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 int[] sortItems(int n, int m, int[] group, List<List<Integer>> beforeItems) {
        int idx = m;
        List<Integer>[] groupItems = new List[n + m];
        int[] itemDegree = new int[n];
        int[] groupDegree = new int[n + m];
        List<Integer>[] itemGraph = new List[n];
        List<Integer>[] groupGraph = new List[n + m];
        Arrays.setAll(groupItems, k -> new ArrayList<>());
        Arrays.setAll(itemGraph, k -> new ArrayList<>());
        Arrays.setAll(groupGraph, k -> new ArrayList<>());
        for (int i = 0; i < n; ++i) {
            if (group[i] == -1) {
                group[i] = idx++;
            }
            groupItems[group[i]].add(i);
        }
        for (int i = 0; i < n; ++i) {
            for (int j : beforeItems.get(i)) {
                if (group[i] == group[j]) {
                    ++itemDegree[i];
                    itemGraph[j].add(i);
                } else {
                    ++groupDegree[group[i]];
                    groupGraph[group[j]].add(group[i]);
                }
            }
        }
        List<Integer> items = new ArrayList<>();
        for (int i = 0; i < n + m; ++i) {
            items.add(i);
        }
        var groupOrder = topoSort(groupDegree, groupGraph, items);
        if (groupOrder.isEmpty()) {
            return new int[0];
        }
        List<Integer> ans = new ArrayList<>();
        for (int gi : groupOrder) {
            items = groupItems[gi];
            var itemOrder = topoSort(itemDegree, itemGraph, items);
            if (itemOrder.size() != items.size()) {
                return new int[0];
            }
            ans.addAll(itemOrder);
        }
        return ans.stream().mapToInt(Integer::intValue).toArray();
    }

    private List<Integer> topoSort(int[] degree, List<Integer>[] graph, List<Integer> items) {
        Deque<Integer> q = new ArrayDeque<>();
        for (int i : items) {
            if (degree[i] == 0) {
                q.offer(i);
            }
        }
        List<Integer> ans = new ArrayList<>();
        while (!q.isEmpty()) {
            int i = q.poll();
            ans.add(i);
            for (int j : graph[i]) {
                if (--degree[j] == 0) {
                    q.offer(j);
                }
            }
        }
        return ans.size() == items.size() ? ans : List.of();
    }
}