<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) {
        vector <int> ans;
        vector <int> adj[(m << 1) + n];
        vector <int> indegree(n + (m << 1));
        for(int i = 0; i < group.size(); ++i) {
            if(group[i] != -1) {
                adj[n + group[i]].push_back(i);
                adj[i].push_back(n + m + group[i]);
                indegree[i]++;
                indegree[n + m + group[i]]++;
            }
            for(int j: beforeItems[i]) {
                if(group[i] == -1 && group[i] == group[j]) {
                    adj[j].push_back(i);
                    indegree[i]++;
                } else {
                    int ig = group[i] == -1 ? i : n + group[i];
                    int jg = group[j] == -1 ? j : n + m + group[j];
                    // cout << i <<  " " << j << endl;
                    // cout << ig << " " << jg << endl;
                    // cout << group[i] << " " << group[j] << endl;
                    if(group[i] == group[j]) {
                        adj[j].push_back(i);
                        indegree[i]++;
                    } else {    
                        adj[jg].push_back(ig);
                        indegree[ig]++;
                    }
                }
                // cout << endl;
            }
        }
        // for(int i = 0; i < (n + (m << 1)); ++i) {
        //     cout << i << " -- ";
        //     for(int child: adj[i]) {
        //         cout << child << " ";
        //     }
        //     cout << endl;
        // }
        for(int i = 0; i < n + (m << 1); ++i) {
            if(indegree[i] == 0)
                dfs(ans, adj, indegree, i, n);
        }
        if(ans.size() != n)
            return vector <int> {};
        return ans;
    }

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

};
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