<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>

class Solution {
public:
    vector<int> sortItems(int n, int m, vector<int>& group, vector<vector<int>>& beforeItems) {
        vector<vector<int>> graph(n+m,vector<int>{});
        vector<int> indegree(n+m,0);
        vector<int> ans;
        for(int i=0; i<group.size();i++)
        {
            if(group[i]==-1)
            {
                continue;
            }
            int group_node = n+group[i];
            graph[group_node].push_back(i); //each group node points to the nodes in its group
            indegree[i]++; //dont forget to increase indegree
        }
        for(int i=0;i<beforeItems.size();i++)
        {
            for (auto before_i:beforeItems[i])
            {
                int i_group_node;
                int before_i_group_node;
                if(group[i]==-1)
                {
                    i_group_node = i;
                }
                else 
                {
                    i_group_node= group[i] + n; //not group[i]
                }
                if(group[before_i]==-1)
                {
                    before_i_group_node=before_i;
                }
                else 
                {
                    before_i_group_node= group[before_i] + n;
                }
                if(i_group_node==before_i_group_node) //they are in same group
                {
                    graph[before_i].push_back(i); //NOT i_group_node
                    indegree[i]++;
                }
                else
                {
                    graph[before_i_group_node].push_back(i_group_node);
                    indegree[i_group_node]++;
                }
                
            }

        }
        for(int i=0;i<indegree.size();i++)
        {
            if(indegree[i]==0)
                dfs(ans, graph, indegree, n, i); //start topological sort from here
        }

        if(ans.size()==n) return ans;
        return vector<int>{};
        
    }
    void dfs(vector<int>&ans, vector<vector<int>>&graph, vector<int>&indegree, int n, int cur)
    {
        if(cur<n) ans.push_back(cur); //dont push back group nodes!
        indegree[cur]=-1; //dont visit this one again
        //we have 'visited' cur
        //now visit curs neighbours 
        for(auto node:graph[cur])
        {
            indegree[node]--;
            if(indegree[node]==0)
            {
                dfs(ans,graph,indegree,n,node);
            }
        }
    }
};
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