<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) {
        vector<int> ans;
        vector<vector<int>> groups(m );
        vector<vector<int>> graph(n);
        vector<int> items_dep(n, 0);
        vector<int> group_dep(m , 0);
        
        for (int i = 0; i < group.size(); i++) {
            if (group[i] == -1) {
            } else {
                groups[group[i]].push_back(i);
            }
        }
        
        for (int i = 0; i < beforeItems.size(); i++) {
            items_dep[i] = beforeItems[i].size();
            for (int j = 0; j < beforeItems[i].size(); j++) {
                graph[beforeItems[i][j]].push_back(i);
                if (group[i] != -1 && group[i] != group[beforeItems[i][j]]) {
                    group_dep[group[i]]++;
                }
            }
        }
        
        queue<int> no_group_val;
        queue<int> valid_groups;
        
        for (int i = 0; i < n; i++) {
            if (group[i] == -1 && items_dep[i] == 0) {
                no_group_val.push(i);
            }
        }
        for (int i = 0; i < m; i++) {
            if (group_dep[i] == 0) {
                valid_groups.push(i);
            }
        }
        
        while (!valid_groups.empty() || !no_group_val.empty()) {
            queue<int> group_items;
            
            if (!valid_groups.empty()) {
                int curr_group = valid_groups.front();
                valid_groups.pop();
                for (int item : groups[curr_group]) {
                    if (items_dep[item] == 0) {
                        group_items.push(item);
                    }
                }
            }
            
            while (!group_items.empty() || !no_group_val.empty()) {
                int curr_item;
                
                if (!group_items.empty()) {
                    curr_item = group_items.front();
                    group_items.pop();
                } else {
                    curr_item = no_group_val.front();
                    no_group_val.pop();
                }
                
                ans.push_back(curr_item);
                
                for (int dep_item : graph[curr_item]) {
                    if (--items_dep[dep_item] == 0) {
                        if (group[dep_item] == group[curr_item]) {
                            group_items.push(dep_item);
                        } else if (group[dep_item] == -1) {
                            no_group_val.push(dep_item);
                        } else {
                            if (--group_dep[group[dep_item]] == 0) {
                                valid_groups.push(group[dep_item]);
                            }
                        }
                    }
                    else if( group[dep_item] != group[curr_item] && group[ dep_item ] != -1 )
                            if (--group_dep[group[dep_item]] == 0) {
                                valid_groups.push(group[dep_item]);
                            }                    
                }
            }
        }
        
        if (ans.size() == n) {
            return ans;
        }
        return {};
    }
};


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