<p>There are&nbsp;<code>n</code>&nbsp;items each&nbsp;belonging to zero or one of&nbsp;<code>m</code>&nbsp;groups where <code>group[i]</code>&nbsp;is the group that the <code>i</code>-th item belongs to and it&#39;s equal to <code>-1</code>&nbsp;if the <code>i</code>-th item belongs to no group. The items and the groups are zero indexed. A group can have no item belonging to it.</p>

<p>Return a sorted list of the items such that:</p>

<ul>
	<li>The items that belong to the same group are next to each other in the sorted list.</li>
	<li>There are some&nbsp;relations&nbsp;between these items where&nbsp;<code>beforeItems[i]</code>&nbsp;is a list containing all the items that should come before the&nbsp;<code>i</code>-th item in the sorted array (to the left of the&nbsp;<code>i</code>-th item).</li>
</ul>

<p>Return any solution if there is more than one solution and return an <strong>empty list</strong>&nbsp;if there is no solution.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<p><strong><img alt="" src="https://assets.leetcode.com/uploads/2019/09/11/1359_ex1.png" style="width: 191px; height: 181px;" /></strong></p>

<pre>
<strong>Input:</strong> n = 8, m = 2, group = [-1,-1,1,0,0,1,0,-1], beforeItems = [[],[6],[5],[6],[3,6],[],[],[]]
<strong>Output:</strong> [6,3,4,1,5,2,0,7]
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> n = 8, m = 2, group = [-1,-1,1,0,0,1,0,-1], beforeItems = [[],[6],[5],[6],[3],[],[4],[]]
<strong>Output:</strong> []
<strong>Explanation:</strong>&nbsp;This is the same as example 1 except that 4 needs to be before 6 in the sorted list.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= m &lt;= n &lt;= 3 * 10<sup>4</sup></code></li>
	<li><code>group.length == beforeItems.length == n</code></li>
	<li><code>-1 &lt;= group[i] &lt;= m - 1</code></li>
	<li><code>0 &lt;= beforeItems[i].length &lt;= n - 1</code></li>
	<li><code>0 &lt;= beforeItems[i][j] &lt;= n - 1</code></li>
	<li><code>i != beforeItems[i][j]</code></li>
	<li><code>beforeItems[i]&nbsp;</code>does not contain&nbsp;duplicates elements.</li>
</ul>

#include <vector>

using namespace std;

class Solution {
    vector<vector<int>> beforeGroups;
    vector<vector<int>> groupItems;
    vector<int> result;
    vector<bool> groupVisited;
    vector<bool> groupVisitedCurrentStack;
    vector<bool> itemVisited;
    vector<bool> itemVisitedCurrentStack;
    bool flag = true;

    void dfsItem(const int item, const int groupNumber, const vector<int>& group, const vector<vector<int>>& beforeItems) {
        if(itemVisitedCurrentStack[item] == true) {
            flag = false;
            return;
        }
        if(flag == false or itemVisited[item]) return;
        itemVisitedCurrentStack[item] = true;
        itemVisited[item] = true;
        for(const int& beforeItem: beforeItems[item]) if(group[beforeItem] == groupNumber) dfsItem(beforeItem, groupNumber, group, beforeItems);
        result.push_back(item);
        itemVisitedCurrentStack[item] = false;
    }

    void dfsGroup(const int groupNumber, const vector<int>& group, const vector<vector<int>>& beforeItems) {
        if(groupVisitedCurrentStack[groupNumber]) {
            flag = false;
            return;
        }
        if(flag == false or groupVisited[groupNumber]) return;
        groupVisitedCurrentStack[groupNumber] = true;
        groupVisited[groupNumber] = true;

        for(const int& beforeGroup: beforeGroups[groupNumber]) dfsGroup(beforeGroup, group, beforeItems);
        groupVisitedCurrentStack[groupNumber] = false;
        for(const int& item: groupItems[groupNumber]) dfsItem(item, groupNumber, group, beforeItems);
    }

public:
    vector<int> sortItems(int n, int m, vector<int>& group, vector<vector<int>>& beforeItems) {
        int groupNumber = m;
        for(int i=0; i<n; i++) if(group[i] == -1) group[i] = groupNumber++;

        beforeGroups.assign(groupNumber, vector<int>());
        groupItems.assign(groupNumber, vector<int>());
        groupVisited.assign(groupNumber, false);
        groupVisitedCurrentStack.assign(groupNumber, false);
        itemVisited.assign(n, false);
        itemVisitedCurrentStack.assign(n, false);

        for(int i=0; i<n; i++) {
            groupItems[group[i]].push_back(i);
            for(const int& beforeItem: beforeItems[i]) if(group[i] != group[beforeItem]) beforeGroups[group[i]].push_back(group[beforeItem]);
        }
        
        for(int i=0; i<groupNumber; i++) {
            if(groupVisited[i] == false) dfsGroup(i, group, beforeItems);
            if(flag == false) return {};
        }

        return result;
    }
};
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