<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) {
        long long cnt = 0;
        int ans = 0;
        unordered_map<long long,int> mp;
        mp[0] = -1;

        int pre = -1;
        for(int i = 0;i < s.length();++i){
            cnt ^= (1 << (s[i] - '0'));
            for(int j = 0;j < 10;++j){
                // if(cnt | (1 << j) != cnt) continue;
                int back = cnt ^ (1 << j);
                if(mp.find(back) != mp.end()){
                    ans = max(ans,i - mp[back]);
                }
            }
            if(mp.find(cnt) != mp.end()){
                ans = max(ans,i - mp[cnt]);
            } else {
                // for(int j = 0;j < 10;++j){
                //     // if(cnt | (1 << j) != cnt) continue;
                //     int back = cnt ^ (1 << j);
                //     if(mp.find(back) != mp.end()){
                //         ans = max(ans,i - mp[back]);
                //     }
                // }
                mp[cnt] = i;
            }
        }
        return max(ans,1);
    }
};
14000