<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<vector<int>> buildMatrix(int k, vector<vector<int>>& rowConditions, vector<vector<int>>& colConditions) {
        unordered_map<int, vector<int>> graph;
        for (const auto& edge : rowConditions) {
            graph[edge[0]].push_back(edge[1]);
        }

        vector<int> row_sorting = topo_sort(graph, k);
        graph.clear();
        for (const auto& edge : colConditions) {
            graph[edge[0]].push_back(edge[1]);
        }
        vector<int> col_sorting = topo_sort(graph, k);

        if (row_sorting.empty() || col_sorting.empty()) {
            return {};
        }

        unordered_map<int, pair<int, int>> value_position;
        for (int i = 0; i < k; ++i) {
            value_position[row_sorting[i]].first = i;
            value_position[col_sorting[i]].second = i;
        }

        vector<vector<int>> res(k, vector<int>(k, 0));
        for (int value = 1; value <= k; ++value) {
            int row = value_position[value].first;
            int column = value_position[value].second;
            res[row][column] = value;
        }

        return res;
    }
    bool dfs(int src, unordered_map<int, vector<int>>& graph, unordered_set<int>& visited, unordered_set<int>& cur_path, vector<int>& res) {
        if (cur_path.count(src)) return false;  // cycle detected
        if (visited.count(src)) return true;  // all okay, but we've already visited this node

        visited.insert(src);
        cur_path.insert(src);

        for (int neighbor : graph[src]) {
            if (!dfs(neighbor, graph, visited, cur_path, res))  // if any child returns false
                return false;
        }

        cur_path.erase(src);  // backtrack path
        res.push_back(src);
        return true;
    }

    vector<int> topo_sort(unordered_map<int, vector<int>>& graph, int k) {
        unordered_set<int> visited;
        unordered_set<int> cur_path;
        vector<int> res;

        for (int src = 1; src <= k; ++src) {
            if (!dfs(src, graph, visited, cur_path, res))
                return {};
        }

        reverse(res.begin(), res.end());  // we will have res as reversed so we need to reverse it one more time
        return res;
    }
};
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