<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>

#include <ranges>

template <typename T> concept Callback = invocable<T, int>;

class Solution {
public:
    template<ranges::range R, ranges::range R2, Callback F>
    bool toposort(R& g, vector<int>& deg, const vector<R2>& edge, F& op) {
        int count = 0, total = 0;
        vector<int> stack;

        for (auto x: g) {
            if (deg[x] == 0) stack.push_back(x);
            ++total;
        }

        while (stack.size()) {
            int x = stack.back();
            stack.pop_back();

            for (auto y: edge[x]) {
                if (--deg[y] == 0) {
                    stack.push_back(y);
                }
            }
            if (!op(x)) return false;
            count++;
        }

        return count == total;
    }

    vector<int> sortItems(int n, int m, vector<int>& group, vector<vector<int>>& beforeItems) {
        int old_m = m;
        for (int i = 0; i < n; i++) {
            if (group[i] == -1) group[i] = m++;
        }
        vector<unordered_set<int>> outer_edge(m);
        vector<int> inner_deg(n), outer_deg(m);
        vector<unordered_set<int>> g(m);

        for (int i = 0; i < n; i++) g[group[i]].insert(i);

        for (int i = 0; i < n; i++) {
            int gi = group[i];
            for (auto j: beforeItems[i]) {
                int gj = group[j];
                if (gi == gj) {
                    inner_deg[j]++;
                } else {
                    auto [_, b] = outer_edge[gi].insert(gj);
                    if (b) outer_deg[gj]++;
                }
            }

            if (gi >= old_m) continue;
            auto pred = [&group, gi](int x) {
                return group[x] == gi;
            };
            auto r = ranges::partition(beforeItems[i], pred);
            beforeItems[i].resize(r.begin() - beforeItems[i].begin());
        }

        auto nonempty_groups = views::iota(0, m) | views::filter([&](int x) { return g[x].size(); });
        vector<int> ans;
        auto op_inner = [&ans](int x) {
            ans.push_back(x);
            return true;
        };
        auto op = [&](int x) {
            if (g[x].size() == 1) {
                ans.push_back(*g[x].begin());
                return true;
            }
            return toposort(g[x], inner_deg, beforeItems, op_inner);
        };

        if (!toposort(nonempty_groups, outer_deg, outer_edge, op)) return {};
        ranges::reverse(ans);
        return ans;
    }
};
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