<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> topSort(unordered_map<int, vector<int>> &adj) {
        int N = adj.size();

        unordered_map<int, int> indegree;
        for(auto entry: adj)
            indegree[entry.first] = 0;
        for(auto entry: adj) {
            int u = entry.first;
            for(int v: adj[u]) {
                indegree[v]++;
            }
        }

        queue<int> Q;
        for(auto entry: indegree) {
            int u = entry.first;
            if(indegree[u] == 0)    
                Q.push(u);
        }

        vector<int> ans(0);
        
        while(!Q.empty()) {
            int u = Q.front(); Q.pop();
            for(int v: adj[u]) {
                indegree[v]--;
                if(indegree[v] == 0)
                    Q.push(v);
            }
            ans.push_back(u);
        }

        if(ans.size() != N)
            return vector<int>(0);

        return ans;
    }

    vector<int> sortItems(int n, int m, vector<int>& group, vector<vector<int>>& beforeItems) {
        // draw edges from parent to child (before element to after element)

        // putting lone elements in new groups
        for(int u=0; u<n; u++) {
            if(group[u] == -1) {
                group[u] = m;
                m++;
            }
        }

        // make group adj list
        unordered_map<int, vector<int>> adjG;
        for(int g=0; g<m; g++)
            adjG[g] = vector<int>(0);
        for(int u=0; u<n; u++) {
            for(int v: beforeItems[u]) {
                if(group[v] != group[u])
                    adjG[group[v]].push_back(group[u]);
            }
        }

        // top sort groups
        vector<int> topSortedG = topSort(adjG);
        if(topSortedG.size() == 0)
            return vector<int>(0);

        // grouping items
        unordered_map<int, set<int>> groupItems;
        for(int u=0; u<n; u++) {
            groupItems[group[u]].insert(u);
        }

        vector<int> ANS(0);

        for(int g: topSortedG) {
            set<int> currentGroup = groupItems[g];

            if(currentGroup.size() == 0)
                continue;
            
            // making adjacency list for current group
            unordered_map<int, vector<int>> adj;
            for(int u: currentGroup)
                adj[u] = vector<int>(0);
            for(int u: currentGroup) {
                for(int v: beforeItems[u]) {
                    if(currentGroup.contains(v))
                        adj[v].push_back(u);
                }
            }

            // running top sort on this
            vector<int> topSortedCurrentGroup = topSort(adj);
            if(topSortedCurrentGroup.size() == 0)
                return vector<int>(0);

            ANS.insert(ANS.end(), topSortedCurrentGroup.begin(), topSortedCurrentGroup.end());
        }

        return ANS;
    }
};
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