<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(vector<vector<int>>& adj, vector<int>& inDegree) {
        queue<int> q;
        for(int i = 0; i < inDegree.size(); i++) {
            if (inDegree[i] == 0) q.push(i);
        }
        vector<int> res;
        while(!q.empty()) {
            int node = q.front();
            q.pop();
            res.push_back(node);
            for(int neighbor: adj[node]) {
                inDegree[neighbor]--;
                if(inDegree[neighbor] == 0) q.push(neighbor);
            }
        }
        return res.size() == inDegree.size() ? res : vector<int>();
    }

    vector<int> sortItems(int n, int m, vector<int>& group, vector<vector<int>>& beforeItems) {
        int groupID = m;
        for(int i = 0; i < n; i++) {
            if (group[i] == -1) group[i] = m;
            m++;
        }
        vector<vector<int>> itemAdj(n);
        vector<int> itemInd(n);
        vector<vector<int>> groupAdj(m);
        vector<int> groupInd(m);
        for(int i = 0; i < n; i++) {
            for(int prev: beforeItems[i]) {
                itemAdj[prev].push_back(i);
                itemInd[i]++;
                if(group[i] != group[prev]) {
                    groupAdj[group[prev]].push_back(group[i]);
                    groupInd[group[i]]++;
                }
            }
        }
        vector<int> itemOrder = topoSort(itemAdj, itemInd);
        vector<int> groupOrder = topoSort(groupAdj, groupInd);

        if(itemOrder.empty() || groupOrder.empty()) return vector<int>();
        unordered_map<int,vector<int>> ordered_groups;
        for(int item: itemOrder) {
            ordered_groups[group[item]].push_back(item);
        }
        vector<int> res;
        for(int group: groupOrder) {
            res.insert( res.end(), ordered_groups[group].begin(), ordered_groups[group].end() );
        }
        return res;
    }
};
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