<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:// easy ni h dekh kr smj kr kiya maine
    int findTheLongestSubstring(string s) {
        int ans =0,j=0;
        map<int,int> M{{0,-1}};
        for(int i=0;i<s.size();++i){
            // M[s[i]]++;
            // while(M['a']>2 or M['e']>2 or M['i']>2 or M['o']>2 or M['u']>2){
            //     M[s[j]]--;++j;
            // }
            j ^= 1 << string("aeiou").find(s[i]) + 1 >> 1;
            if (!M.count(j)) M[j] = i;
            //if(M['a']%2==0 and M['e']%2==0 and M['i']%2==0 and M['o']%2==0 and M['u']%2==0){
            ans=max(ans,i-M[j]);
            //}
        }
        return ans;
    }
};
19800