<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();
        vector<int>state(5,0);
        unordered_map<string,int>mp;
        int result=0;
        string cs="00000";
        mp[cs]=-1;
        for(int i=0;i<n;i++){
            if(s[i]=='a') state[0]=state[0]^1;
            else if(s[i]=='e') state[1]=state[1]^1;
            else if(s[i]=='i') state[2]=state[2]^1;
            else if(s[i]=='o') state[3]=state[3]^1;
            else if(s[i]=='u') state[4]=state[4]^1;
        
        cs="";
        for(int j=0;j<5;j++){
            cs+=to_string(state[j]);
        }
        if(mp.find(cs)!=mp.end()) result = max(result,i-mp[cs]);
        else mp[cs]=i;
        }
        return result;
    
    }
};
20900