<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
{
    int if_vowel(char &c)
    {
        if (c == 'a')
            return 1;
        else if (c == 'e')
            return 2;
        else if (c == 'i')
            return 3;
        else if (c == 'o')
            return 4;
        else if (c == 'u')
            return 5;
        else
            return 0;

        return 0;
    }

    bool if_even_vowel_count(vector<int> arr)
    {
        for (int &x : arr)
            if (x & 1)
                return false;

        return true;
    }

public:
    int findTheLongestSubstring(string &s)
    {
        vector<int> vowels(5, 0);
        for (char &c : s)
        {
            int t = if_vowel(c);
            if (t)
                vowels[t - 1]++;
        }

        int ans = 0;
        for (int len = s.length(); len > 0; len--)
        {
            auto t = vowels;
            for (int start = 0; (start + len - 1) < s.length(); start++)
            {
                if (if_even_vowel_count(t))
                    return len;

                if ((start + len) < s.length())
                {
                    int m = if_vowel(s[start + len]);
                    if (m)
                        t[m - 1]++;
                }

                    int n = if_vowel(s[start]);
                    if (n)
                        t[n - 1]--;
            }
            int k = if_vowel(s[len - 1]);
            if (k)
                vowels[k - 1]--;
        }

        return ans;
    }
};
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