<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) {
        //create a graph for each group
        int indegree[n+m];
        memset(indegree,0,sizeof(indegree));
        vector<vector<int>>graph(n+m);
        for(int i = 0;i<n;i++){
            if(group[i]!=-1){
                graph[n+group[i]].push_back(i);
                indegree[i]++;
            }
        }
        for(int i = 0;i<n;i++){
            int gi = group[i]==-1?i:group[i]+n;
            for(int j:beforeItems[i]){
                int gj = group[j]==-1?j:group[j]+n;
                if(gi==gj){
                    graph[j].push_back(i);
                    indegree[i]++;
                }else{
                    graph[gj].push_back(gi);
                    indegree[gi]++;
                }
            }
        }
        vector<int>ans;
        for(int i =0;i<n+m;i++){
            if(indegree[i]==0)dfs(ans,graph,indegree,n,i);
        }
        return (ans.size()==n)?ans:vector<int>{};
    }

    void dfs(vector<int>&ans, vector<vector<int>>&graph, int*indegree, int n, int curr){
        if(curr<n)ans.push_back(curr);
        indegree[curr] = -1;
        for(int nex:graph[curr]){
            if(--indegree[nex]==0){
                dfs(ans,graph,indegree,n,nex);
            }
        }
    }
};
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