<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<vector<int>> adj;
    vector<int> in, head, tail, order;
    void topsort(int s){
        int n = adj.size();
        queue<int> q;
        q.push(s);
        while(q.size()){
            int u = q.front(); q.pop();
            order.push_back(u);
            for(int v : adj[u]){
                in[v]--;
                if(!in[v]) q.push(v);
            }
        }
    }
    
    vector<int> sortItems(int n, int m, vector<int>& g, vector<vector<int>>& b){
        adj.resize(n + 2 * m, vector<int>()); // 2 * m nodes for head and tail of each grp
        in.resize(n + 2 * m, 0);
        head.resize(m, 0);
        tail.resize(m, 0);
        for(int i = 0, id = n; i < m; i++){
            head[i] = id++; // ith grp ka head id
            tail[i] = id++; // ith grp ka tail id + 1
        }

        // Graph preprocessing
        // T <- (group nodes) <- H
        for(int v = 0; v < n; v++){
            // edge connecting to the node to head and tail of the grp, of which node belongs
            if(g[v] != -1){ // grpHead -> v -> grpTail
                int H = head[g[v]];
                int T = tail[g[v]];
                adj[v].push_back(H); in[H]++;
                adj[T].push_back(v); in[v]++;
            }            
            
            // edge, to represent the actual dependency between grps
            for(int u : b[v]){ // u -> v
                int from = g[u];
                int to = g[v];
                int uu = u, vv = v;
                if(from != to && from != -1) uu = head[from];
                if(from != to && to != -1) vv = tail[to];
                adj[uu].push_back(vv);
                in[vv]++;
            }
        }
        vector<int> node;
        for(int i = 0; i < n + 2 * m; i++){
            if(!in[i]) node.push_back(i);
        }
        for(int s : node){
            topsort(s);
        }
        vector<int> res;
        for(int i : order){
            if(i < n) res.push_back(i);
        }
        if(res.size() != n) res.clear();
        return res;
    }
};
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