<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:
    set<pair<int,int>> order;

    void dfs(vector<vector<int>>& grid, int i,int j, int sx,int sy, int ex,int ey,int val,vector<vector<bool>>& vis){
        vis[i][j] = true;
        if(grid[i][j] != val){
            order.insert({val,grid[i][j]});
        }
        int dx[] = {1,-1,0,0};
        int dy[] = {0,0,1,-1};

        for(int k=0;k<4;k++){
            int ni = i + dx[k];
            int nj = j + dy[k];
            if(ni >= sx && ni <= ex && nj >= sy && nj <= ey && !vis[ni][nj]){
                dfs(grid,ni,nj,sx,sy,ex,ey,val,vis);
            }
        }
        return;
    }


    bool isPrintable(vector<vector<int>>& grid) {
        int n = grid.size(),m= grid[0].size();
        map<int,int> minx,miny,maxx,maxy;
        for(int i=0;i<n;i++){
            for(int j=0;j<m;j++){
                int val = grid[i][j];
                (minx.count(val))? minx[val] = min(minx[val],i) : minx[val] = i;
                (miny.count(val))? miny[val] = min(miny[val],j) : miny[val] = j;
                (maxx.count(val))? maxx[val] = max(maxx[val],i) : maxx[val] = i;
                (maxy.count(val))? maxy[val] = max(maxy[val],j) : maxy[val] = j;
            }
        }

        int maxnum = -1,cnt=0;
        set<int> st;
        for(auto it : minx){
            int val = it.first;
            st.insert(val);
            maxnum = val;
            cnt++;
            int sx = it.second,sy=miny[val];
            int ex = maxx[val],ey = maxy[val];
            vector<vector<bool>> vis(n,vector<bool>(m,false));
            dfs(grid,sx,sy,sx,sy,ex,ey,val,vis);
        }
        vector<vector<int>> adj(maxnum+1);
        vector<int> indegree(maxnum+1,0);
        // cout << "Order : " << endl;
        for(auto it : order){
            int u = it.first,v = it.second;
            // cout << it.first << " " << it.second << endl;
            adj[u].push_back(v);
            indegree[v]++;
        }
        queue<int> q;
        for(int i=1;i<=maxnum;i++) {
            if(indegree[i] == 0 && st.count(i)) q.push(i);
        }
        vector<int> res;
        while(!q.empty()){
            int node = q.front();q.pop();
            res.push_back(node);
            for(auto it : adj[node]){
                indegree[it]--;
                if(indegree[it] == 0) q.push(it);
            }
        }
        for(auto it : res) cout << it << " ";
        cout << endl;

        return res.size()==cnt;


    }
};
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