<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 {
public:
    bool isPrintable(vector<vector<int>>& arr) {
        int n = arr.size(), m = arr[0].size(), N = 0;
        for(int i=0; i<n; i++) {
            for(int j=0; j<m; j++) {
                N = max(arr[i][j], N);
            }
        }
        vector<set<int>> adj(N+1);
        for(int col=1; col<=N; col++) {
            int mnx=1e9, mny = 1e9, mxx=-1, mxy=-1;
            for(int i=0; i<n; i++) {
                for(int j=0; j<m; j++) {
                    if(arr[i][j]==col) {
                        mnx = min(mnx, i);
                        mny = min(mny, j);
                        mxx = max(mxx, i);
                        mxy = max(mxy, j);
                    }
                }
            }
            for(int i=mnx; i<=mxx; i++) {
                for(int j=mny; j<=mxy; j++) {
                    if(arr[i][j]!=col) {
                        adj[col].insert(arr[i][j]);
                    }
                }
            }
        }
        vector<int> ind(N+1);
        for(auto s:adj) {
            for(auto u:s) {
                ind[u]++;
            }
        }
        queue<int> q;
        for(int i=1; i<=N; i++) {
            if(ind[i]==0) q.push(i);
        }
        vector<int> topo;
        while(!q.empty()) {
            int u = q.front();
            q.pop();
            topo.push_back(u);
            for(auto v:adj[u]) {
                ind[v]--;
                if(ind[v]==0) {
                    q.push(v);
                }
            }
        }
        return topo.size()==N;

    }
};
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