<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:
    void dfs(vector<vector<int>>& graph, int i, vector<int>& visited, stack<int>& st) {
        visited[i] = 1;
        for (int j = 0; j < graph[i].size(); j++) {
            if (graph[i][j] == 0 || visited[j]) continue;
            dfs(graph, j, visited, st);
        }
        st.push(i);
    }
    void dfs(vector<vector<int>>& graph, int i, vector<int>& batches, int batch, vector<int>& degree) {
        if (batches[i] < 0) {
            batches[i] = batch;
            for (int j = 0; j < graph.size(); j++) {
                if (graph[j][i] == 0) continue;
                dfs(graph, j, batches, batch, degree);
            }
        } else if (batches[i] != batch) {
            degree[batches[i]]++;
        }
    }
    vector<string> maxNumOfSubstrings(string s) {
        vector<int> mins(26, INT_MAX);
        vector<int> maxs(26, INT_MIN);
        vector<vector<int>> prefix(s.size() + 1, vector<int>(26));
        for (int i = 0; i < s.size(); i++) {
            for (int j = 0; j < 26; j++) {
                prefix[i + 1][j] = prefix[i][j];
            }
            int idx = s[i] - 'a';
            prefix[i + 1][idx]++; 
            mins[idx] = min(mins[idx], i);
            maxs[idx] = max(maxs[idx], i);
        }
        vector<vector<int>> graph(26, vector<int>(26));

        for (int i = 0; i < 26; i++) {
            if (mins[i] == INT_MAX) continue;
            for (int j = 0; j < 26; j++) {
                if (prefix[maxs[i] + 1][j] - prefix[mins[i]][j] > 0) {
                    graph[i][j] = 1;
                }
            }
        }
        stack<int> st;
        vector<int> visited(26, 0);
        for (int i = 0; i < 26; i++) {
            if (mins[i] == INT_MAX) continue;
            if (visited[i]) continue;
                dfs(graph, i, visited, st);
        }
        vector<int> batches(26, -1);
        vector<int> degree(26);
        fill(visited.begin(), visited.end(), 0);
        int batch = 0;
        while (!st.empty()) {
            int cur = st.top();
            st.pop();
            if (batches[cur] != -1) continue;
            dfs(graph, cur, batches, batch, degree);
            batch++;
        }
        vector<string> res;
        for (int i = batch - 1; i >= 0; i--) {
            if (degree[i] == 0) {
                int mini = INT_MAX, maxi = INT_MIN; 
                for (int j = 0; j < 26; j ++) {
                    if (batches[j] == i) {
                        mini = min(mins[j], mini); 
                        maxi = max(maxs[j], maxi); 
                    }
                }
                res.push_back(s.substr(mini, maxi - mini + 1));
            }
        }   

        return res;
    }
};

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