<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>>& tg) {
        int n = tg.size();
        int m = tg[0].size();
        vector<array<int,4>>border(62, {INT_MAX,INT_MAX,INT_MIN,INT_MIN});
        for(int i=0 ; i<n ; i++)
        {
            for(int j=0 ; j<m ; j++)
            {
                int colour = tg[i][j];
                border[colour][0] = min(border[colour][0], i);
                border[colour][1] = min(border[colour][1], j);
                border[colour][2] = max(border[colour][2], i);
                border[colour][3] = max(border[colour][3], j);
            }
        }
        vector<int>adj[62];
        //graph formation
        vector<int>indeg(62);
        for(int col=0 ; col<61 ; col++)
        {
            for(int i=border[col][0] ; i<= border[col][2] ; i++)
            {
               for(int j=border[col][1] ; j<=border[col][3] ; j++)
                {
                    if(tg[i][j] != col)
                    {
                        adj[col].push_back(tg[i][j]);
                        indeg[tg[i][j]]++;
                    }
                } 
            }
        }
        queue<int>q;
        for(int i=0 ; i<61 ; i++)
        {
            if(indeg[i] == 0)q.push(i);
        }
        int topo=0;
        while(!q.empty())
        {
            int node = q.front();
            q.pop();
            topo++;
            for(auto it : adj[node])
            {
                if(!(--indeg[it]))q.push(it);
            }
        }
        return topo == 61;
    }
};
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