<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:
    unordered_map<int, vector<int>> combs;
    void fill(){
        int x = 31;
        for(int i=1;i<=3;i++){
            for(int j=1;j<=3;j++){
                for(int k=1;k<=3;k++){
                    if(j!=i && k!=j){
                        int hash=i*pow(x,1)+j*pow(x,2)+k*pow(x,3);
                        combs[0].push_back(hash);
                        for(int a=1;a<=3;a++){
                            for(int b=1;b<=3;b++){
                                for(int c=1;c<=3;c++){
                                    if(b!=a && c!=b && a!=i && b!=j && c!=k){
                                        int hash2=a*pow(x,1)+b*pow(x,2)+c*pow(x,3);
                                        combs[hash].push_back(hash2);
                                    }
                                }
                            }
                        }
                    }
                }
            }
        }
    }
    long long MOD = 1e9+7;
    int explorePossibilities(int par, int curr, unordered_map<int, unordered_map<int, int>>& dp, int n){
         if(curr==n) return 1;
         if(curr!=0 && dp.count(curr-1) && dp[curr-1].count(par)){
              return dp[curr-1][par];
         }
         int count=0;
         for(int x : combs[par]){
            count = (count+explorePossibilities(x, curr+1, dp, n))%MOD;
         }
         return dp[curr-1][par]=count;
    }
    int numOfWays(int n) {
        fill();
        for(auto& pair: combs){
            cout<<pair.first<<" -> ";
            for(int x : pair.second){
                cout<<x<<" ";
            }
            cout<<endl;
        }
        unordered_map<int, unordered_map<int, int>> dp;
        return explorePossibilities(0, 0, dp, n);
    }
};
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