<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 numOfWays(int n) 
    {
        std::memset(dp_, 0, sizeof(dp_));
        int mod = 1e9 + 7;

        for (int currState = 0; currState < 27; ++currState)
        {
            if (!isCurrentStateOk(currState))
                continue;
            dp_[0][currState] = 1;
        }

        for (int i = 1; i < n; ++i)
        {
            for (int currState = 0; currState < 27; ++currState)
            {
                if (!isCurrentStateOk(currState))
                    continue;

                for (int prevState = 0; prevState < 27; ++prevState)
                {
                    if (!isCurrentStateOk(prevState))
                        continue;

                    if (!noConflictWithPrev(currState, prevState))
                        continue;

                    dp_[i][currState] = (dp_[i][currState] + dp_[i - 1][prevState]) % mod;
                }
            }
        }

        int result = 0;
        for (int currState = 0; currState < 27; ++currState)
        {
            if (!isCurrentStateOk(currState))
                continue;
            result = (result + dp_[n - 1][currState]) % mod;
        }

        return result;
    }

private:
    bool isCurrentStateOk(int currState)
    {
        int temp[3] = { 0 };
        int index = 0;

        while (currState)
        {
            temp[index] = (currState % 3);
            currState /= 3;
            index++;
        }

        return (temp[0] != temp[1] && temp[1] != temp[2]);
    }

    bool noConflictWithPrev(int currState, int prevState)
    {
        int temp1[3] = { 0 }, temp2[3] = { 0 };
        for (int i = 0; i < 3; ++i)
        {
            temp1[i] = (currState % 3);
            temp2[i] = (prevState % 3);

            currState /= 3;
            prevState /= 3;
        }

        return (temp1[0] != temp2[0] && temp1[1] != temp2[1] && temp1[2] != temp2[2]);
    }

private:
    int dp_[5005][27];
};
10175