<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>

int nums[32][2];
int num = [](){
    ios_base::sync_with_stdio(false);
    cin.tie(NULL); cout.tie(NULL);
    return 0;
}();
class Solution {
public:
    int findTheLongestSubstring(const string &s) {
        for(int i = 0; i < 32; i++) nums[i][0] = nums[i][1] = -2;
        nums[0][0] = nums[0][1] = -1;
        num = 0;
        int idx = 0;
        for(const char &c : s){
            switch(c){
                case 'a':
                num ^= 1;
                break;
                case 'e':
                num ^= 2;
                break;
                case 'o':
                num ^= 4;
                break;
                case 'i':
                num ^= 8;
                break;
                case 'u':
                num ^= 16;
                break;
                default:
                break;
            }
            if(nums[num][0] == -2) nums[num][0] = nums[num][1] = idx; else nums[num][1] = idx;
            idx++;
        }
        idx = 0;
        for(int i = 0; i < 32; i++){
            idx = max(idx, nums[i][1] - nums[i][0]);
        }
        return idx;
    }
};
15600