<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 {
int mod = 1e9 + 7;
public:
    vector<vector<vector<vector<int>>>>dp;
    int help(int idx, int n, int col1, int col2, int col3){
        if(idx == n) return 1;
        if(dp[idx][col1][col2][col3] != -1) return dp[idx][col1][col2][col3];
        int ans = 0;
        // 1 -> red, 2 -> greem, 3 -> blue
        for(int i = 1; i<=3; i++){
            if(col1 == i) continue;
            for(int j = 1; j<=3; j++){
                if(j == i || j == col2) continue;
                for(int k = 1; k<=3; k++){
                    if(k == j || k == col3) continue;
                    ans = (ans + help(idx+1, n, i, j, k))%mod;
                }
            }
        }
        return dp[idx][col1][col2][col3] = ans;

        

    }
    int numOfWays(int n) {
        dp.resize(n, vector<vector<vector<int>>>(4, vector<vector<int>>(4, vector<int>(4, -1))));
        help(0, n, 0, 0, 0);
        return dp[0][0][0][0];
    }
};
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