<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:
    std::vector<std::vector<int>> buildMatrix(
        int k, 
        const std::vector<std::vector<int>> &rowConditions, 
        const std::vector<std::vector<int>> &colConditions
    ) {
        static constexpr size_t K = 400;

        static std::vector<int> used;
        static std::vector<int> edges[K];
        static std::vector<int> q;
        
        std::function<bool(int)> dfsCheckCycle = [&dfsCheckCycle](int u) {
            used[u] = 1;
            
            for (const auto v : edges[u])
                if (!used[v] && dfsCheckCycle(v) || used[v] == 1)
                    return true;
            
            used[u] = 2;
            return false;
        };

        const auto tryTopsort = [&](const std::vector<std::vector<int>> &edgeList) {
            for (size_t u = 0; u < k; ++u)
                edges[u].clear();
            for (const auto &e : edgeList)
                edges[e[0] - 1].push_back(e[1] - 1);
            for (size_t u = 0; u < k; ++u) {
                std::sort(edges[u].begin(), edges[u].end());
                edges[u].erase(std::unique(edges[u].begin(), edges[u].end()), edges[u].end());
            }

            used.assign(k, 0);
            for (size_t u = 0; u < k; ++u)
                if (!used[u] && dfsCheckCycle(u))
                    return false;
            
            used.assign(k, 0);
            for (size_t u = 0; u < k; ++u)
                for (const auto v : edges[u])
                    ++used[v];
            
            q.resize(k);
            int qEnd = 0;
            for (size_t u = 0; u < k; ++u)
                if (!used[u])
                    q[qEnd++] = u;
            for (int qStart = 0; qStart < qEnd; ++qStart) {
                const auto u = q[qStart];
                for (const auto v : edges[u])
                    if (!--used[v])
                        q[qEnd++] = v;
            }
            assert(qEnd == k);

            return true;
        };

        const static std::vector<std::vector<int>> empty;

        std::vector<std::vector<int>> out;
        if (!tryTopsort(colConditions))
            return empty;
        
        for (const auto u : q)
            std::cout << u << " ";
        std::cout << "\n";

        out.emplace_back(std::move(q));
        for (auto &u : out[0])
            ++u;
        if (!tryTopsort(rowConditions))
            return empty;
        
        for (const auto u : q)
            std::cout << u << " ";
        std::cout << "\n";

        out.resize(k, std::vector<int>(k, 0));
        for (size_t i = 0; i < k; ++i) {
            const auto j = std::find(out[0].begin(), out[0].end(), q[i] + 1) - out[0].begin();
            assert(j < k);
            
            out[0][j] = 0;
            out[i][j] = q[i] + 1;
        }

        return out;
    }
};
57642