<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:
    void dfs(int ind,vector<vector<int>>&v,vector<int>&visit,bool & ch)
    {
        visit[ind] = 1;
        for(auto &k1 : v[ind])
        {
            if(!visit[k1])
            dfs(k1,v,visit,ch);
            else if(visit[k1] == 1) {ch = 1;}
        }
        visit[ind] = 2;
    }
    bool isPrintable(vector<vector<int>>& targetGrid) {
        vector<vector<int>>v = targetGrid;
        int n= targetGrid.size();
        int m = targetGrid[0].size();
        map<int,vector<int>>m1;
        for(int i = 0;i<n;i++)
        {
            for(int j= 0;j<m;j++)
            {
                if(m1.find(v[i][j]) == m1.end())
                {
                    m1[v[i][j]] = {i,i,j,j};
                }
                else
                {auto &p1 = m1[v[i][j]];
                p1[0] = min(p1[0],i);
                p1[1] = max(p1[1],i);
                p1[2] = min(p1[2],j);
                p1[3] = max(p1[3],j);}
            }
        }
        vector<vector<int>>v1(61);
        for(auto &p1 : m1)
        {
            for(int i = p1.second[0];i<=p1.second[1];i++)
            {
                for(int j = p1.second[2];j<=p1.second[3];j++)
                {
                    if(v[i][j] != p1.first)
                    {
                        v1[v[i][j]].push_back(p1.first);
                    }
                }
            }
        }
        vector<int>visit(61,0);
        bool ch = 0;
        for(int i = 1;i<=60;i++)
        {
            dfs(i,v1,visit,ch);
            if(ch) return 0;
        }
        return 1;
    }
};
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