<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:
    #define fr(i,n) for(int i=0;i<n;i++)
    #define vi vector<int>
    #define vvi vector<vi>

    bool vis[61] , pathvis[61];
    int l[61] , r[61] , u[61] , d[61];
    bool cycle = 0;

    void dfs(int col , vvi &a){
        vis[col] = 1;
        pathvis[col] = 1;

        for(int i=u[col] ; i<=d[col] ; i++){
            for(int j=l[col] ; j<=r[col] ; j++){
                if(a[i][j] != col){

                    if(pathvis[a[i][j]] == 1){
                        // cycle!
                        cycle = 1;
                        return;
                    }

                    if(vis[a[i][j]] == 1)continue; // did not form cycle earlier gtg

                    dfs(a[i][j] , a);
                }
            }
        }
    
        pathvis[col] = 0;
    }   

    bool isPrintable(vector<vector<int>>& a) {
        // hard implem, easy idea : work backwards
        int n = a.size();
        int m = a[0].size();
        
        cycle = 0;

        memset(vis , 0 , sizeof(vis));
        memset(pathvis , 0 , sizeof(pathvis));
        memset(l , 70 , sizeof(l));
        memset(r , -1 , sizeof(r));
        memset(d , -1 , sizeof(d));
        memset(u , 70 , sizeof(u));
        
        fr(i,n){
            fr(j,m){
                l[a[i][j]] = min(l[a[i][j]] , j);
                r[a[i][j]] = max(r[a[i][j]] , j);
                d[a[i][j]] = max(d[a[i][j]] , i);
                u[a[i][j]] = min(u[a[i][j]] , i);
            }
        }

        // need to do the operations in toposort order

        for(int col=1;col<=60;col++){
            if(vis[col] == 0 && l[col] != -1){
                dfs(col , a);
                if(cycle == 1)return false;
            }
        }
        return true;
    }
};
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