<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 col,int arr[][4],vector<vector<int>> grid,int* visited){
        visited[col]=1;
        int t=arr[col][0];
        int r=arr[col][1];
        int b=arr[col][2];
        int l=arr[col][3];
        for(int i=t;i<=b;i++){
            for(int j=l;j<=r;j++){
                int x=grid[i][j];
                if(x==col)  continue;
                if(visited[x]==1) return false;
                bool ans=1;
                if(visited[x]==0){
                    ans=dfs(x,arr,grid,visited);
                }
                if(ans==0) return 0;
            }
        }
        visited[col]=2;
        return true;
    }

    bool isPrintable(vector<vector<int>>& grid) {
        int m = grid.size();
        int n = grid[0].size();
        int arr[61][4];
        int visited[61];
        memset(visited,-1,sizeof(visited));
        for(int i=0;i<m;i++){
            for(int j=0;j<n;j++){
                int col=grid[i][j];
                if(visited[col]==-1){   
                    visited[col]=0;
                    arr[col][0]=i;
                    arr[col][1]=j;
                    arr[col][2]=i;
                    arr[col][3]=j;
                }
                else{  
                    arr[col][0]=min(i,arr[col][0]);
                    arr[col][1]=max(j,arr[col][1]);
                    arr[col][2]=max(i,arr[col][2]);
                    arr[col][3]=min(j,arr[col][3]);
                }
            }
        }
        bool ans=1;
        for(int i=0;i<61;i++){
            if(visited[i]==-1) continue;
            if(visited[i]==0){
                ans=dfs(i,arr,grid,visited);
                if(ans==0) return false;
            }
        }
        return 1;
    }
};
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