<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:
    void DFS(int idx, vector<vector<int>>& g, vector<int>& vis,
             vector<int>& order) {
        vis[idx] = 1;
        for (auto child : g[idx]) {
            if (vis[child])
                continue;
            DFS(child, g, vis, order);
        }
        order.push_back(idx);
    }

    void Top(int idx, vector<vector<int>>& g_rev, vector<int>& vis) {
        vis[idx] = 1;
        for (auto child : g_rev[idx]) {
            if (vis[child])
                continue;
            Top(child, g_rev, vis);
        }
    }
    bool Cycle(int k, vector<vector<int>>& g, vector<vector<int>>& g_rev,
               vector<int>& order) {
        
        vector<int> vis_1(k + 1, 0);
        vector<int> vis_2(k + 1, 0);
        for (int i = 1; i <= k; i++) {
            if (!vis_1[i]) {
                DFS(i, g, vis_1, order);
            }
        }
        vector<int> temp = order;
        int cnt = 0;
        while (temp.size()) {
            if (!vis_2[temp.back()]) {
                Top(temp.back(), g_rev, vis_2);
                cnt++;
            }
            temp.pop_back();
        }
        // 
        // for(auto i : order)
        // {
        //     cout << i << " " ;
        // }
        // cout <<  endl;
        return cnt == k;
    }
    vector<vector<int>> buildMatrix(int k, vector<vector<int>>& rowConditions,
                                    vector<vector<int>>& colConditions) {
        vector<vector<int>> g1(k + 1);
        vector<vector<int>> g1_rev(k + 1);

        vector<vector<int>> h1(k + 1);
        vector<vector<int>> h1_rev(k + 1);

        for (vector<int>& x : rowConditions) {
            g1[x[0]].push_back(x[1]);
            g1_rev[x[1]].push_back(x[0]);
        }

        for (vector<int>& x : colConditions) {
            h1[x[0]].push_back(x[1]);
            h1_rev[x[1]].push_back(x[0]);
        }

        vector<int> vertical;
        vector<int> horizontal;
        if (Cycle(k, g1, g1_rev, vertical) and
            Cycle(k, h1, h1_rev, horizontal)) {
            reverse(vertical.begin(),vertical.end());
            reverse(horizontal.begin(),horizontal.end());
            vector<pair<int, int>> pos(k + 1);
            for (int i = 0; i < k; i++) {
                pos[vertical[i]].second = i;
            }
            for (int i = 0; i < k; i++) {
                pos[horizontal[i]].first = i;
            }
            vector<vector<int>> ans(k, vector<int>(k, 0));
            for (int i = 1; i <= k; i++) {
                int y = pos[i].first;
                int x = pos[i].second;
                ans[x][y] = i;
            }
            return ans;
        }
        return {};
    }
};
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