<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> topologicalSort(vector<int>& inDegrees, vector<vector<int>>& graph) {
        stack<int> st;
        for (int i = 0; i < inDegrees.size(); i++) {
            if (inDegrees[i] == 0) {
                st.push(i);
            }
        }
        vector<int> order;
        while (!st.empty()) {
            int top = st.top();
            st.pop();
            order.push_back(top);
            for (int i = 0; i < graph[top].size(); i++) {
                int node = graph[top][i];
                inDegrees[node]--;
                if (inDegrees[node] == 0) {
                    st.push(node);
                }
            }
        }
        return order;
    }
    vector<int> sortItems(int n, int m, vector<int>& group, vector<vector<int>>& beforeItems) {
        for (int i = 0; i < group.size(); i++) {
            if (group[i] == -1) {
                group[i] = m;
                m++;
            }
        }
        // if there's a cycle or any path that puts groups out of order then it is impossible

        // 2. consolidate edges per group
        // 3. topological sort groups
        vector<vector<int>> groupGraph(m);
        vector<int> groupInDegree(m);
        vector<vector<int>> itemGraph(n); // before -> after
        vector<int> itemInDegree(n); // can add after if indegree == 0
        for (int i = 0; i < beforeItems.size(); i++) {
            for (int j = 0; j < beforeItems[i].size(); j++) {
                int after = i;
                int before = beforeItems[i][j];
                if (group[after] != group[before]) {
                    groupGraph[group[before]].push_back(group[after]);
                    groupInDegree[group[after]]++;
                } else {
                    itemGraph[before].push_back(after);
                    itemInDegree[after]++;
                }
            }
        }

        // we have already topological sorted all items and all groups
        // if we know that we can order the items within each group amongst themselves
        // and we know we can order the groups amongst themselves
        // then we know we can return an answer
        vector<int> groupSort = topologicalSort(groupInDegree, groupGraph);
        if (groupSort.size() != m) {
            return vector<int>();
        }
        vector<int> itemSort = topologicalSort(itemInDegree, itemGraph);
        if (itemSort.size() != n) {
            return vector<int>();
        }

        vector<int> groupSize(m);
        for (int i = 0; i < group.size(); i++) {
            groupSize[group[i]]++;
        }
        vector<int> startIndex(m);
        int soFar = 0;
        for (int i = 0; i < groupSort.size(); i++) {
            startIndex[groupSort[i]] = soFar;
            soFar += groupSize[groupSort[i]];
        }
        
        vector<int> solution(n);
        for (int i = 0; i < itemSort.size(); i++) {
            int grp = group[itemSort[i]];
            int idx = startIndex[grp];
            startIndex[grp]++;
            solution[idx] = itemSort[i];
        }

        
        return solution;
    }
};
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