<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 isPrintable(vector<vector<int>>& a) {
        unordered_map<int, int> mnx, mxx, mny, mxy;
        int n = a.size();
        int m = a[0].size();
        
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                int color = a[i][j];
                if (mnx.find(color) == mnx.end()) {
                    mnx[color] = i;
                    mxx[color] = i;
                    mny[color] = j;
                    mxy[color] = j;
                } else {
                    mnx[color] = min(mnx[color], i);
                    mxx[color] = max(mxx[color], i);
                    mny[color] = min(mny[color], j);
                    mxy[color] = max(mxy[color], j);
                }
            }
        }
        vector<vector<int>> l(61);

        for (auto& c : mnx) {
            int col = c.first;
            int l_x = mnx[col];
            int r_x = mxx[col];
            int u_y = mny[col];
            int d_y = mxy[col];

            for (int i = l_x; i <= r_x; i++) {
                for (int j = u_y; j <= d_y; j++) {
                    if (a[i][j] != col) {
                        l[col].push_back(a[i][j]);
                    }
                }
            }
        }

        vector<int> visited(61, 0);
        vector<int> backtrack(61, 0);

        function<bool(int)> dfs = [&](int i) {
            if (backtrack[i]) return true;
            if (visited[i]) return false;
            visited[i] = 1;
            backtrack[i] = 1;
            bool flag = false;
            for (auto& child : l[i]) {
                flag |= dfs(child);
            }
            backtrack[i] = 0;
            return flag;
        };

        for (auto& c : mnx) {
            if (!visited[c.first]) {
                if (dfs(c.first)) return false;
            }
        }

        return true;
    }
};

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