<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:
    def isPrintable(self, targetGrid: List[List[int]]) -> bool:
        # observation: paint is applied in rectangles
        # if some other colours are in that rectangle, then they must be applied on top of it (overlapped)
        # recursively, then that colour was applied in a rectangle
        # so long as an ancestor color doesnt appear (cycle), then paint order
        # is possible

        # find rectangles for each colour
        # do dfs using visit set to find cycle
        # if no cycle, grid is valid, return True

        # [0,0,1,0]
        # [0,2,2,2]
        # [3,2,3,2]
        # [0,2,2,2]
        # [0,0,1,0]

        rectangle = defaultdict(lambda: [60, 60, 0, 0])
        for i in range(len(targetGrid)):
            for j in range(len(targetGrid[0])):
                # top, left, bottom, right
                # if targetGrid[i][j] not in rectangle:
                #     rectangle[targetGrid[i][j]] = [60,60,0,0]
                r = rectangle[targetGrid[i][j]]
                r[0] = min(r[0], i)
                r[1] = min(r[1], j)
                r[2] = max(r[2], i)
                r[3] = max(r[3], j)

        cycle = set()
        visit = set()

        def dfs(colour):
            others = set()
            r = rectangle[colour]
            for i in range(r[0], r[2] + 1):
                for j in range(r[1], r[3] + 1):
                    if targetGrid[i][j] != colour:
                        others.add(targetGrid[i][j])

            cycle.add(colour)
            visit.add(colour)
            for c in others:
                if c in cycle:
                    return False

                elif c in visit:
                    continue
                if not dfs(c):
                    return False
            # consider [2,1,2]
            # if we mark 1 in cycle, then dfs(2) will return False
            cycle.remove(colour)
            return True

        for i in rectangle:
            cycle = set()
            if i not in visit:
                visit.add(i)
                if not dfs(i):
                    return False
        return True



                    
                    
                    

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