<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();
        int m = targetGrid[0].size();
        vector<vector<int>>adj(61);

             // 61 stuff because total number of colors can be 60
        vector<int>row_min(61,n),row_max(61,-1),col_min(61,m),col_max(61,-1);

            //Getting the Corners of every colors
        for(int i = 0 ; i < n ; i++){
            for(int j = 0 ; j < m ; j++){
                
                int x = targetGrid[i][j];       

                row_min[x] = min(row_min[x],i);
                row_max[x] = max(row_max[x],i);
                col_min[x] = min(col_min[x],j);
                col_max[x] = max(col_max[x],j);
            }
        }
        
            // Generating my Indegree array for Toposort
        vector<int>inDegree(61);
        for(int color = 1 ; color < 61 ; color++){
            for(int i = row_min[color] ; i <= row_max[color] ; i++){
                for(int j = col_min[color] ; j <= col_max[color] ; j++){
                    int x = targetGrid[i][j];
                    if(x != color){
                        adj[color].push_back(x);
                        inDegree[x]++;
                    }
                }
            }
        }

            // Simple TopoSort Algorithm
        queue<int>q;
        for(int color = 1 ; color < 61 ; color++){
            if(inDegree[color]==0) q.push(color);
        }

        int cnt =0;
        while(!q.empty()){
            int node=q.front();
            q.pop();
            cnt++;
            for(auto it :adj[node]){
                inDegree[it]--;
                if(inDegree[it]==0) q.push(it);
            }
        }
        if(cnt!=60) return false;
    
        return true;
    }
};
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