<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) {
        unordered_map<string, int> mp;
    
        // Initialize an array to track the count of each vowel
        vector<int> vowelCount(5, 0); // Count of 'a', 'e', 'i', 'o', 'u'

        string currentState = "00000";  // Initial state where all vowels have even counts
        mp[currentState] = -1;
        
        int maxLength = 0;

        for (int i = 0; i < s.length(); ++i) {
            // Update count for the vowel encountered
            if (s[i] == 'a')      vowelCount[0] = (vowelCount[0] + 1) % 2;
            else if (s[i] == 'e') vowelCount[1] = (vowelCount[1] + 1) % 2;
            else if (s[i] == 'i') vowelCount[2] = (vowelCount[2] + 1) % 2;
            else if (s[i] == 'o') vowelCount[3] = (vowelCount[3] + 1) % 2;
            else if (s[i] == 'u') vowelCount[4] = (vowelCount[4] + 1) % 2;

            // Create a string representing the current parity state of vowels
            currentState = "";
            for (int j = 0; j < 5; ++j) {
                currentState += to_string(vowelCount[j]);
            }

            // Check if we've seen this state before
            if (mp.find(currentState) != mp.end()) {
                // Calculate the length of the substring
                maxLength = max(maxLength, i - mp[currentState]);
            } else {
                // Store the first occurrence of this state
                mp[currentState] = i;
            }
        }

        return maxLength;
    }
};
20800