<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 u, vector<int> &vis, vector<int> &dfsVis, vector<vector<int>> &adj){
        vis[u] = 1;
        dfsVis[u] = 1;
        bool ans = false;

        for(auto &v : adj[u]){
            if(!vis[v]){
                ans |= dfs(v, vis, dfsVis, adj);
            }
            else if(dfsVis[v]){
                return true;
            }
        }

        dfsVis[u] = 0;
        return ans;
    }

    bool hasCycle(int n, vector<vector<int>> &adj){
        vector<int> vis(n+1, 0);
        bool ans = false;
        for(int i=1; i<n+1; i++){
            if(!vis[i]){
                vector<int> dfsVis(n+1, 0);
                ans |= dfs(i, vis, dfsVis, adj);
            }
        }

        return ans;
    }

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

        unordered_map<int, int> minRow, maxRow, minCol, maxCol;
        vector<int> colors;
        int mxN = 0;
        for(int i=0; i<m; i++){
            for(int j=0; j<n; j++){
                int cur = targetGrid[i][j];
                if(minRow.find(cur) == minRow.end()){
                    mxN = max(mxN, cur);
                    colors.push_back(cur);
                    minRow[cur] = i;
                }
                else if(minRow[cur] > i){
                    minRow[cur] = i;
                }

                if(minCol.find(cur) == minCol.end()){
                    minCol[cur] = j;
                }
                else if(minCol[cur] > j){
                    minCol[cur] = j;
                }

                if(maxRow.find(cur) == maxRow.end()){
                    maxRow[cur] = i;
                }
                else{
                    maxRow[cur] = max(maxRow[cur], i);
                }

                if(maxCol.find(cur) == maxCol.end()){
                    maxCol[cur] = j;
                }
                else{
                    maxCol[cur] = max(maxCol[cur], j);
                }
            }
        }

        vector<vector<int>> adj(mxN+1);
        for(auto &c : colors){
            int r1 = minRow[c], r2 = maxRow[c], c1 = minCol[c], c2 = maxCol[c];

            for(int i=r1; i<=r2; i++){
                for(int j=c1; j<=c2; j++){
                    if(targetGrid[i][j] != c){
                        adj[c].push_back(targetGrid[i][j]);
                    }
                }
            }
        }

        if(hasCycle(mxN, adj)){
            return false;
        }
        else{
            return true;
        }
    }
};
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