Painting a Grid With Three Different Colors
<p>You are given two integers <code>m</code> and <code>n</code>. Consider an <code>m x n</code> grid where each cell is initially white. You can paint each cell <strong>red</strong>, <strong>green</strong>, or <strong>blue</strong>. All cells <strong>must</strong> be painted.</p>

<p>Return<em> the number of ways to color the grid with <strong>no two adjacent cells having the same color</strong></em>. Since the answer can be very large, return it <strong>modulo</strong> <code>10<sup>9</sup> + 7</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/1900-1999/1931.Painting%20a%20Grid%20With%20Three%20Different%20Colors/images/colorthegrid.png" style="width: 200px; height: 50px;" />
<pre>
<strong>Input:</strong> m = 1, n = 1
<strong>Output:</strong> 3
<strong>Explanation:</strong> The three possible colorings are shown in the image above.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/1900-1999/1931.Painting%20a%20Grid%20With%20Three%20Different%20Colors/images/copy-of-colorthegrid.png" style="width: 321px; height: 121px;" />
<pre>
<strong>Input:</strong> m = 1, n = 2
<strong>Output:</strong> 6
<strong>Explanation:</strong> The six possible colorings are shown in the image above.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> m = 5, n = 5
<strong>Output:</strong> 580986
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= m &lt;= 5</code></li>
	<li><code>1 &lt;= n &lt;= 1000</code></li>
</ul>
Dynamic Programming
class Solution:
    def colorTheGrid(self, m: int, n: int) -> int:
        def f1(x: int) -> bool:
            last = -1
            for _ in range(m):
                if x % 3 == last:
                    return False
                last = x % 3
                x //= 3
            return True

        def f2(x: int, y: int) -> bool:
            for _ in range(m):
                if x % 3 == y % 3:
                    return False
                x, y = x // 3, y // 3
            return True

        mod = 10**9 + 7
        mx = 3**m
        valid = {i for i in range(mx) if f1(i)}
        d = defaultdict(list)
        for x in valid:
            for y in valid:
                if f2(x, y):
                    d[x].append(y)
        f = [int(i in valid) for i in range(mx)]
        for _ in range(n - 1):
            g = [0] * mx
            for i in valid:
                for j in d[i]:
                    g[i] = (g[i] + f[j]) % mod
            f = g
        return sum(f) % mod
