<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();
        unordered_map<int, vector<int>> border; // t, b, l, r
        for(int i=0; i<m; i++){
            for(int j=0; j<n; j++){
                int c = targetGrid[i][j];
                if(border.find(c) == border.end()){
                    border[c] = {i, i, j, j};
                }
                else{
                    border[c] = {min(border[c][0], i), max(border[c][1], i), min(border[c][2], j), max(border[c][3], j)};
                }
            }
        }
        unordered_map<int, vector<int>> graph;
        unordered_map<int, int> indegree;
        for(auto it: border){
            int t = it.second[0];
            int b = it.second[1];
            int l = it.second[2];
            int r = it.second[3];
            for(int i=t; i<=b; i++){
                for(int j=l; j<=r; j++){
                    int c = targetGrid[i][j];
                    if(c!=it.first){
                        graph[it.first].push_back(c);
                        indegree[c]++;
                    }
                }
            }
        }

        queue<int> q;
        for(auto it: border){
            if(indegree[it.first]==0)
            q.push(it.first);
        }
        int cnt = 0;
        while(!q.empty()){
            int c = q.front();
            q.pop();
            cnt++;
            for(auto it: graph[c]){
                indegree[it]--;
                if(indegree[it]==0) q.push(it);
            }
        }
        return cnt==border.size();
    }
};
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