<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>

#define fast ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL)
#define ull unsigned long long int
#define yes cout << "YES" << endl
#define no cout << "NO" << endl
#define ll long long int
#define mii unordered_map<int,int>
#define loop(i,s, e) for (int i = s; i < e; i++)
#define show(x) cout<<x<<" ";
#define pii pair<int,int>
#define vi vector<int>
#define pb push_back
#define rr return
#define s second
#define f first
#define inf 1e9
const int mod = 1e9+7;
const int N = 2e5+5;

class Solution {
    bool dfs(vector<int> adj[] , vector<int>& vis , vector<int>& pathvis ,int n){
        vis[n]=1;
        pathvis[n]=1;
        for(auto it:adj[n])
        if(!vis[it]){
            if(dfs(adj , vis , pathvis , it)== true){
                return true;
            }
        }
        else if (pathvis[it]){
            return true;
        }  
        pathvis[n]=0;
        return false;
    }
bool cycle(int V, vector<int> adj[]) {
        vector<int> vis(V+1,0);
        vector<int> pathvis(V+1,0);
        
        for(int i = 1 ; i<=V ; i ++){
            if(!vis[i]){
                if(dfs(adj , vis , pathvis , i)==true){
                    return true;
                }
            }
        }
        return false;
    }
vector<int> decide(vector<int> adj[],vector<int>& degree, int k){
    queue<int> q;
    vector<int> ans(k+1);
    loop(i,1,k+1){
        if (degree[i]==0)q.push(i);
    }
    int tt = 0;
    while(!q.empty()){
        int id = q.front();
        q.pop();
        ans[id]=tt++;
        for (auto x:adj[id]){
            degree[x]--;
            if (degree[x]==0)q.push(x);
        }
    }
    return ans;
}
public:
    vector<vector<int>> buildMatrix(int k, vector<vector<int>>& rowConditions, vector<vector<int>>& colConditions) {
        vector<int> adj[k+1];
        vector<int> degree(k+1);
        for (auto x:rowConditions){
            adj[x[0]].pb(x[1]);
            degree[x[1]]++;
        }
        if (cycle(k,adj)){return {};}
        vector<int> row = decide(adj,degree,k);
        loop(i,1,k+1){
            adj[i].clear();
            degree[i]=0;
        }
        for (auto x:colConditions){
            adj[x[0]].pb(x[1]);
            degree[x[1]]++;
        }
        if (cycle(k,adj)){return {};}
        vector<int> col= decide(adj,degree,k);
        vector<vector<int>> ans(k,vector<int>(k));
        loop(i,1,k+1){
            ans[row[i]][col[i]] = i;
        }
        return ans;
    }
};
auto init = []() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    return 'c';
}();
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