<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 check(vector<pair<int,int>>& v, int val, vector<vector<int>>& targetGrid) {
        int mini = v[0].first, minj = v[0].second;
        int maxi = v[0].first, maxj = v[0].second;
        for(int i=0; i<v.size(); i++) mini = min(mini, v[i].first), minj = min(minj, v[i].second), maxi = max(maxi, v[i].first), maxj = max(maxj, v[i].second);
        for(int i=mini; i<=maxi; i++) {
            for(int j=minj; j<=maxj; j++) {
                if(targetGrid[i][j] != val and targetGrid[i][j] != 0) return false;
            }
        }
        for(int i=mini; i<=maxi; i++) {
            for(int j=minj; j<=maxj; j++) {
                targetGrid[i][j] = 0;
            }
        }
        return true;
    }
    bool isPrintable(vector<vector<int>>& targetGrid) {
        int n = targetGrid.size(), m = targetGrid[0].size();
        vector<vector<pair<int,int>>> adj(61);
        for(int i=0; i<n; i++) {
            for(int j=0; j<m; j++) {
                adj[targetGrid[i][j]].push_back({i,j});
            }
        }
        vector<vector<int>> vis(n, vector<int>(m, 0));
        queue<pair<int,int>> q;
        unordered_set<int> s;
        for(int i=1; i<=60; i++) {
            if(adj[i].size()) s.insert(i);
        }
        while(s.size()) {
            unordered_set<int> next;
            for(auto it: s) {
                if(check(adj[it], it, targetGrid)) continue;
                else next.insert(it);
            }
            if(next == s) return false;
            s = next;
        }
        return true;
    }
};
23151