<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:
    using LL = long long;
    LL M = 1e9+7;

    vector<int>temp1;
    vector<int>temp2;
    int numOfWays(int n) {

        vector<int>dp(27, 0);

        temp1.resize(3);
        temp2.resize(3);

        for (int p=0; p<27; p++)
            if (selfOK(p))
                dp[p] = 1;

        for (int row=1; row<n; row++) {
            auto dp_prev = dp;
            for (int p=0; p<27; p++) {
                if (!selfOK(p)) continue;

                dp[p] = 0;
                for (int q=0; q<27; q++) {
                    if (!selfOK(q)) continue;
                    if (crossOK(p,q))
                        dp[p] = (dp[p] + dp_prev[q])%M;
                }
            }
        }

        LL ret = 0;
        for (int p=0; p<27; p++) {
            if (selfOK(p))
                ret = (ret + dp[p])%M;
        }

        return ret;
    }

    int selfOK(int p) {
        for (int i=0; i<3; i++) {
            temp1[i] = p%3;
            p/=3;
        }
        return (temp1[0] != temp1[1]) && (temp1[1] != temp1[2]);
    }

    int crossOK(int p, int q) {
        for (int i=0; i<3; i++) {
            temp1[i] = p%3;
            temp2[i] = q%3;
            p/=3;
            q/=3;
        }
        return (temp1[0] != temp2[0]) && 
               (temp1[1] != temp2[1]) && 
               (temp1[2] != temp2[2]);
    }
};

/*
27
0
9
00
3
000
1
1000
*/
18785