<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) {
            // Create two hashmaps
    unordered_map<string, int> indexes;
    unordered_map<char, int> chars(
        { { 'a', 0 }, { 'e', 1 }, 
          { 'i', 2 }, { 'o', 3 }, 
          { 'u', 4 } });
 
    // Keep the track of frequencies
    // of the vowels
    string evenOdd = "00000";
    indexes[evenOdd] = -1;
 
    // Stores the maximum length
    int length = 0;
 
    // Traverse the given string S
    for (int i = 0; i < s.size(); ++i) {
 
        char c = s[i];
 
        // Find character in the map
        auto it = chars.find(c);
 
        // If it is a vowel, then update
        // the frequency
        if (it != chars.end()) {
            evenOdd[it->second]
                = evenOdd[it->second]
                          == '0'
                      ? '1'
                      : '0';
        }
 
        // Find the index of occurrence
        // of the string evenOdd in map
        auto lastIndex = indexes.find(evenOdd);
 
        if (lastIndex == indexes.end()) {
            indexes[evenOdd] = i;
        }
 
        // Update the maximum length
        else {
            length = max(
                length, i - lastIndex->second);
        }
    }
 
        // Print the maximum length
        cout << length;
        return length;
    }
};
18100