<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 ans = 0;
        map<map<char, int>, int> mp;
        map<char, int> mp1;
        mp1['a'] = 0;
        mp1['e'] = 0;
        mp1['i'] = 0;
        mp1['o'] = 0;
        mp1['u'] = 0;

        // Initialize with a starting point for empty prefix.
        mp[mp1] = -1;

        for (int i = 0; i < s.size(); i++) {
            if (s[i] == 'a' || s[i] == 'e' || s[i] == 'i' || s[i] == 'o' ||
                s[i] == 'u') {
                mp1[s[i]]++; // Update vowel count
                if (mp1[s[i]] % 2 == 0) {
                    mp1[s[i]] = 0; // Reset to even parity (0)
                }
            }

            // Check if the current state of vowels has been seen before
            if (mp.find(mp1) != mp.end()) {
                ans = max(ans, i - mp[mp1]);
            } else {
                mp[mp1] = i; // Store the first occurrence of this state
            }
        }
        return ans;
    }
};

19700