<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 {
    void solve(int k, vector<vector<int>>& conditions, vector<int>& idx) {
        queue<int> queue;
        vector<int> degree(k + 1);
        vector<vector<int>> E(k + 1);
        for (auto& condition: conditions) {
            int num0 = condition[0];
            int num1 = condition[1];
            degree[num1]++;
            E[num0].push_back(num1);
        }
        for (int num = 1; num <= k; num++)
            if (degree[num] == 0) idx[num] = 0, queue.push(num);
        while (queue.size() != 0) {
            int num0 = queue.front(); queue.pop();
            for (int num1: E[num0]) if (--degree[num1] == 0)
                idx[num1] = idx[num0] + 1, queue.push(num1);
        }
        return;
    }
public:
    vector<vector<int>> buildMatrix(int k,
            vector<vector<int>>& rowConditions,
            vector<vector<int>>& colConditions) {
        vector<int> row(k + 1, -1);
        vector<int> col(k + 1, -1);
        solve(k, rowConditions, row);
        solve(k, colConditions, col);
        for (int num = 1; num <= k; num++) {
            if (row[num] == -1) return {};
            if (col[num] == -1) return {};
        }
        vector<pair<int, int>> row_num;
        vector<pair<int, int>> col_num;
        vector<pair<int, int>> coords(k + 1);
        vector<vector<int>> matrix(k, vector<int>(k));
        for (int num = 1; num <= k; num++) {
            row_num.push_back({row[num], num});
            col_num.push_back({col[num], num});
        }
        sort(row_num.begin(), row_num.end());
        sort(col_num.begin(), col_num.end());
        for (int i = 0; i < k; i++) coords[row_num[i].second]. first = i;
        for (int j = 0; j < k; j++) coords[col_num[j].second].second = j;
        for (int num = 1; num <= k; num++)
            matrix[coords[num].first][coords[num].second] = num;
        return matrix;
    }
};
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