<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();
        int n = targetGrid[0].size();

        int left[61];
        int top[61];
        int bottom[61];
        int right[61];
        for(int i=0; i<61; i++){
            left[i] = n;
            top[i] = m;
            bottom[i] = -1;
            right[i] = -1;
        }

        for(int r=0; r<m; r++){
            for(int c=0; c<n; c++){
                int color = targetGrid[r][c];
                left[color] = min(left[color], c);
                right[color] = max(right[color], c);
                top[color] = min(top[color], r);
                bottom[color] = max(bottom[color], r);
            }
        }
        set<int> colors;
        vector<int> g[61];
        int indegree[61];
        memset(indegree, 0, sizeof(indegree));
        for(int r=0; r<m; r++){
            for(int c=0; c<n; c++){
                int cur_color = targetGrid[r][c];
                colors.insert(cur_color);
                for(int color=1; color<=60; color++){
                    if(cur_color!=color && left[color]<=c && c<=right[color] && top[color]<=r && r<=bottom[color]){
                        g[color].push_back(cur_color);
                        indegree[cur_color]++;
                    }
                }
            }
        }

        int rest = colors.size();
        queue<int> q;
        for(auto color: colors) if(indegree[color]==0) q.push(color);
        while(!q.empty()){
            auto cur = q.front(); q.pop();
            rest--;
            for(auto neighbor: g[cur]){
                indegree[neighbor]--;
                if(indegree[neighbor]==0) q.push(neighbor);
            }
        }
        return rest==0;
    }
};
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