<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<int>graph[61] ; 
        int r = targetGrid.size() ;
        int c = targetGrid[0].size() ;
        vector<int>r1(61 , -1),r2(61 , -1),c1(61 ,-1),c2(61,-1);
        for(int i = 0 ; i < r ; i++){
            for(int j = 0 ; j < c ; j++){
                if(r1[targetGrid[i][j]] == -1){
                    r1[targetGrid[i][j]] = i;
                }
                r2[targetGrid[i][j]] = i ; 
            }
        }
        for(int j = 0 ; j < c ; j++){
            for(int i = 0 ;  i < r ; i++){
                 if(c1[targetGrid[i][j]] == -1){
                    c1[targetGrid[i][j]] = j;
                }
                c2[targetGrid[i][j]] = j ; 
            }
        }
        map<pair<int,int>,bool>seen ; 
        vector<int>indeg(61,0);
      for(int k = 1 ; k<=60 ; k++){
        if(c1[k] == -1)continue;
        for(int i = r1[k] ; i<=r2[k] ; i++){
            for(int j = c1[k] ; j<=c2[k] ; j++){
                if(targetGrid[i][j] != k and seen[{k ,targetGrid[i][j]}] == 0){
                    seen[{k,targetGrid[i][j]}] = 1;
                    graph[k].push_back(targetGrid[i][j]) ;
                    indeg[targetGrid[i][j]]++; 
                }
            }
        }
      }
      queue<int>q;
      vector<int>topo ; 
    //  for(int i = 1 ; i<=5 ; i++){
    //     for(auto x : graph[i]){
    //         cout<<i<<" "<<x<<'\n';
    //     }
    //  }
      for(int i = 1 ; i<=60 ; i++){
        if(indeg[i] == 0)q.push(i) ; 
      }
      while(q.size()){
        int f = q.front() ;
        topo.push_back(f);
        q.pop() ;
        for(auto x  : graph[f]){
            indeg[x]--;
            if(indeg[x] == 0)q.push(x);
        }
      }

      return (topo.size() == 60) ; 
    }
};
28198