<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 mins[26], maxs[26];
        fill(mins, mins + 26, INT_MAX);
        fill(maxs, maxs + 26, -1);
        bool exists[26] = {false};
        vector<vector<int>> prefixSum(S.size()+1, vector<int>(26,0));
        for (int i = 0; i < S.length(); i++) {
            prefixSum[i+1]= prefixSum[i];
            prefixSum[i + 1][S[i] - 'a'] += 1;
            mins[S[i] - 'a'] = min(mins[S[i] - 'a'], i);
            maxs[S[i] - 'a'] = max(maxs[S[i] - 'a'], i);
            exists[S[i] - 'a'] = true;
        }

        bool graph[26][26] = {false};
        for (int i = 0; i < 26; i++) {
            if (exists[i]) {
                for (int j = 0; j < 26; j++) {
                    if (prefixSum[maxs[i] + 1][j] - prefixSum[mins[i]][j] > 0) {
                        graph[i][j] = true;
                    }
                }
            }
        }

        stack<int> stack;
        bool visited[26] = {false};
        for (int i = 0; i < 26; i++) {
            if (exists[i] && !visited[i]) {
                dfs(i, graph, stack, visited);
            }
        }

        int batch = 0;
        int batches[26];
        int degree[26] = {0};
        fill(batches, batches + 26, -1);

        while (!stack.empty()) {
            int v = stack.top();
            stack.pop();
            if (batches[v] < 0) {
                dfs(v, graph, batches, batch, degree);
                batch++;
            }
        }

        vector<string> res;
        for (int i = batch - 1; i >= 0; i--) {
            if (degree[i] == 0) {
                int min = INT_MAX, max = -1;
                for (int j = 0; j < 26; j++) {
                    if (batches[j] == i) {
                        min = std::min(mins[j], min);
                        max = std::max(maxs[j], max);
                    }
                }
                res.push_back(S.substr(min, max - min + 1));
            }
        }

        return res;
    }

private:
    void dfs(int v, bool graph[26][26], std::stack<int>& stack, bool visited[26]) {
        if (!visited[v]) {
            visited[v] = true;
            for (int i = 0; i < 26; i++) {
                if (graph[v][i] && !visited[i]) {
                    dfs(i, graph, stack, visited);
                }
            }
            stack.push(v);
        }
    }

    void dfs(int v, bool graph[26][26], int batches[26], int batch, int degree[26]) {
        if (batches[v] < 0) {
            batches[v] = batch;
            for (int i = 0; i < 26; i++) {
                if (graph[i][v]) {
                    dfs(i, graph, batches, batch, degree);
                }
            }
        } else {
            if (batches[v] != batch) {
                degree[batches[v]] += 1;
            }
        }
    }
};
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