<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> t, res(n), state(n + 2 * m);
        vector<vector<int>> g(n + 2 * m);
        for (int i = 0; i < n; ++i) {
            if (group[i] != -1) {
                g[n + group[i]].push_back(i);
                g[i].push_back(n + m + group[i]);
            }
            for (int j : beforeItems[i]) {
                if (group[i] != -1 && group[i] == group[j]) {
                    g[j].push_back(i);
                } else {
                    int p = group[i] == -1 ? i : n + group[i];
                    int q = group[j] == -1 ? j : n + m + group[j];
                    g[q].push_back(p);
                }
            }
        }
        for (int i = (int)g.size() - 1; i >= 0; --i) {
            if (!helper(g, i, state, t)) return {};
        }
        reverse(t.begin(), t.end());
        copy_if(t.begin(), t.end(), res.begin(), [&](int i) {return i < n;});
        return res;
    }
    bool helper(vector<vector<int>>& g, int i, vector<int>& state, vector<int>& res) {
        if (state[i] != 0) return state[i] == 2;
        state[i] = 1;
        for (int next : g[i]) {
            if (!helper(g, next, state, res)) return false;
        }
        state[i] = 2;
        res.push_back(i);
        return true;
    }
};

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