<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:
    int nope = 0;
    void ts(vector<int> &ans, vector<vector<int>> &edges){
        vector<int> adj[402];
        vector<int> in(402,0);
        vector<int> vis(402,0);
        unordered_map<int,int> mp;
        queue<int> q;
        for(int i = 0; i<edges.size(); i++){
            adj[edges[i][0]].push_back(edges[i][1]);
            in[edges[i][1]]++;
            mp[edges[i][0]] = mp[edges[i][1]] = 1;
        }
        for(auto it : mp){
            if(in[it.first] == 0)
                q.push(it.first);
        }
        while(!q.empty()){
            int x = q.front(); q.pop();
            vis[x] = 1;
            ans.push_back(x);
            for(auto it : adj[x]){
                in[it]--;
                if(!in[it] && !vis[it]){
                    vis[it] = 1, q.push(it);
                }
            }
        }
        if(!q.empty() || ans.size() < mp.size()){
            nope = 1;
        }
    }
    vector<vector<int>> buildMatrix(int k, vector<vector<int>>& rowConditions, vector<vector<int>>& colConditions) {
        vector<int> rs,cs;
        vector<int> covered(402,0);
        ts(rs, rowConditions); ts(cs, colConditions);
        vector<vector<int>> empty;
        if(nope){
            return empty;
        }
        vector<vector<int>> grid(k,vector<int>(k,0));
        unordered_map<int,int> cmp,rmp;
        for(int i = 0; i<cs.size(); i++)
            cmp[cs[i]] = i;
        for(int i = 0; i<rs.size(); i++)
            rmp[rs[i]] = i;
        for(int i = 0; i<rs.size(); i++){
            if(cmp.find(rs[i]) != cmp.end()){
                grid[i][cmp[rs[i]]] = rs[i];
                covered[rs[i]] = 1;
                cmp[rs[i]] = rmp[rs[i]] = -1;
            }
        }
        for(auto it : cmp){
            if(it.second >= 0){
                int f = 0;
                for(int i = 0; i<k; i++){
                    if(grid[i][it.second] == 0){
                        covered[it.first] = 1;
                        grid[i][it.second] = it.first;
                        f = 1;
                        break;
                    }
                }
                if(!f){
                    return empty;
                }      
            }
        }
        for(auto it : rmp){
            if(it.second >= 0){
                int f = 0;
                for(int i = 0; i<k; i++){
                    if(grid[it.second][i] == 0){
                        covered[it.first] = 1;
                        grid[it.second][i] = it.first;
                        f = 1;
                        break;
                    }
                }
                if(!f){
                    return empty;
                }      
            }
        }

        int c = 0;
        for(int i = 1; i<=k; i++){
            int prevJ = 0, prevK = 0;
            if(!covered[i]){
                int f = 0;
                for(int j = prevJ; j<k; j++){
                    for(int l = prevK; l<k; l++){
                        if(grid[j][l] == 0){
                            grid[j][l] = i;
                            f = 1;
                            prevJ = j, prevK = l;
                            break;
                        }
                    }
                    if(f)
                        break;
                }
                if(!f)
                    return empty;
            }
        }
        return grid;
    }
};
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