<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 dfs2(int node, vector<vector<int>>& transpose, vector<int>& vis2, vector<int>& temp){
        vis2[node]=1;
        temp.push_back(node);
        for(auto it: transpose[node]){
            if(!vis2[it]) dfs2(it,transpose,vis2,temp);
        }
    }
    void dfs(int node, vector<int>& vis, vector<vector<int>>& graph ,stack<int>& st){
        vis[node]=1;
        for(auto it: graph[node]){
            if(!vis[it]) dfs(it,vis,graph,st);
        }
        st.push(node);
    }
    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>> graph(26);
        for (int i = 0; i < 26; i++) {
            if (l[i] != INT_MAX) { 
                for (int j = l[i]; j <= r[i]; j++) {
                    if (s[j] - 'a' != i) {
                        graph[i].push_back(s[j] - 'a');
                    }
                }
            }
        }


        
        stack<int> st;
        vector<int> vis(26,0);
        for(int i=0;i<26;i++){
            if(!vis[i]){
                dfs(i,vis,graph,st);
            }
        }
 
        vector<vector<int>> transpose(26);
        for(int i=0;i<26;i++){
            for(auto it : graph[i]){
                transpose[it].push_back(i);
            }
        }
        //kosaraju dfs acc to finish time on transpose graph
        vector<int> vis2(26,0);
        vector<vector<int>> scc;
        while(!st.empty()){
            int node = st.top();
            st.pop();
            if(!vis2[node]){
                vector<int> temp;
                dfs2(node,transpose,vis2,temp);
                scc.push_back(temp);
            }
        }
        //finding scc with outdegree 0: scc that has no other connected component 
        //this will ensure smallest length scc
        
        vector<int> sccIndex(26,-1);
        for(int i=0;i<scc.size();i++){
            for(auto ele: scc[i]){
                sccIndex[ele]=i;
                //mapping each node ele to its corresponding SCC index i
            }
        }
        vector<int> outdegree(scc.size(),0);
        for(int u=0;u<26;u++){
            for(auto v: graph[u]){
                if(sccIndex[u]!=sccIndex[v]) outdegree[sccIndex[u]]++;
                //This count represents how many SCCs this particular SCC has edges going out to.
            }
        }
        vector<string> result;
        for (int i = 0; i < scc.size(); i++) {
            if (outdegree[i] == 0) {
                int left = n, right = -1;
                for (auto node : scc[i]) {
                    left = min(left, l[node]);
                    right = max(right, r[node]);
                }
                string substr = s.substr(left, right - left + 1);
                if (!substr.empty()) result.push_back(substr);
            }
        }
        return result;
    }
};
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