length(), m = -1;
        int f = 0, fl = 0;
        vector<int> v = {-1};
        int itP = 0, itQ = 1, it;
        while (q < l) {
            if (flip(s[q], f))
                v.push_back(q);
            q++;
            if (!f) {
                m = q;
                itQ = v.size();
            }
        }
        v.push_back(l);
        if (m == l)
            return m;
        f = 0;
        // delta = *itQ - *itP - 1;
        // s....v.....v...v...v...v..e
        //   itP^~~~~delta~~~~^itQ
        while (1) {
            itP++;
            if (m >= l - v[itP] - 1)
                break;
            flip(s[v[itP]], f);
            if (itQ == itP) {
                f = 0;
                itQ = itP + 1;
                if (itQ == v.size())
                    break;
                m = max(m, v[itQ] - v[itP] - 1);
            }
            it = itQ;
            fl = f;
            while (it + 1 < v.size()) {
                flip(s[v[it]], fl);
                it++;
                if (!fl) {
                    f = fl;
                    itQ = it;
                    m = max(m, v[itQ] - v[itP] - 1);
                }
            }
        }
        return m;
    }
};
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