<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:
    vector<int> topologicalSort(vector<vector<int> >& adj)
{
    int V=adj.size();
    // Vector to store indegree of each vertex
    vector<int> indegree(V);
    for (int i = 0; i < V; i++) {
        for (auto it : adj[i]) {
            indegree[it]++;
        }
    }

    // Queue to store vertices with indegree 0
    queue<int> q;
    for (int i = 0; i < V; i++) {
        if (indegree[i] == 0) {
            q.push(i);
        }
    }
    vector<int> result;
    while (!q.empty()) {
        int node = q.front();
        q.pop();
        result.push_back(node);

        // Decrease indegree of adjacent vertices as the
        // current node is in topological order
        for (auto it : adj[node]) {
            indegree[it]--;

            // If indegree becomes 0, push it to the queue
            if (indegree[it] == 0)
                q.push(it);
        }
    }

    // Check for cycle
    // if (result.size() != V) {
    //     cout << "Graph contains cycle!" << endl;
    //     return {};
    // }

    return result;
}
    vector<int> toposort(vector<vector<int>> &adj) {
        int n=adj.size(); 
        vector<vector<int>> copy(n); 
        vector<int> in(n,0); 
        for(int i=0;i<n;i++) {
            for(auto x:adj[i]) {
                // copy[adj[i][j]].push_back(i); 
                // in[adj[i][j]]++; 
                in[x]++;
            }
        }
        vector<int> topo; 
        queue<int> q; 
        for(int i=0;i<n;i++) {
            if(in[i]==0) q.push(i); 
        }
        while(!q.empty()) {
            int node=q.front(); 
            topo.push_back(node); 
            q.pop(); 
            for(auto x:adj[node]) {
                in[x]--; 
                if(in[x]==0) {
                    q.push(x); 
                    // ans.push_back(x); 
                }
            }
        }
        return topo; 
    }
    vector<vector<int>> buildMatrix(int k, vector<vector<int>>& rowConditions, vector<vector<int>>& colConditions) {
        vector<vector<int>> adj1(k+1),adj2(k+1); 
        for(auto x:rowConditions) adj1[x[0]].push_back(x[1]);
        for(auto x:colConditions) adj2[x[0]].push_back(x[1]);
        vector<int> topo1=toposort(adj1); 
        vector<int> topo2=toposort(adj2); 
        for(auto x:topo1) cout<<x<<" ";
        cout<<endl;
        for(auto x:topo2) cout<<x<<" ";
        if(topo1.size() != k+1 || topo2.size() != k+1) return {}; 
        vector<vector<int>> matrix(k, vector<int>(k, 0));
        for (int i = 1; i <= k; i++) {
            for (int j = 1; j <= k; j++) {
                if (topo1[i] == topo2[j]) {
                    matrix[i-1][j-1] = topo1[i];
                }
            }
        }
        return matrix;
        return {};

    }
};
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