<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 n = targetGrid.size(), m = targetGrid[0].size();
        unordered_map<int, vector<int>> mp;
        for(int i = 0; i < n; i++) {
            for(int j = 0; j < m; j++) {
                if(mp.find(targetGrid[i][j]) == mp.end()) mp[targetGrid[i][j]] = {(int)1e8, (int)1e8, (int)-1e8, (int)-1e8};
                mp[targetGrid[i][j]][0] = min(mp[targetGrid[i][j]][0], i);
                mp[targetGrid[i][j]][1] = min(mp[targetGrid[i][j]][1], j);
                mp[targetGrid[i][j]][2] = max(mp[targetGrid[i][j]][2], i);
                mp[targetGrid[i][j]][3] = max(mp[targetGrid[i][j]][3], j);
            }
        }
        vector<vector<int>> adj(61);
        vector<int> indeg(61, 0);
        for(auto it : mp) {
            int node = it.first;
            for(int i = mp[node][0]; i <= mp[node][2]; i++) {
                for(int j = mp[node][1]; j <= mp[node][3]; j++) {
                    if(targetGrid[i][j] != node) {
                        auto it = find(adj[targetGrid[i][j]].begin(), adj[targetGrid[i][j]].end(), node);
                        if(it == adj[targetGrid[i][j]].end()) {
                            adj[targetGrid[i][j]].push_back(node);
                            indeg[node]++;
                        }
                    }
                }
            }
        }
        // for(auto it : mp) {
        //     int node = it.first;
        //     cout << node << " : ";
        //     for(auto v : adj[node]) {
        //         cout << v << " ";
        //     }
        //     cout << endl;
        // }
        queue<int> q;
        for(auto it : mp) {
            if(indeg[it.first] == 0) {
                q.push(it.first);
            }   
        }
        while(!q.empty()) {
            int u = q.front(); q.pop();
            for(int v : adj[u]) {
                indeg[v]--;
                if(indeg[v] == 0) q.push(v);
            }
        }
        for(int in : indeg) {
            if(in != 0) return false;
        }
        return true;
    }
};
20858