<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>

#include<bits/stdc++.h>
using namespace std;
#define subS1(v, i) string(v.begin() + i, v.end())
#define subS2(v, i, j) string(v.begin() + i, v.begin() + j + 1)
#define e_(vals, val) (vals.find(val) != vals.end())
#define g_(val) ((exists(memo, val)) ? dic[val] : 1)
#define _c(x, y) ((x < n && x >= 0 && y < n && y >= 0) ? true : false)
#define forr(i, n) for(int i=0;i<n;i++)
int IO = []{
    ios::sync_with_stdio(false); cin.tie(nullptr); return 0;
}();


class Solution {
public:
    vector<vector<int>> buildMatrix(int k, vector<vector<int>>& rowConditions, vector<vector<int>>& colConditions) {
        int n = rowConditions.size(), m = colConditions.size();
        vector<int> row(k, -1);
        vector<vector<int>> pre(k);
        // cout << "Good";
        forr(i, n){
            pre[rowConditions[i][1]-1].push_back(rowConditions[i][0]-1);
        }
        forr(i, k){
            bool done = false;
            forr(j, k){
                if(row[j]==-1 && pre[j].size()==0){
                    row[j] = i;
                    done = true;
                    break;
                }
            }
            if(done)continue;
            forr(j, k){
                if(row[j]!=-1)continue;
                for(int x=pre[j].size()-1;x>=0;x--){
                    if(row[pre[j][x]]!=-1)pre[j].erase(x + pre[j].begin());
                }
                if(pre[j].size() == 0){
                    row[j] = i;
                    break;
                }
                if(j == k-1)return {};
            }
        }
        // for(int i=0;i<k;i++)cout << row[i] << " ";
        
        vector<int> col(k, -1);
        vector<vector<int>> tmp(k);
        pre = tmp;
        forr(i, m){
            pre[colConditions[i][1]-1].push_back(colConditions[i][0]-1);
        }
        forr(i, k){
            bool done = false;
            forr(j, k){
                if(col[j]==-1 && pre[j].size()==0){
                    col[j] = i;
                    done = true;
                    break;
                }
            }
            if(done)continue;
            forr(j, k){
                if(col[j]!=-1)continue;
                for(int x=pre[j].size()-1;x>=0;x--){
                    if(col[pre[j][x]]!=-1)pre[j].erase(x + pre[j].begin());
                }
                if(pre[j].size() == 0){
                    col[j] = i;
                    break;
                }
                if(j == k-1)return {};
            }
        }
        // cout << endl;
        // for(int i=0;i<k;i++)cout << col[i] << " ";
        vector<vector<int>> ans(k, vector<int>(k, 0));
        forr(i, k){
            ans[row[i]][col[i]] = i+1;
        }
        return ans;
    }
};
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