<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:
    long long mod = 1e9 + 7;
    long long ans(int n,int curr,int u,int mid,int l,vector<vector<long long>> &dp){
         if(n<=0)return 1;
         long long  get = 0;
         if(dp[curr][n]!=-1)return dp[curr][n];
         for(int i=1;i<=3;i++){
            for(int j=1;j<=3;j++){
                if(i==j)continue;
                for(int k=1;k<=3;k++){
                 if(j==k)continue;
                 if(i!=u && j!=mid && l!=k){
                    int cur1 = 4*i + 5*j + 6*k;
                    get = (get + ans(n-1,cur1,i,j,k,dp))%mod;
                 }
                }
            }
         }
         return dp[curr][n] = get%mod;
    }
    int numOfWays(int n) {
        // if(n==1)return 12;
        vector<vector<long long>> dp(50,vector<long long> (n+1,-1));
        return ans(n,0,0,0,0,dp);
    }
};
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