<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) {
        const int m = targetGrid.size(), n = targetGrid[0].size();
        // Dependency graph: dep[i] = j means color j covers color i;
        vector<unordered_set<int>> dep(61);
        for (int i = 1; i <= 60; ++i) {
            // Determine the rectangle of the current color i.
            int minx = m, miny = n, maxx = -1, maxy = -1;
            for (int x = 0; x < m; ++x) {
                for (int y = 0; y < n; ++y) {
                    if (targetGrid[x][y] == i) {
                        minx = min(x, minx);
                        miny = min(y, miny);
                        maxx = max(x, maxx);
                        maxy = max(y, maxy);
                    }
                }
            }
            // Add any color covering the current color as dependence.
            for (int tx = minx; tx <= maxx; ++tx) {
                for (int ty = miny; ty <= maxy; ++ty) {
                   if (targetGrid[tx][ty] != i) dep[i].insert(targetGrid[tx][ty]);
                }
            }
        }
        // Standard DFS to check the existence of a cycle.
        vector<int> vis(61, 0);
        for (int i = 1; i <= 60; ++i) {
            if (!vis[i] && hasCircle(i, dep, vis)) return false;
        }
        return true;
    }
    
private:
    bool hasCircle(int curr, const vector<unordered_set<int>>& dep, vector<int>& vis) {
        vis[curr] = 1;
        for (const int d : dep[curr]) {
            if (vis[d] == 2) continue;
            if (vis[d] == 1) return true;
            if (hasCircle(d, dep, vis)) return true;
        }
        vis[curr] = 2;
        return false;
    }
};
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