<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>

#include <bits/stdc++.h>
using namespace std;

class Solution {
public:
    int numOfWays(int N) {
        const int MOD = 1e9 + 7;

        vector<vector<int>> patterns;
        for(int c1 = 0; c1 < 3; ++c1){
            for(int c2 = 0; c2 < 3; ++c2){
                if(c2 == c1) continue;
                for(int c3 = 0; c3 < 3; ++c3){
                    if(c3 == c2) continue;
                    patterns.emplace_back(vector<int>{c1, c2, c3});
                }
            }
        }

        int P = patterns.size();
        vector<vector<int>> compatible(P, vector<int>());
        for(int i = 0; i < P; ++i){
            for(int j = 0; j < P; ++j){
                bool conflict = false;
                for(int k = 0; k < 3; ++k){
                    if(patterns[i][k] == patterns[j][k]){
                        conflict = true;
                        break;
                    }
                }
                if(!conflict){
                    compatible[i].push_back(j);
                }
            }
        }

        vector<long long> prev_dp(P, 1);
        for(int row = 2; row <= N; ++row){
            vector<long long> current_dp(P, 0);
            for(int curr = 0; curr < P; ++curr){
                for(auto &prev : compatible[curr]){
                    current_dp[curr] = (current_dp[curr] + prev_dp[prev]) % MOD;
                }
            }
            prev_dp = current_dp;
        }

        long long total = 0;
        for(int i = 0; i < P; ++i){
            total = (total + prev_dp[i]) % MOD;
        }

        return (int)total;
    }
};

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