<p>You have a <code>grid</code> of size <code>n x 3</code> and you want to paint each cell of the grid with exactly one of the three colors: <strong>Red</strong>, <strong>Yellow,</strong> or <strong>Green</strong> while making sure that no two adjacent cells have the same color (i.e., no two cells that share vertical or horizontal sides have the same color).</p>

<p>Given <code>n</code> the number of rows of the grid, return <em>the number of ways</em> you can paint this <code>grid</code>. As the answer may grow large, the answer <strong>must be</strong> computed modulo <code>10<sup>9</sup> + 7</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/03/26/e1.png" style="width: 400px; height: 257px;" />
<pre>
<strong>Input:</strong> n = 1
<strong>Output:</strong> 12
<strong>Explanation:</strong> There are 12 possible way to paint the grid as shown.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> n = 5000
<strong>Output:</strong> 30228214
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>n == grid.length</code></li>
	<li><code>1 &lt;= n &lt;= 5000</code></li>
</ul>

const int MOD = 1e9 + 7;

class Solution {
public:
  int numOfWays(int n) {
    vector<array<int,3>> valid_rows; // valid_row_configs
    
    for (int i = 0; i < 3; ++i) {
      for (int j = 0; j < 3; ++j) if (j != i) {
        for (int k = 0; k < 3; ++k) if (k != j) {
          valid_rows.push_back(array<int,3>{ i, j, k });
        }
      }
    }
    
    int m = valid_rows.size();
    vector<vector<bool>> adj(m, vector<bool>(m, false));

    for (int i = 0; i < m; ++i) {
      for (int j = 0; j < m; ++j) {
        adj[i][j] = true;
        for (int k = 0; k < 3; ++k) if (valid_rows[i][k] == valid_rows[j][k]) adj[i][j] = false;
      }
    }

    vector<vector<int>> f(n + 1, vector<int>(m, 0));
    
    for (int i = 1; i <= n; ++i) {
      for (int j = 0; j < m; ++j) {
        if (i == 1) f[i][j] = 1;
        else {
          for (int k = 0; k < m; ++k) if (adj[j][k]) {
            f[i][j] += f[i-1][k];
            f[i][j] %= MOD;
          }
        }
      }
    }
    
    // for (int i = 1; i <= n; ++i) {
    //   for (int j = 0; j < m; ++j) cout << f[i][j] << " ";
    //   cout << endl;
    // }

    int res = 0;
    for (int j = 0; j < m; ++j) {
      res += f[n][j];
      res %= MOD;
    }
    return res;
  }
};
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