Sort Items by Groups Respecting Dependencies
<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://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/1200-1299/1203.Sort%20Items%20by%20Groups%20Respecting%20Dependencies/images/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>
Depth-First Search; Breadth-First Search; Graph; Topological Sort
class Solution {
public:
    vector<int> sortItems(int n, int m, vector<int>& group, vector<vector<int>>& beforeItems) {
        int idx = m;
        vector<vector<int>> groupItems(n + m);
        vector<int> itemDegree(n);
        vector<int> groupDegree(n + m);
        vector<vector<int>> itemGraph(n);
        vector<vector<int>> groupGraph(n + m);
        for (int i = 0; i < n; ++i) {
            if (group[i] == -1) {
                group[i] = idx++;
            }
            groupItems[group[i]].push_back(i);
        }
        for (int i = 0; i < n; ++i) {
            for (int j : beforeItems[i]) {
                if (group[i] == group[j]) {
                    ++itemDegree[i];
                    itemGraph[j].push_back(i);
                } else {
                    ++groupDegree[group[i]];
                    groupGraph[group[j]].push_back(group[i]);
                }
            }
        }
        vector<int> items(n + m);
        iota(items.begin(), items.end(), 0);
        auto topoSort = [](vector<vector<int>>& graph, vector<int>& degree, vector<int>& items) -> vector<int> {
            queue<int> q;
            for (int& i : items) {
                if (degree[i] == 0) {
                    q.push(i);
                }
            }
            vector<int> ans;
            while (!q.empty()) {
                int i = q.front();
                q.pop();
                ans.push_back(i);
                for (int& j : graph[i]) {
                    if (--degree[j] == 0) {
                        q.push(j);
                    }
                }
            }
            return ans.size() == items.size() ? ans : vector<int>();
        };
        auto groupOrder = topoSort(groupGraph, groupDegree, items);
        if (groupOrder.empty()) {
            return vector<int>();
        }
        vector<int> ans;
        for (int& gi : groupOrder) {
            items = groupItems[gi];
            auto itemOrder = topoSort(itemGraph, itemDegree, items);
            if (items.size() != itemOrder.size()) {
                return vector<int>();
            }
            ans.insert(ans.end(), itemOrder.begin(), itemOrder.end());
        }
        return ans;
    }
};