<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>

/*
step = 1 find the lowest starting row,column,ending row,column for all  possible color with grid
stpe = 2 find the all grid that i can remove from the matrix by the help of erase by checking for each color max that we can erase
step = 3 if cannot erasable for push in next and check for curr there make a new queue for next 
tc - O(k * n * m)

*/

class Solution {
public:
    int n,m;

    bool erase(vector<vector<int>>& targetGrid,int color,vector<int>&coord){ 
        
        for(int i = coord[0] ; i <= coord[2] ; i ++ ){ 
            for(int j = coord[1] ; j <= coord[3] ; j ++ ){ 
                if(targetGrid[i][j] != 0 && targetGrid[i][j] != color){ 
                    return false;
                }
            }
        }

        for(int i = coord[0] ; i <= coord[2] ; i ++ ){ 
            for(int j = coord[1] ; j <= coord[3] ; j ++ ){ 
                targetGrid[i][j] = 0;
            }
        }

        return true;
    }


    bool isPrintable(vector<vector<int>>& targetGrid) {
        
        unordered_map<int,vector<int>>mp;
        n = targetGrid.size();
        m = targetGrid[0].size();


        for(int i = 0 ; i < n ; i ++ ){ 
            for(int j = 0 ; j < m ; j ++ ){ 
                int color = targetGrid[i][j];
                vector<int>coord;
                if(mp.find(color) == mp.end()){ 
                    coord.resize(4,0);
                    coord[0] = i;
                    coord[1] = j;
                    coord[2] = i;
                    coord[3] = j;
                    mp[color] = coord;
                    continue;
                }

                coord = mp[color];
                coord[0] = min(coord[0],i);
                coord[1] = min(coord[1],j);
                coord[2] = max(coord[2],i);
                coord[3] = max(coord[3],j);
                mp[color] = coord;
            }
        }


        vector<int>color;
        for(auto clr : mp){ 
            color.push_back(clr.first);
        }

        vector<int>next;
        while(color.size() != 0){ 
            for(auto clr : color){ 
                bool erases = erase(targetGrid,clr,mp[clr]);
                if(!erases){ 
                    next.push_back(clr);
                }
            }

            if(next.size() == color.size()) return false;

            color = next;
            next.clear();
        }

        return true;
    }
    
};
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