<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>>& v) {
        // Bounds for each color (min row, min col, max row, max col)
        vector<vector<int>> w(61, {INT_MAX, INT_MAX, INT_MIN, INT_MIN});
        int n = v.size();
        int m = v[0].size();
        
        // Find the bounding rectangle for each color
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                int a = v[i][j];
                w[a][0] = min(w[a][0], i); // min row
                w[a][1] = min(w[a][1], j); // min col
                w[a][2] = max(w[a][2], i); // max row
                w[a][3] = max(w[a][3], j); // max col
            }
        }

        vector<vector<int>> g(61); // Graph adjacency list
        vector<int> ind(61, 0); // In-degree array
        
        // Build the graph based on dependencies between colors
        for (int i = 1; i <= 60; i++) {
            int sx = w[i][0];
            int sy = w[i][1];
            int ex = w[i][2];
            int ey = w[i][3];
            for (int x = sx; x <= ex; x++) {
                for (int y = sy; y <= ey; y++) { // Corrected to loop from sy to ey
                    if (v[x][y] != i) { // If the cell contains a different color
                        g[v[x][y]].push_back(i); // Color v[x][y] must be printed before color i
                        ind[i]++; // Increase in-degree of color i
                    }
                }
            }
        }

        // Topological sort using BFS (Kahn's algorithm)
        queue<int> q;
        vector<int> res;
        
        // Initialize the queue with colors having zero in-degree
        for (int i = 1; i <= 60; i++) {
            if (ind[i] == 0) {
                q.push(i);
            }
        }
        
        while (!q.empty()) {
            int node = q.front(); q.pop();
            res.push_back(node);
            for (int fnode : g[node]) {
                ind[fnode]--;
                if (ind[fnode] == 0) {
                    q.push(fnode);
                }
            }
        }

        // If all colors are processed, return true, otherwise false
        // Check if we were able to process all colors that were used
        for (int i = 1; i <= 60; i++) {
            if (ind[i] != 0) return false; // If any color is left unprocessed, return false
        }

        return true;
    }
};

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