<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 dfs(int node, vector<set<int>> &adj, vector<int> &vis) {
        vis[node] = 1;
        for(auto &v: adj[node]) {
            if(vis[v]==1 || dfs(v, adj, vis)) {
                // cout << node << " : " << v << endl;
                return true;
            }
        }
        vis[node] = 2;
        return false;
    }

    bool isPrintable(vector<vector<int>>& targetGrid) {
        int n = targetGrid.size(), m = targetGrid[0].size();
        map<int,int> l, r, u, d;
        for(int i = 0; i < n; i++) for(int j = 0; j < m; j++) {
            int num = targetGrid[i][j];
            if(l.count(num)==0) {
                l[num] = j, r[num] = j;
                u[num] = i, d[num] = i;
            }
            else {
                l[num] = min(l[num], j); r[num] = max(r[num], j);
                u[num] = min(u[num], i); d[num] = max(d[num], i);
            }
        }
        vector<set<int>> adj(61);
        for(auto &p: l) {
            int i1 = u[p.first], i2 = d[p.first], j1 = l[p.first], j2 = r[p.first];
            // cout << p.first << " => " << i1 << " : " << j1 << " || " << i2 << " : " << j2 << endl;
            for(int i = i1; i <= i2; i++) for(int j = j1; j <= j2; j++) {
                if(targetGrid[i][j]!=p.first) adj[p.first].insert(targetGrid[i][j]);
            }
        }
        vector<int> vis(61, 0);
        for(int i = 1; i <= 60; i++) {
            if(vis[i]) continue;
            if(dfs(i, adj, vis)) return false;
        }
        return true;
    }
};
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