<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>

#include <unordered_map>
#include <set>
#include <string>

class Solution {

std::unordered_map<int, size_t> location_map{{0, -1}};

//for each character we want to compute how many even or odds we have seen of each vowel so far so we use 
// a 5 bit key 00000 where the lsb is for a, second least for e nd etc
inline static int computeBitMask(char c, int n) {
    static const std::set<char> vowels{'a', 'e', 'i', 'o', 'u'};
    if(vowels.contains(c)){
        switch(c) {
            case 'a':
                n ^= 1;
                break;
            case 'e':
                n ^= 2;
                break;
            case 'i':
                n ^= 4;
                break;
            case 'o':
                n ^= 8;
                break;
            case 'u':
                n ^= 16;
                break;                
        }

    }
    return n;    
}

public:
    int findTheLongestSubstring(string s) {
        int bit_mask_value = 0, longest_substr = 0;
        for(size_t i = 0; i < s.size(); i++) {
            bit_mask_value = computeBitMask(s.at(i), bit_mask_value);
            if(!location_map.contains(bit_mask_value)) {
                location_map[bit_mask_value] = i;
            }
            else{
                longest_substr = std::max(longest_substr, static_cast<int>(i - location_map[bit_mask_value]));
            }
        }
        return longest_substr;
    }
};
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