<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 ev=5;
        string str="";
        for(int i=0; i<26;i++)
        {
            str+='0';
        }
        int l=-1;
        int h=-1;
        int ans=0;
        unordered_map<int,unordered_map<string,int>>mp;
        mp[5][str]=0;
        for(int i=0; i<s.size();i++)
        {
            if(s[i]=='a'|| s[i]=='e'||s[i]=='o'|| s[i]=='u'|| s[i]=='i')
            {
                if(str[s[i]-'a']=='0')
                {
                    str[s[i]-'a']='1';
                    ev--;
                }
                else 
                {
                     str[s[i]-'a']='0';
                     ev++;
                }
                if(!mp[ev].count(str))
                {
                    mp[ev][str]=i+1;
                }
                else
                {
                    if(i-mp[ev][str]+1>ans)
                    {
                        ans=i-mp[ev][str]+1;
                        l=mp[ev][str];
                        h=i;
                    }
                }
        
            }
            else
            {
                         if(i==13)
            {
                cout<<mp[ev][str]<<endl;
                 cout<<l<<" "<<h;
            }
                if(ev==5)
                {
                   ans=i+1;
                   l=0;
                   h=i;
                }
                else if(mp[ev].count(str))
                {
                    if(i-mp[ev][str]+1>ans)
                    {
                        ans=i-mp[ev][str]+1;
                        l=mp[ev][str];
                        h=i;
                    }
                    
                }
            }
           
        }
       
        
        return ans;
        
        
        
    }
};
20000