<p>Given a string <code>s</code> of lowercase letters, you need to find the maximum number of <strong>non-empty</strong> substrings of <code>s</code> that meet the following conditions:</p>

<ol>
	<li>The substrings do not overlap, that is for any two substrings <code>s[i..j]</code> and <code>s[x..y]</code>, either <code>j &lt; x</code> or <code>i &gt; y</code> is true.</li>
	<li>A substring that contains a certain character <code>c</code> must also contain all occurrences of <code>c</code>.</li>
</ol>

<p>Find <em>the maximum number of substrings that meet the above conditions</em>. If there are multiple solutions with the same number of substrings, <em>return the one with minimum total length. </em>It can be shown that there exists a unique solution of minimum total length.</p>

<p>Notice that you can return the substrings in <strong>any</strong> order.</p>

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

<pre>
<strong>Input:</strong> s = &quot;adefaddaccc&quot;
<strong>Output:</strong> [&quot;e&quot;,&quot;f&quot;,&quot;ccc&quot;]
<b>Explanation:</b>&nbsp;The following are all the possible substrings that meet the conditions:
[
&nbsp; &quot;adefaddaccc&quot;
&nbsp; &quot;adefadda&quot;,
&nbsp; &quot;ef&quot;,
&nbsp; &quot;e&quot;,
  &quot;f&quot;,
&nbsp; &quot;ccc&quot;,
]
If we choose the first string, we cannot choose anything else and we&#39;d get only 1. If we choose &quot;adefadda&quot;, we are left with &quot;ccc&quot; which is the only one that doesn&#39;t overlap, thus obtaining 2 substrings. Notice also, that it&#39;s not optimal to choose &quot;ef&quot; since it can be split into two. Therefore, the optimal way is to choose [&quot;e&quot;,&quot;f&quot;,&quot;ccc&quot;] which gives us 3 substrings. No other solution of the same number of substrings exist.
</pre>

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

<pre>
<strong>Input:</strong> s = &quot;abbaccd&quot;
<strong>Output:</strong> [&quot;d&quot;,&quot;bb&quot;,&quot;cc&quot;]
<b>Explanation: </b>Notice that while the set of substrings [&quot;d&quot;,&quot;abba&quot;,&quot;cc&quot;] also has length 3, it&#39;s considered incorrect since it has larger total length.
</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> contains only lowercase English letters.</li>
</ul>

class Solution {
public:
    vector<string> maxNumOfSubstrings(string s) {
        int n = s.size();
        vector<int>left(26, INT_MAX), right(26, -1);
        for(int i = 0 ; i < n ; i++){
            char ch = s[i];
            left[s[i]-'a'] = min(left[s[i]-'a'], i);
            right[s[i]-'a'] = max(right[s[i]-'a'], i);
        }
        vector<set<int>>adj(26);
        for(int i = 0 ; i < n; i++){
            for(int j = 0 ; j < 26; j++){
                if(left[j]<i&&right[j]>i){
                    adj[j].insert(s[i]-'a');
                }
            }
        }
        vector<string>ans;
        int lt = -1, rt = n, lasttaken = -1;
        for(int i = 0 ; i < n; i++){
            if(i==left[s[i]-'a']){
                lt = i;
                rt = right[s[i]-'a'];
                vector<int>vis(26, 0);
                queue<int>q;
                q.push(s[i]-'a');
                while(!q.empty()){
                    int a = q.front(); q.pop();
                    lt = min(lt, left[a]);
                    rt = max(rt, right[a]);
                    for(auto &child : adj[a]){
                        if(vis[child]==0){
                            vis[child] = 1;
                            q.push(child);
                        }
                    }                    
                }
                if(lt<lasttaken){
                    lt = -1;
                    rt = -1;
                }
            }
            else {
                lt = min(lt, left[s[i]-'a']);
                if(lt<lasttaken){
                    lt = -1;
                    rt = -1;
                }
                rt = max(rt, right[s[i]-'a']);
            }
            if(i>=rt&&rt!=-1&&lt!=-1){
                lasttaken = i;
                ans.push_back(s.substr(lt, rt-lt+1));
                rt = n;
            }
        }
        return ans;
    }
};
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