<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:
    int findTheLongestSubstring(string s) {
        int length = 0;
        int n = s.length();
        map<string, int> indexes;
        string evenOdd = "00000";
        indexes[evenOdd] = -1;
        map<char, int> vowels;
        vowels['a'] = 0;
        vowels['e'] = 1;
        vowels['i'] = 2;
        vowels['o'] = 3;
        vowels['u'] = 4;
        for (int i = 0; i < n; i++) {
            char current = s[i];
            auto it = vowels.find(current);
            if (it != vowels.end()) {
                evenOdd[it->second] = evenOdd[it->second] == '0' ? '1' : '0';
            }

            auto lastIndex = indexes.find(evenOdd);
            if (lastIndex == indexes.end()) {
                indexes[evenOdd] = i;
            } else {
                length = max(length, i - lastIndex->second);
            }
        }
        return length;
    }
};
18200