<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>

import queue
class Solution:
        
    def topologicalSort(self, condition:List[List[int]], k:int) -> List[int]:
        gr_map = {}
        n_degree = [0]*(k+1)
        for sr, dest in condition:

            if sr in gr_map:
                gr_map[sr].append(dest)
                
            else:
                gr_map[sr] = [dest] 
            n_degree[dest]+=1
           
        qu = queue.Queue()  
        count = 0
        for i in range(1, k+1):
            if  n_degree[i] == 0:
                qu.put(i)
                count+=1
        ans = []
        # for key, value in visited.items():
        #     if not value:
        #         qu.put(key)
        #         vis[key] = True
        #         ans.append(key)  
        # print(ans)  
        # print(gr_map)
        # while not qu.empty():
        #     print(gr_map[qu.get()])   
        while not qu.empty():
            val = qu.qsize()
            while val>0:
                val-=1

                vlu = qu.get()
                ans.append(vlu)
                if vlu in gr_map:
                    for valu in gr_map[vlu]:
                        n_degree[valu]-=1
                        if n_degree[valu] == 0:
                            qu.put(valu)
                            count+=1  
        if count != k :
            return []
        return ans                            


                   

    def buildMatrix(self, k: int, rowConditions: List[List[int]], colConditions: List[List[int]]) -> List[List[int]]:
        rows = self.topologicalSort(rowConditions, k)
        cols = self.topologicalSort(colConditions, k)
        print(rows)
        print(cols)
        if not rows or not cols: return []
        ans = [[0 for _ in range(k)] for _ in range(k)]
        clm = {}
        i_used = {}

        i=0
        for i in range(k):
            for j in range(k):
                if i<len(rows)and j<len(cols):
                    if rows[i] == cols[j]:
                        ans[i][j] = rows[i]
        # for cl in cols:
        #     clm[cl] = i
        #     i+=1
        # for i in range(k):
        #     val = rows[i]
        #     if val in clm:
        #         j = clm[val]
        #         ans[i][j] = val

            
        return ans        


        
24600