<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:
    #define ll long long int
    const int M = 1e9 + 7;
    
    
    ll dp[5005][3][4][4][4];

    ll fun(ll i, ll j, ll pr2, ll pr1, ll pr, ll n) {
        if (i >= n) return 1;  
        if (dp[i][j][pr2][pr1][pr] != -1) return dp[i][j][pr2][pr1][pr];  

        ll ans = 0;
        if (i == 0) {
            for (int new_pr = 1; new_pr <= 3; new_pr++) {
                if (new_pr != pr) {
                    if (j == 2) {
                        ans = (ans + fun( 1, 0, pr1, pr, new_pr, n)) % M;
                    } else {  
                        ans = (ans + fun(0, j + 1, pr1, pr, new_pr, n)) % M;
                    }
                }
            }
        } else {
           
            for (int new_pr = 1; new_pr <= 3; new_pr++) {
                if ((new_pr != pr||j==0) && (new_pr != pr2)) {  
                    if (j == 2) {  
                        ans = (ans + fun(i + 1, 0, pr1, pr, new_pr, n)) % M;
                    } else {  
                        ans = (ans + fun(i, j + 1, pr1, pr, new_pr, n)) % M;
                    }
                }
            }
        }

       
        return dp[i][j][pr2][pr1][pr] = ans;  
    }

    int numOfWays(int n) {
        memset(dp, -1, sizeof(dp));  
      
        ll result = (fun(0, 0, 0, 0, 0, n) ) % M;

        return result;
    }
};

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