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

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

    void dfs_to_check_outdegree(int v, vector<vector<int>>& g,
                                map<int, int>& mp, vector<int>& visted,
                                bool& flg) {
        visted[v] = 1;
        for (auto ch : g[v]) {
            if (mp[ch] != 1) {
                flg = false;
                return;
            }
            if (!visted[ch]) {
                dfs_to_check_outdegree(ch, g, mp, visted, flg);
            }
        }
    }

    vector<string> maxNumOfSubstrings(string s) {
        int n = s.size();
        vector<int> l(26, INT_MAX), r(26, INT_MIN);
        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[s[i] - 'a'] != INT_MAX) {
                    if (l[j] <= i && r[j] >= i) {
                        g[j].push_back(s[i] - 'a');
                    }
                }
            }
        }
        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);
            }
        }
        vector<vector<int>> gr(n);
        for (int i = 0; i < n; i++) {
            for (auto ch : g[i]) {
                gr[ch].push_back(i);
            }
        }
        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();
            if (v[node])
                continue;
            vector<int> temp;
            dfs2(node, gr, v, temp);
            scc.push_back(temp);
        }

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

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