<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 {
private:
    void dfs(int node, vector<vector<int>> &g, vector<int> &vis, stack<int> &st) {
        vis[node] = 1;
        for(auto it : g[node]) {
            if(!vis[it]) {
                dfs(it, g, vis, st);
            }
        }
        st.push(node);
    }

    void dfs2(int node, vector<vector<int>> &gr, vector<int> &vis, vector<int> &temp) {
        vis[node] = 1;
        if(gr[node].size() == 0) return;
        temp.push_back(node);
        for(auto it : gr[node]) {
            if(!vis[it]) {
                dfs2(it, gr, vis, temp);
            }
        }
    }

    void dfs_to_check_outdegree(int node, vector<vector<int>> &g, map<int, int> &mp, vector<int> &visited, bool &flag) {
        visited[node] = 1;

        for(auto it : g[node]) {
            if(mp[it] != 1) {
                flag = false;
                return;
            }
            if(!visited[it]) {
                dfs_to_check_outdegree(it, g, mp, visited, flag);
            }
        }
    }

public:
    vector<string> maxNumOfSubstrings(string s) {
        int n = s.size();

        vector<int> l(26, INT_MAX), r(26, -1);
        for(int i=0; i<n; i++) {
            l[s[i] - 'a'] = min(l[s[i] - 'a'], i);
            r[s[i] - 'a'] = i;
        }

        vector<vector<int>> g(27);
        for(int i=0; i<n; i++) {
            for(int j=0; j<26; j++) {
                if(l[j] <= i && r[j] >= i) {
                    g[j].push_back(s[i] - 'a');
                }
            }
        }

        // Step 1 -> Topological Sorting
        n  = 26;
        stack<int> end_time;
        vector<int> vis(n, 0);
        for(int i=0; i<n; i++) {
            if(!vis[i]) {
                dfs(i, g, vis, end_time);
            }
        }

        // Step 2 -> Transpose 
        vector<vector<int>> gr(n);
        for(int i=0; i<n; i++) {
            for(auto it : g[i]) {
                gr[it].push_back(i);
            }
        }

        // Step 3
        vector<int> v(n, 0);
        vector<int> degree(n, 0);
        vector<vector<int>> scc;
        for(int i=0; i<n; i++) {
            int node = end_time.top(); 
            end_time.pop();

            vector<int> temp;
            dfs2(node, gr, v, temp);
            scc.push_back(temp);
        }

        vector<string> ans;
        for(auto it : scc) {
            vector<int> visited(n, 0);

            if(it.size() != 0) {
                map<int, int> mp;
                for(auto ele : it) mp[ele] = 1;

                bool flag = true;
                dfs_to_check_outdegree(it[0], g, mp, visited, flag);

                if(flag) {
                    int min_l = 1e9, max_r = -1;
                    for(auto ele : it) {
                        min_l = min(min_l, l[ele]);
                        max_r = max(max_r, r[ele]);
                    }

                    ans.push_back(s.substr(min_l, max_r - min_l + 1));
                }
            }
        }

        return ans;
    }
};
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