<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) {
        unordered_map<int, vector<int>> groups;
        int id = 30001;
        for (int i = 0; i < n; ++i) {
            if (group[i] == -1) {
                groups[id].push_back(i);
                group[i] = id;
                ++id;
            } else {
                groups[group[i]].push_back(i);
            }
        }

        // build graph and degrees
        vector<int> deg(n, 0);
        unordered_map<int, int> gdeg;
        vector<vector<int>> graph(n);
        for (int i = 0; i < n; ++i) {
            for (auto x: beforeItems[i]) {
                graph[x].push_back(i);
                ++deg[i];

                if (group[i] != group[x]) {
                    ++gdeg[group[i]];
                }
            }
        }

        // top sort
        queue<int> gq;
        vector<int> ans;
        vector<int> seen(n, 0);

        // order group first
        for (auto &[k, v]: groups) {
            if (gdeg[k] == 0) {
                gq.push(k);
            }
        }

        while (!gq.empty()) {
            int g = gq.front();
            gq.pop();

            // order item in group
            queue<int> q;
            
            for (auto u: groups[g]) {
                if (deg[u] == 0) {
                    q.push(u);
                }
            }

            int count = 0;

            while (!q.empty()) {
                int node = q.front();
                q.pop();
                seen[node] = 1;

                ans.push_back(node);
                ++count;
                
                for (auto &u: graph[node]) {
                    --deg[u];

                    if (group[u] != g) {
                        --gdeg[group[u]];
                        
                        if (gdeg[group[u]] == 0) {
                            gq.push(group[u]);
                        }
                    } else {
                        if (seen[u] == 1) {
                            return {};
                        }

                        if (deg[u] == 0) {
                            q.push(u);
                        }
                    }
                }
            }

            if (count != groups[g].size()) {
                return {};
            }
        }

        if (ans.size() != n) {
            return {};
        }

        return ans;
    }
};
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