<p>You are given a string <code>s</code>. An <strong>awesome</strong> substring is a non-empty substring of <code>s</code> such that we can make any number of swaps in order to make it a palindrome.</p>

<p>Return <em>the length of the maximum length <strong>awesome substring</strong> of</em> <code>s</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;3242415&quot;
<strong>Output:</strong> 5
<strong>Explanation:</strong> &quot;24241&quot; is the longest awesome substring, we can form the palindrome &quot;24142&quot; with some swaps.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;12345678&quot;
<strong>Output:</strong> 1
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;213123&quot;
<strong>Output:</strong> 6
<strong>Explanation:</strong> &quot;213123&quot; is the longest awesome substring, we can form the palindrome &quot;231132&quot; with some swaps.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= s.length &lt;= 10<sup>5</sup></code></li>
	<li><code>s</code> consists only of digits.</li>
</ul>

class Solution {
public:
    int longestAwesome(string s) {
        int n = s.length();
        int maxLength = 0;
        unordered_map<int, int> firstSeen; // store the first occurrence index of each bitmask 
        firstSeen[0] = -1;  // initialize for the case where the first substring is valid 
        int mask = 0; // initialize bitmask (all digit counts are even initially) 

        for (int i = 0; i < n; ++i) {
            int digit = s[i] - '0'; 
            mask ^= (1 << digit); // toggle the bit corresponding to the current digit 

            // check if we've seen this bitmask before
            if (firstSeen.count(mask)) { 
                maxLength = max(maxLength, i - firstSeen[mask]); // update max length
            } else {
                firstSeen[mask] = i; // store the first occurrence of this bitmask
            }

            // check for palindromes with one odd digit count 
            for (int j = 0; j < 10; ++j) {
                int tempMask = mask ^ (1 << j); // toggle each digit bit to check for odd count
                if (firstSeen.count(tempMask)) { 
                    maxLength = max(maxLength, i - firstSeen[tempMask]);
                }
            }
        }
        return maxLength;
    }
};
13600