<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:
    void dfs(int v,vector<bool>& vis,vector<int>& ans,vector<vector<int>>& g) {
        vis[v] = true;
        for (int u : g[v]) {
            if (!vis[u])
                dfs(u,vis,ans,g);
        }
        ans.push_back(v);
    }
    vector<int> topoSort(vector<vector<int>>& g,bool& f) {
        vector<bool> visited;
        vector<int> ans;
        visited.assign(g.size(), false);
        for (int i = 0; i < g.size(); i++) {
            if (!visited[i]) {
                dfs(i,visited,ans,g);
            }
        }
        reverse(ans.begin(), ans.end());
        return ans;
    }
    vector<int> topoSort(vector<vector<int>>& g){
        int n=g.size();
        vector<int> inDegree(n,0);
        for(int i=0;i<n;i++){
            for(int j=0;j<g[i].size();j++){
                inDegree[g[i][j]]++;
            }
        }
        stack<int> st;
        for(int i=0;i<n;i++){
            if(inDegree[i]==0)st.push(i);
        }
        vector<int> ans;
        while(!st.empty()){
            int node=st.top();
            st.pop();
            ans.push_back(node);
            for(int i=0;i<g[node].size();i++){
                inDegree[g[node][i]]--;
                if(inDegree[g[node][i]]==0)st.push(g[node][i]);
            }
        }
        return ans;
    }
    void show(vector<int>& v){
        for(int i=0;i<v.size();i++){
            cout<<v[i]<<" ";
        }
        cout<<endl;
    }
    vector<int> sortItems(int n, int m, vector<int>& group, vector<vector<int>>& beforeItems) {
        unordered_map<int,int> gp;
        int freeGp=0;
        for(int i=0;i<group.size();i++){
            if(group[i]!=-1)gp[i]=group[i];
            else{
                gp[i]=m+freeGp;
                freeGp++;
            }
        }
        vector<vector<int>> gpDepend(m+freeGp);
        vector<vector<int>> graphWithoutInterDep(n);
        for(int i=0;i<beforeItems.size();i++){
            for(int j=0;j<beforeItems[i].size();j++){
                if(gp[i]!=gp[beforeItems[i][j]]){
                    gpDepend[gp[beforeItems[i][j]]].push_back(gp[i]);
                }
                else{
                    graphWithoutInterDep[beforeItems[i][j]].push_back(i);
                }
            }
        }
        bool f=true;
        vector<int> topoSortedGpDepend=topoSort(gpDepend);
        // if(!f)return {};
        vector<int> topoSortedInternalGraph=topoSort(graphWithoutInterDep);
        // if(!f)return {};
        show(topoSortedGpDepend);
        show(topoSortedInternalGraph);
        if(topoSortedGpDepend.size()!=m+freeGp)return {};
        if(topoSortedInternalGraph.size()!=n)return {};
        unordered_map<int,vector<int>> mp;
        vector<int> ans;
        for(int i=0;i<topoSortedInternalGraph.size();i++){
            mp[gp[topoSortedInternalGraph[i]]].push_back(topoSortedInternalGraph[i]);
        }
        for(int i=0;i<topoSortedGpDepend.size();i++){
            ans.insert(ans.end(),mp[topoSortedGpDepend[i]].begin(),mp[topoSortedGpDepend[i]].end());
        }
        return ans;
    }
};
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