<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 n = s.size();
        unordered_map<char,int> vowelToBit = {{'a', 1}, {'e', 2}, {'i', 4}, {'o', 8}, {'u', 16}};
        
        int mask = 0; // Current bitmask representing the state of vowel counts
        unordered_map<int, int> maskIndex; // Maps each mask to the earliest index where it occurred
        maskIndex[0] = -1; // Initialize with mask 0 at index -1 (no vowels seen)
        
        int ans = 0;
        for (int i = 0; i < n; ++i) {
            if (vowelToBit.count(s[i])) {
                // Toggle the bit corresponding to the current vowel
                mask ^= vowelToBit[s[i]];
            }
            // Check if this mask was seen before
            if (maskIndex.count(mask)) {
                // If yes, calculate the distance from the previous occurrence
                ans = max(ans, i - maskIndex[mask]);
            } else {
                // If not, store the current index for this mask
                maskIndex[mask] = i;
            }
        }
        return ans;
    }
};

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