<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>>& grid) {
        vector<int> max_i(61, -1), min_i(61, 1e9);
        vector<int> max_j(61, -1), min_j(61, 1e9);

        int n = grid.size(), m = grid[0].size();
        for (int i = 0; i < n; i++)
            for (int j = 0; j < m; j++) {
                int x = grid[i][j];

                max_i[x] = max(max_i[x], i);
                min_i[x] = min(min_i[x], i);

                max_j[x] = max(max_j[x], j);
                min_j[x] = min(min_j[x], j);
            }
        vector<int> adj[61], deg(61);
        for (int c = 1; c <= 60; c++)
            if (max_i[c] >= 0) {
                long long mask = 0;
                for (int i = min_i[c]; i <= max_i[c]; i++) {
                    for (int j = min_j[c]; j <= max_j[c]; j++) {
                        if (grid[i][j] == c)
                            continue;

                        if (mask & (1ll << grid[i][j]))
                            continue;

                        mask |= (1ll << grid[i][j]);
                        adj[grid[i][j]].push_back(c);
                        deg[c]++;
                    }
                }
            }
        
        queue<int> q;
        for(int i = 1; i <= 60; i++)
            if(deg[i] == 0)
                q.push(i);
        
        int taken = 0;
        while(!q.empty()) {
            int u = q.front();
            q.pop();
            taken++;

            for(int v: adj[u]) {
                deg[v]--;
                if(deg[v] == 0)
                    q.push(v);
            }
        }

        return taken == 60;
    }
};
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