<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 findcycle(int node,vector<vector<int>> &adj,vector<int> &vis,vector<int> &pathvis,int n){
    vis[node]=1;
    pathvis[node]=1;
    for(auto it: adj[node]){
        if(!vis[it]){
            if(findcycle(it,adj,vis,pathvis,n))return true;
        }else if(pathvis[it])return true;
    }
    pathvis[node]=0;
    return false;
}
    bool isPrintable(vector<vector<int>>& targetGrid) {
        int n=targetGrid.size();
        int m=targetGrid[0].size();
        vector<int> left(61,INT_MAX);
        vector<int> right(61,INT_MIN);
        vector<int> up(61,INT_MAX);
        vector<int> down(61,INT_MIN);
        set<int> st;
        for(int i=0;i<n;i++){
            for(int j=0;j<m;j++){
                st.insert(targetGrid[i][j]);
               left[targetGrid[i][j]]=min(left[targetGrid[i][j]],j);
                right[targetGrid[i][j]]=max(right[targetGrid[i][j]],j);
                 down[targetGrid[i][j]]=max(down[targetGrid[i][j]],i);
                  up[targetGrid[i][j]]=min(up[targetGrid[i][j]],i);
            }
        }

        vector<vector<int>> adj(61);
       
        for(auto it: st){
            for(int i=up[it];i<=down[it];i++){
                for(int j=left[it];j<=right[it];j++){
                    if(targetGrid[i][j]!=it){
                        adj[it].push_back(targetGrid[i][j]);
                        
                    }
                }
            }
        }
       
        vector<int> vis(61,0);
        vector<int> pathvis(61,0);
        
        for(auto it:st){
            if(!vis[it]){
           if(findcycle(it,adj,vis,pathvis,n))return false;
            }
        }
        
        return true;
    }
};
41960