<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> min_x(61, INT_MAX);
        vector<int> min_y(61, INT_MAX);
        vector<int> max_x(61, INT_MIN);
        vector<int> max_y(61, INT_MIN);

        // Find min and max bounds for each color
        for(int i = 0; i < targetGrid.size(); i++){
            for(int j = 0; j < targetGrid[0].size(); j++){
                int color = targetGrid[i][j];
                
                min_x[color] = min(min_x[color], i);
                min_y[color] = min(min_y[color], j);
                max_x[color] = max(max_x[color], i);
                max_y[color] = max(max_y[color], j);
            }
        }

        vector<int> Indegree(61, 0);
        vector<int> adj[61];

        // Create adjacency list and calculate indegrees
        for(int i = 1; i <= 60; i++){
            if(min_x[i] == INT_MAX) continue; // Skip unused colors
            for(int x = min_x[i]; x <= max_x[i]; x++){
                for(int y = min_y[i]; y <= max_y[i]; y++){
                    int node = targetGrid[x][y];
                    if(i != node){
                        Indegree[i]++;
                        adj[node].push_back(i);
                    }
                }
            }
        }

        // Topological sort using BFS
        queue<int> q;
        for(int i = 1; i <= 60; i++){
            if(min_x[i] != INT_MAX && Indegree[i] == 0){
                q.push(i);
            }
        }

        int cnt = 0;
        while(!q.empty()){
            int node = q.front();
            q.pop();
            cnt++;
            for(int neighbor : adj[node]){
                Indegree[neighbor]--;
                if(Indegree[neighbor] == 0){
                    q.push(neighbor);
                }
            }
        }

        // Check if all used colors were processed
        int totalColors = 0;
        for(int i = 1; i <= 60; i++){
            if(min_x[i] != INT_MAX){
                totalColors++;
            }
        }

        return cnt == totalColors;
    }
};

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