<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:
    bool topoSort(vector<vector<int>> adj ,vector<int> &topo,vector<int> &deg){
        queue<int> q;
        for(int i=0;i<deg.size();i++){
            if(deg[i]==0)
                q.push(i);
        }
        while(!q.empty()){
            int curr=q.front();
            q.pop();
            topo.push_back(curr);
            for(int v : adj[curr]){
                if(--deg[v]==0)
                    q.push(v);
            }
        }
        return topo.size()==deg.size();
    }
    vector<int> sortItems(int n, int m, vector<int>& group, vector<vector<int>>& bef) {
        for(auto &i: group)   
            if(i == -1) i = m++;

        vector<int> ideg(n,0), gdeg(m,0);
        vector<vector<int>> iadj(n) , gadj(m);
        for(int i = 0; i< n ; i++){
            for(auto j : bef[i]){
                iadj[j].push_back(i);
                ideg[i]++;
                if(group[i]!=group[j])
                    gadj[group[j]].push_back(group[i]),gdeg[group[i]]++;
            }
        }
        vector<int> grp, it;
        vector<int> ans;
        if(topoSort(iadj, it, ideg) && topoSort(gadj, grp, gdeg)){
             unordered_map<int,vector<int>> mp;
            for(int i : it){
                mp[group[i]].push_back(i);
            }
            vector<int> ans;
            for(int g : grp){
                vector<int> temp = mp[g];
                for(auto i : temp){
                    ans.push_back(i);
                }
            }
            return ans;
        }
        return {};

    }
};

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