<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>& groups, vector<vector<int>>& beforeItems) {
      using i32 = int;
      vector <vector <i32>> A (n + m + m);
      vector <i32> deg (n + m + m);
      for (i32 i = 0; i < groups.size(); ++i) if (groups[i] != -1) {
        i32 L = groups[i] + n;
        i32 R = groups[i] + n + m;
        A[L].push_back(i);
        A[i].push_back(R);
        deg[i]++;
        deg[R]++;
      }
      for (i32 i = 0; i < n; ++i) {
        for (i32 x : beforeItems[i]) {
          if (groups[i] != -1 && groups[i] == groups[x]) {
            A[x].push_back(i);
            deg[i]++;
          }
          else {
            i32 R = groups[x] == -1 ? x : groups[x] + n + m;
            i32 L = groups[i] == -1 ? i : groups[i] + n;
            A[R].push_back(L);
            deg[L]++;
          }
        }
      }
      vector <i32> Ans;
      auto dfs = [&] (auto &&dfs, i32 u) -> void {
        if (u < n) Ans.push_back(u);
        for (i32 v : A[u]) if (deg[v] != -1) {
          deg[v]--;
          if (deg[v] == 0) {
            deg[v] = -1;
            dfs(dfs, v);
          }
        }
      };
      for (i32 i = 0; i < n + m + m; ++i) {
        if (deg[i] == 0 && deg[i] != -1) {
          dfs(dfs, i);
        }
      }
      return Ans.size() == n ? Ans : vector <i32> {};
    }
};
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