<p>There is a strange printer with the following two special requirements:</p>

<ul>
	<li>On each turn, the printer will print a solid rectangular pattern of a single color on the grid. This will cover up the existing colors in the rectangle.</li>
	<li>Once the printer has used a color for the above operation, <strong>the same color cannot be used again</strong>.</li>
</ul>

<p>You are given a <code>m x n</code> matrix <code>targetGrid</code>, where <code>targetGrid[row][col]</code> is the color in the position <code>(row, col)</code> of the grid.</p>

<p>Return <code>true</code><em> if it is possible to print the matrix </em><code>targetGrid</code><em>,</em><em> otherwise, return </em><code>false</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/12/23/print1.jpg" style="width: 600px; height: 175px;" />
<pre>
<strong>Input:</strong> targetGrid = [[1,1,1,1],[1,2,2,1],[1,2,2,1],[1,1,1,1]]
<strong>Output:</strong> true
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/12/23/print2.jpg" style="width: 600px; height: 367px;" />
<pre>
<strong>Input:</strong> targetGrid = [[1,1,1,1],[1,1,3,3],[1,1,3,4],[5,5,1,4]]
<strong>Output:</strong> true
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> targetGrid = [[1,2,1],[2,1,2],[1,2,1]]
<strong>Output:</strong> false
<strong>Explanation:</strong> It is impossible to form targetGrid because it is not allowed to print the same color in different turns.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>m == targetGrid.length</code></li>
	<li><code>n == targetGrid[i].length</code></li>
	<li><code>1 &lt;= m, n &lt;= 60</code></li>
	<li><code>1 &lt;= targetGrid[row][col] &lt;= 60</code></li>
</ul>

class Solution {
private:
    bool isBlocking(const vector<int> &rectA, const vector<int> &rectB, vector<vector<int>>& targetGrid) {
        // Check if rectA blocks rectB
        for (int i = rectA[0]; i <= rectA[2]; i++) {
            for (int j = rectA[1]; j <= rectA[3]; j++) {
                if (targetGrid[i][j] == rectB[4]) return true;
            }
        }
        return false;
    }

    vector<int> topologicalSort(vector<int> adj[], int V) {
        vector<int> in_degree(V, 0);
        for (int u = 0; u < V; u++) {
            for (int x : adj[u]) in_degree[x]++;
        }

        queue<int> q;
        for (int i = 0; i < V; i++) {
            if (in_degree[i] == 0) q.push(i);
        }

        vector<int> topo;
        while (!q.empty()) {
            int u = q.front();
            q.pop();

            topo.push_back(u);

            for (int x : adj[u]) {
                in_degree[x]--;
                if (in_degree[x] == 0) q.push(x);
            }
        }

        return topo;
    }

    bool isCyclic(int V, vector<int> adj[]) {
        vector<int> topoSort = topologicalSort(adj, V);
        return topoSort.size() != V;
    }

    void removeSameAdjNode(vector<int> adj[]) {
        for (int i = 0; i <= 60; i++) {
            if (adj[i].size() == 0) continue;
            sort(adj[i].begin(), adj[i].end());
            adj[i].erase(unique(adj[i].begin(), adj[i].end()), adj[i].end());
        }
    }

public:
    bool isPrintable(vector<vector<int>>& targetGrid) {
        int m = targetGrid.size(), n = targetGrid[0].size();
        vector<vector<int>> idx(61, {INT_MAX, INT_MAX, INT_MIN, INT_MIN, 0});

        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                int color = targetGrid[i][j];
                idx[color][0] = min(idx[color][0], i);
                idx[color][1] = min(idx[color][1], j);
                idx[color][2] = max(idx[color][2], i);
                idx[color][3] = max(idx[color][3], j);
                idx[color][4] = color; // Store the color itself for easier reference
            }
        }

        vector<int> adj[61];

        for (int i = 0; i <= 60; i++) {
            if (idx[i][4] == 0) continue; // Skip colors not in use
            for (int j = 0; j <= 60; j++) {
                if (i != j && idx[j][4] != 0) {
                    if (isBlocking(idx[i], idx[j], targetGrid)) {
                        adj[i].push_back(j);
                    }
                }
            }
        }

        // removeSameAdjNode(adj);
        return !isCyclic(61, adj);
    }
};

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