<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 kmax = 1e9 + 7;
    int rec(vector<vector<int>>& dp, int i, int n, vector<vector<int>>& possible, int p) {
        if(i == n) return 1;
        if(p > -1 && dp[i][p] > -1) return dp[i][p];
        int ans = 0;
        for(int j=0; j<possible.size(); j++) {
            if(p != -1 && (possible[p][0] == possible[j][0] || possible[p][1] == possible[j][1] || possible[p][2] == possible[j][2])) continue;
            ans = (ans + (rec(dp, i+1, n, possible, j))%kmax)%kmax;
        }
        if(p > -1) return dp[i][p] = ans%kmax;
        return ans%kmax;
    }
public:
    int numOfWays(int n) {
        vector<vector<int>> possible;
        vector<vector<int>> dp(5004, vector<int>(27, -1));
        for(int i=0; i<3; i++)
        for(int j=0; j<3; j++)
        for(int k=0; k<3; k++)
        {
            if(i == j || j == k) continue;
            possible.push_back({i,j,k});
        }

        return rec(dp, 0, n, possible, -1);
    }
};
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