<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 {
    const int mod = 1e9 + 7;
public:
    int numOfWays(int n) {
        vector<string> state;
        // enumerate the state
        for(char i = '0'; i < '3'; i++){
            for(char j = '0'; j < '3'; j++){
                if(i == j) continue;
                for(char k = '0'; k < '3'; k++){
                    if(j == k) continue;
                    string cur{i,j,k};
                    state.push_back(cur);
                }
            }
        }
        int s = state.size();
        if(n == 1) return s;
        // check the connections between status
        vector<vector<int>> adj(s);
        for(int i = 0; i < s; i++){
            for(int j = 0; j < s; j++){
                bool compat = true;
                for(int k = 0; k < 3; k++){
                    if(state[i][k] == state[j][k]){
                        compat = false;
                        break;
                    }
                }
                if(compat) adj[i].push_back(j);
            }
        }
        // repeat the dp process for n times
        vector<int> dp(s, 1);
        for(int k = 2; k <= n; k++){
            vector<int> dp_next(s, 0);
            for(int i = 0; i < s; i++){
                for(auto x : adj[i]){
                    dp_next[i] += dp[x];
                    dp_next[i] %= mod;
                }
            }
            dp = dp_next;
        }
        int res = 0;
        for(int i = 0; i < s; i++){
            res += dp[i];
            res %= mod;
        }
        return res;
    }
};
12635