<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:

    bool adj(string& s1, string& s2)
    {
        for (int i = 0; i <= 2; i++)
        {
            if (s1[i] == s2[i])
                return false;
        }
        return true;
    }

    int numOfWays(int n) {
        vector<map<string, int>> dp(n+1);
        const int mod = 1e9+7;
        vector<string> tot = {"ABC", "ACB", "BAC", "BCA", "CAB", "CBA", "ABA", "ACA", "BAB", "BCB", "CAC", "CBC"};

        for (int i = 1; i <= n; i++)
        {
            for (auto s1 : tot)
            {
                if (i == 1)
                {
                    dp[i][s1] = 1;
                    continue;
                }

                for (auto s2 : tot)
                {
                    if (adj(s1, s2))
                    {
                        dp[i][s1] += dp[i-1][s2];
                        dp[i][s1] %= mod;
                    }
                }
            }
        }

        int ans = 0;
        for (auto [_, t] : dp[n])
        {
            ans += t;
            ans %= mod;
        }
        return ans;
    }
};
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