<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> topoSort(unordered_map<int,list<int>>& adjList,vector<int>& inDegree)
    {
        int n = inDegree.size();
        queue<int> q;
        for(int i = 0 ; i < n ; i++)
            if(inDegree[i] == 0)
                q.push(i);

        vector<int> ans;
        while(!q.empty())
        {
            int src = q.front();
            q.pop();
            ans.push_back(src);

            for(auto& neigh : adjList[src])
            {
                inDegree[neigh]--;
                if(inDegree[neigh] == 0)
                    q.push(neigh);
            }
        }

        return (ans.size() != n) ? vector<int>{} : ans;
    }

    vector<int> sortItems(int n, int m, vector<int>& grp, vector<vector<int>>& beforeItems) {
        unordered_map<int,list<int>> depGraph;
        for(int i = 0 ; i < n ; i++)
        {            
            for(int x : beforeItems[i])
                depGraph[x].push_back(i);
        }

        vector<int> depGraphInDegree(n,0);
        for(auto& [src,neighList] : depGraph)
        {
            for(auto& neigh : neighList)
                depGraphInDegree[neigh]++;
        }

        vector<int> depGraphTopoSort = topoSort(depGraph,depGraphInDegree);
        if (depGraphTopoSort.empty()) 
            return {};

        // Modify the group array to ensure independent nodes
        for(int& x : grp)
        {
            if(x == -1)
                x = m++;
        }

        unordered_map<int,list<int>> grpGraph;
        for(int i = 0 ; i < n ; i++)
        {
            for(int x : beforeItems[i])
            {
                if(grp[x] != grp[i])
                    grpGraph[grp[x]].push_back(grp[i]);
            }
        }

        vector<int> grpGraphInDegree(m,0);
        for(auto& [src,neighList] : grpGraph)
        {
            for(auto& neigh : neighList)
                grpGraphInDegree[neigh]++;
        }

        vector<int> grpGraphTopoSort = topoSort(grpGraph,grpGraphInDegree);
        if (grpGraphTopoSort.empty()) 
            return {};
            
        vector<vector<int>> combinedTopoSort(m);
        for(int& x : depGraphTopoSort)
            combinedTopoSort[grp[x]].push_back(x);
        
        vector<int> ans;
        for(int& x : grpGraphTopoSort)
        {
            for(int y : combinedTopoSort[x])
                ans.push_back(y);
        }

        return ans;
    }
};
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