<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<array<int, 3>> possible;
        for (int i = 1; i<=3; ++i)
            for (int j = 1; j<=3; ++j)
                for (int k = 1; k<=3; ++k)
                    if (i != j && j != k)
                        possible.push_back({i, j, k});
        
        int sz = (int)possible.size();
        vector<vector<long long>> dp(n+1, vector<long long>(sz, 0));
        for (int i = 0; i<sz; ++i)
            dp[1][i] = 1;
        long long MOD = (long long) 1e9+7;
        
        for (int i = 2; i<=n; ++i) {
            for (int j = 0; j<sz; ++j)
                for (int k = 0; k<sz; ++k) {
                    if (dp[i-1][k] > 0 && (possible[j][0] != possible[k][0]) && (possible[j][1] != possible[k][1]) && (possible[j][2] != possible[k][2])) {
                        dp[i][j] += dp[i-1][k];
                        dp[i][j] %= MOD;
                    }
                }
        }
        long long ans = 0;
        for (int i = 0; i<sz; ++i) {
            ans += dp[n][i];
            ans %= MOD;
        }
        return ans;
        
    }
};

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