<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>>& targetGrid) {
        int n = targetGrid.size(), m = targetGrid[0].size();

        vector<vector<int>> adj(61);
        for(int i = 1; i<=60; i++){
            int minx = n, miny = m, maxx = -1, maxy = -1;
            for(int x = 0; x<n; x++){
                for(int y = 0; y<m; y++){
                    if(targetGrid[x][y] == i){
                        minx = min(minx, x);
                        miny = min(miny, y);
                        maxx = max(maxx, x);
                        maxy = max(maxy, y);
                    }
                }
            }

            for(int r = minx; r<=maxx; r++)
                for(int c = miny; c<=maxy; c++)
                    if(targetGrid[r][c] != i)
                        adj[i].push_back(targetGrid[r][c]);
        }

        vector<int> vis(61, 0);
        for(int i = 1; i<=60; i++){
            if(!vis[i] && dfs(i, adj, vis))
                return false;
        }

        return true;
    }

    bool dfs(int node, vector<vector<int>> &adj, vector<int> &vis){
        vis[node] = 1;

        for(auto nbgh: adj[node]){
            if(vis[nbgh] == 2)  continue;
            if(vis[nbgh] == 1)  return true;
            if(dfs(nbgh, adj, vis)) return true;
        }

        vis[node] = 2;
        return false;
    }
};
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