<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) {
        int m = targetGrid.size(), n = targetGrid[0].size();
        // colorCordinates[color] = {start_row, end_row, start_col, end_col}
        vector<vector<int>> colorCoordinates(61, {65, -1, 65, -1});
        
        for (int row = 0; row < m; ++row) {
            for (int col = 0; col < n; ++col) {
                int color = targetGrid[row][col];
                colorCoordinates[color][0] = min(colorCoordinates[color][0], row);
                colorCoordinates[color][1] = max(colorCoordinates[color][1], row);
                colorCoordinates[color][2] = min(colorCoordinates[color][2], col);
                colorCoordinates[color][3] = max(colorCoordinates[color][3], col);
            }
        }

        vector<vector<int>> removeColors(61);
        vector<int> vis(61, 0), dfsVis(61, 0);

        for (int color = 1; color <= 60; color++) {
            if (colorCoordinates[color][0] == 65) continue;
            for (int row = colorCoordinates[color][0]; row <= colorCoordinates[color][1]; row++) {
                for (int col = colorCoordinates[color][2]; col <= colorCoordinates[color][3]; col++) {
                    if (targetGrid[row][col] != color) {
                        removeColors[color].push_back(targetGrid[row][col]);
                    }
                }
            }
        }

        
        function<bool(int)> dfs = [&] (int curColor) {
            dfsVis[curColor] = vis[curColor] = 1;
            for (int nextColor : removeColors[curColor]) {
                if (dfsVis[nextColor]) {
                    return false;
                }
                if (vis[nextColor]) {
                    continue;
                }
                if (!dfs(nextColor)) {
                    return false;
                }
            }
            dfsVis[curColor] = 0;
            return true;
        };

        for (int color = 1; color <= 60; color++) {
            if (colorCoordinates[color][0] == 65) continue;
            if (!dfs(color)) {
                return false;
            }
        }
        return true;
    }
};
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