<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> topoSort(vector<vector<int>>& edges, int n) {
        unordered_map<int, vector<int>> adj;
        stack<int> st;
        vector<int> order;
        vector<int> visited(n + 1, 0);
        bool hasCycle = false;

        for (vector<int>& edge : edges) {
            int u = edge[0];
            int v = edge[1];
            adj[u].push_back(v);
        }

        for (int i = 1; i <= n; i++) {
            if (visited[i] == 0) {
                dfs(i, adj, visited, st, hasCycle);
                if (hasCycle) 
                    return {}; //no ordering possible
            }
        }
        
        while(!st.empty()) {
            order.push_back(st.top());
            st.pop();
        }
        return order;
    }

    void dfs(int node, unordered_map<int, vector<int>>& adj, vector<int>& visited,
             stack<int>& st, bool& hasCycle) {
        
        visited[node] = 1;  // Mark node as visiting
        //First, visit node's children
        for (int& nbr : adj[node]) {
            if (visited[nbr] == 0) {
                dfs(nbr, adj, visited, st, hasCycle);
            } else if (visited[nbr] == 1) {
                // Cycle detected
                hasCycle = true;
                return;
            }
        }

        visited[node] = 2; //visited
        st.push(node); //Now node can be added
    }

    vector<vector<int>> buildMatrix(int k, vector<vector<int>>& rowConditions,
                                    vector<vector<int>>& colConditions) {        
        vector<int> orderRows    = topoSort(rowConditions, k);
        vector<int> orderColumns = topoSort(colConditions, k);

        // We might have found cycle. That's why topo order is empty
        if (orderRows.empty() or orderColumns.empty())
            return {};

        vector<vector<int>> matrix(k, vector<int>(k, 0));
        for (int i = 0; i < k; i++) {
            for (int j = 0; j < k; j++) {
                if (orderRows[i] == orderColumns[j]) {
                    matrix[i][j] = orderRows[i];
                }
            }
        }
        return matrix;
    }
};


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