<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<vector<int>> member(m, vector<int>());
        unordered_set<int> member[m];
        for (int i = 0; i < n; i++) {
            if (group[i] > -1) {
                member[group[i]].insert(i);
            }
        }

        //vector<vector<int>> beforeGroup(m, vector<int>());
        unordered_set<int> beforeGroup[m];
        for (int i = 0; i < m; i++) {
            for (auto j : member[i]) {
                for (auto k : beforeItems[j]) {
                    if (member[i].count(k) == 0) {
                        beforeGroup[i].insert(k);
                    }
                }
            }

            for (auto j : member[i]) {
                unordered_set<int> s(beforeItems[j].begin(), beforeItems[j].end());
                for (auto k : beforeGroup[i]) {
                    if (s.count(k) == 0) {
                        beforeItems[j].push_back(k);
                    }
                }
            }
        }

        queue<int> q;
        vector<vector<int>> out(n, vector<int>());
        vector<int> in(n, 0);
        queue<int> g;
        for (int i = 0; i < n; i++) {
            in[i] = beforeItems[i].size();
            for (auto k : beforeItems[i]) {
                out[k].push_back(i);
            }
            if (in[i] == 0) {
                q.push(i);
            }
        }

        vector<int> v, ret;
        while (!q.empty()) {
            v.push_back(q.front());
            for (auto k : out[q.front()]) {
                in[k]--;
                if (in[k] == 0) {
                    q.push(k);
                }
            }
            q.pop();
        }
        
        if (v.size() != n) {
            return {};
        }

        vector<vector<int>> groupOrder(m, vector<int>());
        for (auto k : v) {
            if (group[k] != -1) {
                groupOrder[group[k]].push_back(k);
            }
        }

        vector<bool> useGroup(m, false);
        for (auto k : v) {
            if (group[k] == -1) {
                ret.push_back(k);
            } else {
                if (useGroup[group[k]]) continue;
                useGroup[group[k]] = true;
                for (auto l : groupOrder[group[k]]) {
                    ret.push_back(l);
                }
            }
        }

        return ret;
    }
};
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