<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;

    int dp1[5001][65] = {};
    vector<int> dp2[65] = {};

    int getColor(int r,int mask) {
        return (mask >> (2*r)) & 3;
    }

    void fn(int r,int currmask,int prevmask,vector<int> &ans) {
        if(r == 3) {
            ans.push_back(currmask);
            return;
        }

        for(int color=1;color<=3;color++) {
            if(color != getColor(r,prevmask) && (r==0 || color != getColor(r-1,currmask))) {
                int tempmask = currmask | (color << (2*r));
                fn(r+1,tempmask,prevmask,ans);
            }
        }
    }

    vector<int> getPossibilities(int prevmask) {
        if(!dp2[prevmask].empty()) return dp2[prevmask];
        vector<int> ans;
        fn(0,0,prevmask,ans);
        return dp2[prevmask] = ans;
    }

    int solve(int col,int prevmask,int n) {
        if(col == n) return 1;
        if(dp1[col][prevmask]) return dp1[col][prevmask];

        int ans = 0;

        auto vec = getPossibilities(prevmask);
        for(auto currmask : vec) {
            ans += solve(col+1,currmask,n);
            ans %= mod;
        }

        return dp1[col][prevmask] = ans;
    }

public:
    int numOfWays(int n) {

        return solve(0,0,n);
        
    }
};
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