<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) {
        int mod = 1e9 + 7;
        vector<vector<vector<vector<int>>>> dp(n + 1, vector<vector<vector<int>>>(4, vector<vector<int>>(4, vector<int>(4))));
        for(int i=0; i<4; i++) {
            for(int j=0; j<4; j++) {
                for(int k=0; k<4; k++) dp[n][i][j][k] = 1;
            }
        }

        for(int i=n-1; i>=0; i--) {
            for(int a=0; a<4; a++) {
                for(int b=0; b<4; b++) {
                    for(int c=0; c<4; c++) {
                        int ans = 0;
                        for(int c1=1; c1<=3; c1++) {
                            if(c1 == a) continue;
                            for(int c2=1; c2<=3; c2++) {
                                if(c2 == b || c2 == c1) continue;
                                for(int c3=1; c3<=3; c3++) {
                                    if(c3 == c || c3 == c2) continue;
                                    ans = (ans + dp[i + 1][c1][c2][c3]) % mod;
                                }
                            }
                        }
                        dp[i][a][b][c] = ans;

                    }
                }
            }
        }
        return dp[0][0][0][0];
    }
};
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