<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>>& targetGrid) {
      int n = targetGrid.size();
      int m = targetGrid[0].size();
      unordered_map<int, vector<int>> mp;
      for (int i = 0; i < n; i++) {
        for (int j = 0; j < m; j++) {
          int tar = targetGrid[i][j];
          if (!mp.count(tar)) {
            mp[tar] = {i, j, i, j};
          } else {
            vector<int> &a = mp[tar];
            a[0] = min(a[0], i);
            a[1] = min(a[1], j);
            a[2] = max(a[2], i);
            a[3] = max(a[3], j);
          }
        }
      }
      unordered_map<int, vector<int>> graph;
      unordered_map<int, int> indegree;
      for (auto &[val, v] : mp) {
        indegree[val] = 0;
        for (int i = v[0]; i <= v[2]; i++) {
          for (int j = v[1]; j <= v[3]; j++) {
            if (targetGrid[i][j] != val) {
              graph[targetGrid[i][j]].push_back(val);
              indegree[val]++;
            } 
          }
        }
      }
      int total_elements = indegree.size();
      queue<int> q;
      for (auto &[u, degree] : indegree) {
        if (degree == 0) {
          q.push(u);
        }
      }
      int cnt = 0;
      while (!q.empty()) {
        int u = q.front();
        q.pop();
        cnt++;
        for (auto &v: graph[u]) {
          indegree[v]--;
          if (indegree[v] == 0) {
            q.push(v);
          }
        }
      }
      return cnt == total_elements;
    }
};
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