<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:
    int unpaint(vector<vector<int>>& targetGrid, int x1, int x2, int y1, int y2, int val)
    {
        int count=0;
        for(int i =x1; i<=x2; i++)
        {
            for(int j=y1; j<=y2; j++)
            {
                if(targetGrid[i][j]!=val && targetGrid[i][j]!=0)
                  return 0;
            }
        }
        for(int i =x1; i<=x2; i++)
        {
            for(int j=y1; j<=y2; j++)
            {
                if(targetGrid[i][j]==val)
                  {
                      targetGrid[i][j]=0;
                      count++;
                  }
            }
        }
        return count;
    }
    int calc(vector<vector<int>>& targetGrid, unordered_map<int, int>& L1, unordered_map<int, int>& L2, unordered_map<int, int> B1, unordered_map<int, int> B2)
    {
        int m = targetGrid.size();
        int n = targetGrid[0].size();
         int res=0;
        for(auto it: L1)
        {
            int col = it.first;
           res+=unpaint(targetGrid, L1[col], L2[col], B1[col], B2[col], col);
        }
        if(res==0)return false;
        return res+calc(targetGrid, L1, L2, B1, B2);
    }
    bool isPrintable(vector<vector<int>>& targetGrid) {
        unordered_map<int, int> L1;
        unordered_map<int, int> L2;
        unordered_map<int, int> B1;
        unordered_map<int, int> B2;
        int m = targetGrid.size();
        int n = targetGrid[0].size();
        for(int i=0; i<m; i++)
        {
            for(int j=0; j<n; j++)
            {
                int val = targetGrid[i][j];
                if(L1.find(val)==L1.end())
                  L1[val]=i;
               else   
                  L1[val]= min(L1[val], i);
                L2[val]= max(L2[val], i);
                if(B1.find(val)==L1.end())
                  B1[val]=j;
               else   
                B1[val]= min(B1[val], j);
                B2[val]= max(B2[val], j);
            }
        }  
      

        return calc(targetGrid, L1, L2, B1, B2) == m*n; 
    }
};
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