<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) {
        vector<set<int>>vec(61);
        int n=targetGrid.size(),m=targetGrid[0].size();

        for(int val=1; val<=60; val++){
            int mnx=n,mny=m,mxx=-1,mxy=-1;
            for(int i=0; i<n; i++){
                for(int j=0; j<m; j++){
                    if(targetGrid[i][j]==val){
                        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(targetGrid[i][j]!=val){
                        vec[val].insert(targetGrid[i][j]);
                    }
                }
            }
        }
        vector<int>vis(61,0); 
        for(int i=1; i<=60; i++){
            if(!vis[i]){ 
                int val=findCycle(i,vec,vis);
                if(val)
                    return false;
            }
        }
        return true;    
    }

    bool findCycle(int idx, vector<set<int>>&vec, vector<int>& vis) {
        // cout<<idx<<" "<<endl;
        vis[idx]=true;
        for(auto x: vec[idx]) {
            if(!vis[x] and findCycle(x,vec,vis)) return true;
            else if(vis[x]==1) return true;
        }
        vis[idx]=2;
        return false;
    }
    
};


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