<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>

from collections import deque

class Solution:
    def buildMatrix(self, k: int, rowConditions: List[List[int]], colConditions: List[List[int]]) -> List[List[int]]:
        matrix = [[0 for i in range(k)] for j in range(k)]

        class Node:
            def __init__(self, val):
                self.val = val
                self.incomming = set()
                self.outcomming = set()

        rows = {i: Node(i) for i in range(1, k+1)}
        cols = {i: Node(i) for i in range(1, k+1)}
        row, col = {}, {}
        
        rows_potential = set(range(1, k+1))
        for left, right in rowConditions:
            rows[left].outcomming.add(right)
            rows[right].incomming.add(left)
            if right in rows_potential:
                rows_potential.remove(right)
        if not rows_potential:
            return []
        del rowConditions
        rows_potential = deque(rows_potential)
        
        cols_potential = set(range(1, k+1))
        for left, right in colConditions:
            cols[left].outcomming.add(right)
            cols[right].incomming.add(left)
            if right in cols_potential:
                cols_potential.remove(right)
        if not cols_potential:
            return []
        del colConditions
        cols_potential = deque(cols_potential)
        
        i = 0
        while rows_potential:
            x = rows_potential.pop()
            outcomming = rows[x].outcomming
            for y in outcomming:
                incomming = rows[y].incomming
                incomming.remove(x)
                if not incomming:
                    rows_potential.append(y)
            row[x] = i
            i += 1
        if i != k:
            return []
        
        i = 0
        while cols_potential:
            x = cols_potential.pop()
            outcomming = cols[x].outcomming
            for y in outcomming:
                incomming = cols[y].incomming
                incomming.remove(x)
                if not incomming:
                    cols_potential.append(y)
            col[x] = i
            i += 1
        if i != k:
            return []
        
        for i in range(1, k+1):
            matrix[row[i]][col[i]] = i
        
        return matrix
            
590