<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>

using namespace std;

// Custom hash function for vector<int>
struct VectorHash {
    size_t operator()(const vector<int>& v) const {
        size_t hash = 0;
        for (int i : v) {
            hash ^= hash * 31 + std::hash<int>()(i);  // Combine the hash values of all elements
        }
        return hash;
    }
};

class Solution {
public:
    int findTheLongestSubstring(string s) {
                // Array to store vowel counts: a -> 0, e -> 1, i -> 2, o -> 3, u -> 4
        unordered_map<char, int> vowelToIndex = {{'a', 0}, {'e', 1}, {'i', 2}, {'o', 3}, {'u', 4}};
        vector<int> vowelCount(5, 0);  // Array to track counts of vowels
        
        // Map to track the first occurrence of vowel counts with custom hash
        unordered_map<vector<int>, int, VectorHash> firstOccurrence;
        
        // Initialize the first occurrence map with the empty state (all counts 0)
        firstOccurrence[vowelCount] = -1;
        
        int maxLength = 0;
        
        for (int i = 0; i < s.length(); i++) {
            // If the character is a vowel, update the vowel count
            if (vowelToIndex.count(s[i])) {
                vowelCount[vowelToIndex[s[i]]] ^= 1;  // Toggle between odd and even count
            }
            
            // Check if we have seen this count combination before
            if (firstOccurrence.count(vowelCount)) {
                maxLength = max(maxLength, i - firstOccurrence[vowelCount]);
            } else {
                firstOccurrence[vowelCount] = i;
            }
        }
        
        return maxLength;
    }

    // bool isValid(const string &s, int start, int end) {
    //     vector<int> vowels(5, 0);
    //     for (int i = start; i < end; i++) {
    //         if (s[i] == 'a') vowels[0]++;
    //         else if (s[i] == 'e') vowels[1]++;
    //         else if (s[i] == 'i') vowels[2]++;
    //         else if (s[i] == 'o') vowels[3]++;
    //         else if (s[i] == 'u') vowels[4]++;
    //     }
    //     for (int i = 0; i < 5; i++) {
    //         if (vowels[i] % 2 != 0) return false;
    //     }
    //     return true;
    // }

    // void backtrack(const string &s, int ind, int &maxLen) {
    //     if (ind == s.length()) {
    //         return;
    //     }

    //     for (int i = ind; i < s.length(); i++) {
    //         if (isValid(s, ind, i)) {
    //             maxLen = max(maxLen, i = ind + 1);
    //         }
    //     }
    //     if (ind+1 < s.length()) {
    //         backtrack(s, ind+1, maxLen);
    //     }  
    // }

    // int findTheLongestSubstring(string s) {
    //     int maxLen(0);
    //     backtrack(s, 0, maxLen);
    //     return maxLen;
    // }
};
20200