<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>

static const int _ = [] {
    cin.tie(0);
    ios::sync_with_stdio(0);
    return 0;
}();

class Solution {
public:
    bool isPrintable(vector<vector<int>>& targetGrid) {
        int m = targetGrid.size(), n = targetGrid[0].size();
        vector<int> minRow(61, m), maxRow(61, -1), minCol(61, n),
            maxCol(61, -1);

        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                int color = targetGrid[i][j];
                minRow[color] = min(minRow[color], i);
                maxRow[color] = max(maxRow[color], i);
                minCol[color] = min(minCol[color], j);
                maxCol[color] = max(maxCol[color], j);
            }
        }

        vector<unordered_set<int>> adj(61);
        vector<int> indegree(61, 0);

        for (int color = 1; color <= 60; ++color) {
            if (maxRow[color] == -1)
                continue; // Color does not exist

            for (int i = minRow[color]; i <= maxRow[color]; ++i) {
                for (int j = minCol[color]; j <= maxCol[color]; ++j) {
                    int otherColor = targetGrid[i][j];
                    if (otherColor != color &&
                        adj[color].find(otherColor) == adj[color].end()) {
                        adj[color].insert(otherColor);
                        indegree[otherColor]++;
                    }
                }
            }
        }

        queue<int> q;
        for (int color = 1; color <= 60; ++color) {
            if (indegree[color] == 0 && maxRow[color] != -1) {
                q.push(color);
            }
        }

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

        return printedColors == count_if(maxRow.begin(), maxRow.end(),
                                         [](int val) { return val != -1; });
    }
};
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