<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 {
    bool topoSort(vector<int>g[], vector<int>&topo,int k, vector<int>&indegree){
        queue<int>q;
        for(int i=1;i<=k;i++){
            if(indegree[i]==0){
                q.push(i);
            }
        }
        while(!q.empty()){
            int first = q.front();
            q.pop();
            topo.push_back(first);
            for(auto x:g[first]){
                indegree[x]--;
                if(indegree[x]==0){
                    q.push(x);
                }
            }
        }
        return indegree.size() == topo.size()+1;
    }
public:
    vector<vector<int>> buildMatrix(int k, vector<vector<int>>& rowConditions, vector<vector<int>>& colConditions) {
        vector<int>row[k+1];
        vector<int>col[k+1];
        vector<int>rows,cols;
        vector<int>rowDegree(k+1,0), colDegree(k+1,0);
        for(auto rc:rowConditions){
            int u = rc[0];
            int v = rc[1];
            row[u].push_back(v);
            rowDegree[v]++;
        }
        for(auto cc: colConditions){
            int u = cc[0];
            int v = cc[1];
            col[u].push_back(v);
            colDegree[v]++;
        }
        if(!topoSort(col,cols,k,colDegree)) return {};
        if(!topoSort(row,rows,k,rowDegree)) return {};
        // for(int i=0;i<rows.size();i++) cout<<rows[i]<<" "; cout<<endl;
        // for(int i=0;i<cols.size();i++) cout<<cols[i]<<" "; cout<<endl;
        vector<vector<int>> ans(k,vector<int>(k,0));
        for(int i=0;i<k;i++){
            for(int j=0;j<k;j++){
                if(rows[i]==cols[j])
                    ans[i][j] = rows[i];
            }
        }
        return ans;
    }
};

/*
0 0 1
3 0 0
0 2 0
*/
64737