<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:

  auto topsort(vector<vector<int>>& conditions, int k, const set<int>& unique) {
    vector<vector<int>> adj(k + 1);
    vector<int> deg(k + 1);
    for (auto x: conditions) {
      adj[x[1]].push_back(x[0]);
      deg[x[0]]++;
    }
    vector<int> ans;
    stack<int> curr;
    for (int i = 1; i <= k; i++)
      if (!deg[i] && unique.contains(i)) curr.push(i);

    while (!curr.empty()) {
      int vert = curr.top();
      curr.pop();
      ans.push_back(vert);
      for (auto x: adj[vert])
        if (!(--deg[x]))
          curr.push(x);
    }
    return unique.size() == ans.size();
  }

  bool isPrintable(vector<vector<int>>& targetGrid) {
    int n = targetGrid.size(), m = targetGrid[0].size();
    unordered_map<int, pair<pair<int,int>, pair<int,int>>> a;
    set<int> unique;
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < m; j++) {
        if (!a.contains(targetGrid[i][j])) {
          a[targetGrid[i][j]].first.first = i;
          a[targetGrid[i][j]].first.second = j;
          a[targetGrid[i][j]].second.first = i;
          a[targetGrid[i][j]].second.second = j;
          unique.insert(targetGrid[i][j]);
          continue;
        }
        a[targetGrid[i][j]].first.first = min(a[targetGrid[i][j]].first.first, i);
        a[targetGrid[i][j]].first.second = min(a[targetGrid[i][j]].first.second, j);
        a[targetGrid[i][j]].second.first = max(a[targetGrid[i][j]].second.first, i);
        a[targetGrid[i][j]].second.second = max(a[targetGrid[i][j]].second.second, j);
      }
    }

    vector<vector<int>> conditions;
    for (auto [color, coords]: a) {
      unordered_set<int> used;
      for (int i = coords.first.first; i <= coords.second.first; i++) {
        for (int j = coords.first.second; j <= coords.second.second; j++) {
          if (!used.contains(targetGrid[i][j]) && targetGrid[i][j] != color) {
            used.insert(targetGrid[i][j]);
            conditions.push_back({targetGrid[i][j], color});
          }
        }
      }
    }
    return topsort(conditions, 60, unique);
  }
};
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