<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>

class Solution {
public:
    const static int NUM = 1e9 + 7;
    int numOfWays(int n) {
        vector<vector<int>> v(27);
        for (int i = 0; i < 27; ++i) {
            int f = i % 3;
            int s = (i / 3) % 3;
            int t = (i / 9) % 3;
            if (f == s || s == t) {
                continue;
            }
            for (int j = 0; j < 27; ++j) {
                int ff= (j % 3);
                int ss = (j / 3) % 3;
                int tt = (j / 9) % 3;
                if (ff == ss || ss == tt) {
                    continue;
                }
                if (f == ff || s == ss || t == tt) {
                    continue;
                }
                v[i].push_back(j);
            }
        }
        vector<vector<int>> dp(n, vector<int>(27, 0));
        for (int i = 0; i < 26; ++i) {
            if (v[i].size()) {
                dp[0][i] = 1;
            }
        }
        for (int i = 0; i < n - 1; ++i) {
            for (int j = 0; j < 27; ++j) {
                if (dp[i][j] == 0) {
                    continue;
                }
                for (int a : v[j]) {
                    dp[i + 1][a] = (dp[i + 1][a] + dp[i][j]) % NUM;
                }
            }
        }
        int ans = 0;
        for (int i = 0; i < 27; ++i) {
            ans = (ans + dp[n - 1][i]) % NUM;
        }
        return ans;
    }
};
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