<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:
    vector<vector<int>> cross;
    bool selfOk(int p) {
        int temp[3];
        for (int i = 0; i < 3; i++) {
            temp[i] = p % 3;
            p /= 3;
        }
        return temp[0] != temp[1] && temp[1] != temp[2];
    }
    bool crossOk(int p, int q) {
        int temp1[3];
        for (int i = 0; i < 3; i++) {
            temp1[i] = p % 3;
            p /= 3;
        }
        int temp2[3];
        for (int i = 0; i < 3; i++) {
            temp2[i] = q % 3;
            q /= 3;
        }
        return temp1[0] != temp2[0] && temp1[1] != temp2[1] &&
               temp1[2] != temp2[2];
    }
    int numOfWays(int n) {
        cross.resize(27);

        for (int i = 0; i < 27; i++) {
            if (!selfOk(i)) continue;
            for (int j = 0; j < 27; j++) {
                if (!selfOk(j)) continue;
                if (crossOk(i, j))
                    cross[i].push_back(j);
            }
        }
        vector<int> dp(27);
        int M = 1e9 + 7;
        for (int i = 0; i < 27; i++) {
            if (selfOk(i)) {
                dp[i] = 1;
            }
        }

        for (int k = 1; k < n; k++) {
            auto dp_prev = dp;
            for (int p = 0; p < 27; p++) {
                dp[p] = 0;
                for (int q : cross[p]) {
                    dp[p] = (dp[p] + dp_prev[q]) % M;
                }
            }
        }
        int ans = 0;
        for (int i = 0; i < 27; i++) {
            ans = (ans + dp[i]) % M;
        }
        return ans;
    }
};
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