<p>You are given a <strong>positive</strong> integer <code>k</code>. You are also given:</p>

<ul>
	<li>a 2D integer array <code>rowConditions</code> of size <code>n</code> where <code>rowConditions[i] = [above<sub>i</sub>, below<sub>i</sub>]</code>, and</li>
	<li>a 2D integer array <code>colConditions</code> of size <code>m</code> where <code>colConditions[i] = [left<sub>i</sub>, right<sub>i</sub>]</code>.</li>
</ul>

<p>The two arrays contain integers from <code>1</code> to <code>k</code>.</p>

<p>You have to build a <code>k x k</code> matrix that contains each of the numbers from <code>1</code> to <code>k</code> <strong>exactly once</strong>. The remaining cells should have the value <code>0</code>.</p>

<p>The matrix should also satisfy the following conditions:</p>

<ul>
	<li>The number <code>above<sub>i</sub></code> should appear in a <strong>row</strong> that is strictly <strong>above</strong> the row at which the number <code>below<sub>i</sub></code> appears for all <code>i</code> from <code>0</code> to <code>n - 1</code>.</li>
	<li>The number <code>left<sub>i</sub></code> should appear in a <strong>column</strong> that is strictly <strong>left</strong> of the column at which the number <code>right<sub>i</sub></code> appears for all <code>i</code> from <code>0</code> to <code>m - 1</code>.</li>
</ul>

<p>Return <em><strong>any</strong> matrix that satisfies the conditions</em>. If no answer exists, return an empty matrix.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2022/07/06/gridosdrawio.png" style="width: 211px; height: 211px;" />
<pre>
<strong>Input:</strong> k = 3, rowConditions = [[1,2],[3,2]], colConditions = [[2,1],[3,2]]
<strong>Output:</strong> [[3,0,0],[0,0,1],[0,2,0]]
<strong>Explanation:</strong> The diagram above shows a valid example of a matrix that satisfies all the conditions.
The row conditions are the following:
- Number 1 is in row <u>1</u>, and number 2 is in row <u>2</u>, so 1 is above 2 in the matrix.
- Number 3 is in row <u>0</u>, and number 2 is in row <u>2</u>, so 3 is above 2 in the matrix.
The column conditions are the following:
- Number 2 is in column <u>1</u>, and number 1 is in column <u>2</u>, so 2 is left of 1 in the matrix.
- Number 3 is in column <u>0</u>, and number 2 is in column <u>1</u>, so 3 is left of 2 in the matrix.
Note that there may be multiple correct answers.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> k = 3, rowConditions = [[1,2],[2,3],[3,1],[2,3]], colConditions = [[2,1]]
<strong>Output:</strong> []
<strong>Explanation:</strong> From the first two conditions, 3 has to be below 1 but the third conditions needs 3 to be above 1 to be satisfied.
No matrix can satisfy all the conditions, so we return the empty matrix.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>2 &lt;= k &lt;= 400</code></li>
	<li><code>1 &lt;= rowConditions.length, colConditions.length &lt;= 10<sup>4</sup></code></li>
	<li><code>rowConditions[i].length == colConditions[i].length == 2</code></li>
	<li><code>1 &lt;= above<sub>i</sub>, below<sub>i</sub>, left<sub>i</sub>, right<sub>i</sub> &lt;= k</code></li>
	<li><code>above<sub>i</sub> != below<sub>i</sub></code></li>
	<li><code>left<sub>i</sub> != right<sub>i</sub></code></li>
</ul>

class Solution {
public:
    //topological sort using BFS (Kahn's algorithm)
    static vector<int> topo_sort(int k, vector<vector<int>>& conditions){
        vector<int> deg(k+1, 0);
        vector<vector<int>> adj(k+1);
        for(auto& edge: conditions){
            int v=edge[0], w=edge[1];
            adj[v].push_back(w);
            deg[w]++;
        }
        queue<int> q;
        for (int i = 1; i<=k; i++) {
            if (deg[i] == 0) 
                q.push(i);
        }

        int count=0;
        vector<int> ans;
        ans.reserve(k);
        while(!q.empty()){
            int j=q.front();
            q.pop();
            ans.push_back(j);
            count++;
            for(int k:adj[j]){
                deg[k]--;
                if (deg[k] == 0) {
                    q.push(k);
                }
            }
        }
        if(count != k) return {};// has cycle
        else return ans ;

    }

    static void print(auto& c){
        cout<<"[";
        for(int x: c){
            cout<<x;
            if (x!=c.back()) cout<<", ";
        }
        cout<<"]\n";
    }

    static vector<vector<int>> buildMatrix(int k, vector<vector<int>>& rowConditions, vector<vector<int>>& colConditions) {
        auto order_row=topo_sort(k, rowConditions);
        auto order_col=topo_sort(k, colConditions);

    //  print out the ordering in row & in col
    //    print(order_row), print(order_col);

        if (order_row.empty()|| order_col.empty())
            return {};// some conflict

        vector<vector<int>> arr(k, vector<int>(k));
        // Find pos for x where 1<=x<=k
        vector<int> pos_i(k+1, -1), pos_j(k+1, -1);
        for(int i=0; i<k; i++){
            pos_i[order_row[i]]=i;
            pos_j[order_col[i]]=i;
        }

        for(int x=1; x<=k; x++)
            arr[pos_i[x]][pos_j[x]]=x;
        
        return arr;
    }
};

auto init = []() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.tie(nullptr);
    return 'c';
}();
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