<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:
    void calc(int i, vector<int>&mir, vector<int>&mar, vector<int>&mic, vector<int>&mac, vector<vector<int>> & grid){
        for(int k=0; k<grid.size(); k++){
            for(int j=0; j<grid[0].size();j++){
                if(grid[k][j] == i){
                    mir[i] = min(mir[i],k);
                    mar[i] = max(mar[i],k);
                    mic[i] = min(mic[i],j);
                    mac[i] = max(mac[i],j);
                }
            }
        }
    }
    bool chk(int i, vector<int>&vis, vector<int>&pathvis, vector<vector<int>> &adj){
        vis[i]=1;
        pathvis[i]=1;
        for(auto it: adj[i]){
            if(!vis[it]){
                if(chk(it,vis,pathvis,adj)){
                    return true;
                }
            }else if(pathvis[it]){
                return true;
            }
        }
        pathvis[i]=0;
        return false;
    }

    bool isPrintable(vector<vector<int>>& grid) {
        vector<int> mir(65,(int)1e9);
        vector<int> mar(65,-1);
        vector<int> mac(65,-1);
        vector<int> mic(65,(int)1e9);
        int n = grid.size(), m = grid[0].size();
        vector<vector<int>> adj(65);
        for(int i=0;i<65;i++){
            calc(i,mir,mar,mic,mac,grid);
        }
        for(int i=0; i<65; i++){
            if(mar[i]!=-1){
                for(int j=mir[i]; j<= mar[i]; j++){
                    for(int k=mic[i]; k<= mac[i]; k++){
                        if(grid[j][k] != i){
                            adj[i].push_back(grid[j][k]);
                        }
                    }
                }
            }
        }

        for(int i=0; i<65; i++){
            if(mar[i]!=-1){
                vector<int> vis(65,0),pathvis(65,0);
                if(chk(i,vis,pathvis,adj)){return false;}
            }
        }
        return true;

    }
};
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