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
func sortItems(n int, m int, group []int, beforeItems [][]int) []int {
	idx := m
	groupItems := make([][]int, n+m)
	itemDegree := make([]int, n)
	groupDegree := make([]int, n+m)
	itemGraph := make([][]int, n)
	groupGraph := make([][]int, n+m)
	for i, g := range group {
		if g == -1 {
			group[i] = idx
			idx++
		}
		groupItems[group[i]] = append(groupItems[group[i]], i)
	}
	for i, gi := range group {
		for _, j := range beforeItems[i] {
			gj := group[j]
			if gi == gj {
				itemDegree[i]++
				itemGraph[j] = append(itemGraph[j], i)
			} else {
				groupDegree[gi]++
				groupGraph[gj] = append(groupGraph[gj], gi)
			}
		}
	}
	items := make([]int, n+m)
	for i := range items {
		items[i] = i
	}
	topoSort := func(degree []int, graph [][]int, items []int) []int {
		q := []int{}
		for _, i := range items {
			if degree[i] == 0 {
				q = append(q, i)
			}
		}
		ans := []int{}
		for len(q) > 0 {
			i := q[0]
			q = q[1:]
			ans = append(ans, i)
			for _, j := range graph[i] {
				degree[j]--
				if degree[j] == 0 {
					q = append(q, j)
				}
			}
		}
		return ans
	}
	groupOrder := topoSort(groupDegree, groupGraph, items)
	if len(groupOrder) != len(items) {
		return nil
	}
	ans := []int{}
	for _, gi := range groupOrder {
		items = groupItems[gi]
		itemOrder := topoSort(itemDegree, itemGraph, items)
		if len(items) != len(itemOrder) {
			return nil
		}
		ans = append(ans, itemOrder...)
	}
	return ans
}