<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:
int M = 1e9 + 7;
    int numOfWays(int n) {
        vector<int> dp(27, 0);
        for(int i=0; i<27; i++) {
            if(selfOK(i))
                dp[i] = 1;
        }
        for(int i=1; i<n; i++) {
            auto dp_prev = dp;
            
            for(int j=0; j<27; j++) {
                dp[j] = 0;
                if(!selfOK(j))
                    continue;
                for(int k=0; k<27; k++) {
                    // if(!selfOK(k)) continue;
                    if(crossOK(j, k)) {
                        dp[j] += dp_prev[k] % M;
                        dp[j] %= M;
                    }
                }
                // cout << dp[j] << " ";
            }
            // cout<< endl;
        }
        int ret = 0;
        for(int j=0; j<27; j++) {
            // if(!selfOK(j)) continue;
            ret += dp[j];
            ret %= M;
        }
        return ret;

    }
    bool selfOK(int x) {
        int tmp[3];
        for(int i=0; i<3; i++) {
            tmp[i] = x % 3;
            x /=3;
        }
        if(tmp[0] == tmp[1] || tmp[1] == tmp[2])
            return false;
        return true;
    }
    bool crossOK(int x, int y) {
        int tmp[3], tmp2[3];
        for(int i=0; i<3; i++) {
            tmp[i] = x % 3;
            tmp2[i] = y % 3;
            x /=3;
            y /=3;
        }
        if(tmp[0] == tmp2[0] || tmp[1] == tmp2[1] || tmp[2] == tmp2[2])
            return false;
        return true;
    }
};
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