<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:
    template<typename P>
    bool topoIter(int n, vector<vector<int>> out, P&& job) {
        vector<int> in(n);
        for(int i = 0; i < n; ++i) 
            for(const auto& j : out[i])
                ++in[j];
        queue<int> q{};
        for(int i = 0; i < n; ++i) if(in[i] == 0) q.push(i);
        while(!q.empty()) {
            auto now = q.front(); q.pop();
            job(now);
            --n;
            for(const int& nxt : out[now]) 
                if(--in[nxt] == 0)
                    q.push(nxt);
        }
        return n == 0;
    }

    vector<int> sortItems(int n, int m, vector<int>& group, vector<vector<int>>& beforeItems) {
        for(auto& id : group) if(id == -1) id = m++;
        
        vector<int> sz_grp(m,0);
        vector<vector<int>> beforeGroups(m);
        for(int i = 0; i < n; ++i) {
            ++sz_grp[group[i]];
            beforeGroups[group[i]].reserve(beforeItems[i].size()/4);
            for(const int& j : beforeItems[i])
                if(group[i] != group[j]) 
                    beforeGroups[group[i]].push_back(group[j]);
        }
        
        vector<int> idx = std::move(sz_grp);
        if(!topoIter(m, std::move(beforeGroups),
            [&,acc=0](const int& g) mutable { acc += exchange(idx[g], acc); })) {
            return {};
        }

        vector<int> ans(n);
        if(!topoIter(n, std::move(beforeItems),
            [&](const int& i) { ans[idx[group[i]]++] = i; })) {
            return {};
        }
        reverse(ans.begin(), ans.end());
        return ans;
    }
};
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