<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:
    int numOfWays(int n) {
        vector<vector<int>> cnt(27);
        vector<long long> dp(27);
        for (int i = 0; i < 27; i++) {
            int a = (i / 9) % 3;
            int b = (i / 3) % 3;
            int c = i % 3;
            if (a == b || b == c) continue;
            dp[i] = 1;
            for (int j = 0; j < 27; j++) {
                int na = (j / 9) % 3;
                int nb = (j / 3) % 3;
                int nc = j % 3;
                if (na == nb || nb == nc) continue;
                if (a != na && b != nb && c != nc) {
                    cnt[i].emplace_back(j);
                }
            }
        }

        long long mod = 1e9 + 7;
        for (int i = 1; i < n; i++) {
            vector<long long> ndp(27);
            for (int j = 0; j < 27; j++) {
                for (auto v : cnt[j]) {
                    ndp[j] = (ndp[j] + dp[v]) % mod; 
                }
            }
            dp = move(ndp);
        }
        long long res = 0;
        for (auto v : dp) {
            res = (res + v) % mod;
        }
        return res;
    }
};
24935