<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>

static const int fastIO = [] {
	std::ios_base::sync_with_stdio(false), std::cin.tie(nullptr), std::cout.tie(nullptr);
	return 0;
}();

class Solution {
public:
	vector<vector<int>> buildMatrix(int k, vector<vector<int>>& rowConditions, vector<vector<int>>& colConditions) {
		vector<vector<int>> cells(k);
		vector<vector<int>> graph(k);
		vector<int> inGrade(k);
		queue<int> q;
		vector<int> topoSort;
		for (int t = 0; t < 2; t++) {
			graph.assign(k, vector<int>());
			inGrade.assign(k, 0);
			vector<vector<int>>& conditions = (t == 0) ? rowConditions : colConditions;
			for (auto& condition: conditions) {
				graph[condition[0] - 1].push_back(condition[1] - 1);
				inGrade[condition[1] - 1]++;
			}
			for (int i = 0; i < k; i++)
				if (inGrade[i] == 0)
					q.push(i);
			topoSort.clear();
			topoSort.reserve(k);
			while (!q.empty()) {
				int u = q.front();
				q.pop();
				topoSort.push_back(u);
				for (int v: graph[u]) {
					inGrade[v]--;
					if (inGrade[v] == 0)
						q.push(v);
				}
			}
			if (topoSort.size() < k) return {};
			for (int i = 0; i < k; i++)
				cells[topoSort[i]].push_back(i);
		}
		vector<vector<int>> ans(k, vector<int>(k, 0));
		for (int i = 0; i < k; i++)
			ans[cells[i][0]][cells[i][1]] = i + 1;
		return ans;
	}
};
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