<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:
    bool checkVowel(char ch){
        return ch=='a' || ch=='e' || ch=='o'||ch=='i'||ch=='u';
    }
    int findTheLongestSubstring(string s) {
        unordered_map<char , int>position={{'a', 0}, {'e', 1}, {'i', 2}, {'o', 3}, {'u', 4}};
        map<int,int>last_occured;
        last_occured[0]=-1;

        int cur = 0;
        int result=0;

        for(int i=0 ; i<s.size() ; i++){
            if(checkVowel(s[i])){
                cur^= (1<<position[s[i]]);
            }

            if(last_occured.find(cur )!=last_occured.end()){
                result = max(result , i - last_occured[cur]);
            }
            else{
                last_occured[cur] = i;
            }
        }

        return result;
        
    }
    //refer:https://youtu.be/rXedJboEcxU?si=TM6biW5W8PG51twZ;
};
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