<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();
        s = '@' + s;
        vector<string> ret;
        unordered_map<int, vector<int>> psa;
        unordered_map<int, int> l, r;
        for(int i = 'a';i<='z';++i) {
            psa[i].resize(s.size());
            for(int j = 1;j<=n;++j) {
                psa[i][j] = psa[i][j - 1] + (s[j] == i);
            }
        }
        for(int i = 1;i<=n;++i) {
            r[s[i]] = i;
        }
        for(int i = n;i;--i) {
            l[s[i]] = i;
        }
        for(int _ = 0;_<=100;++_) {
            for(int i = 'a';i<='z';++i) if(l.count(i)) {
                for(int j = 'a';j<='z';++j) if(l.count(j)) {
                    if(psa[j][r[i]] - psa[j][l[i] - 1]) {
                        l[i] = min(l[i], l[j]);
                        r[i] = max(r[i], r[j]);
                    }
                }
            }
        }
        vector<pair<int, int> > a;
        for(int i = 'a';i<='z';++i) if(l.count(i)) {
            a.push_back({l[i], r[i]});
        }
        sort(a.begin(), a.end(), [](pair<int, int>& x, pair<int, int>& y) {
            return x.second < y.second;
        });
        vector<pair<int, int> > ans;
        for(auto[l, r] : a) {
            if(ans.empty()) ans.push_back({l, r});
            else {
                auto[ll, rr] = ans.back();
                if(rr > l) {
                    if(rr - ll > r - l) {
                        ans.pop_back();
                        ans.push_back({l, r});
                    }
                } else ans.push_back({l, r});
            }
        }
        for(auto[l, r] : ans) ret.push_back(s.substr(l, r - l + 1));
        return ret;
    }
};
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