You have a grid of size n x 3 and you want to paint each cell of the grid with exactly one of the three colors: Red, Yellow, or Green 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).
Given n the number of rows of the grid, return the number of ways you can paint this grid. As the answer may grow large, the answer must be computed modulo 109 + 7.
Example 1:
Input: n = 1
Output: 12
Explanation: There are 12 possible way to paint the grid as shown.
Example 2:
Input: n = 5000
Output: 30228214
Constraints:
n == grid.length
1 <= n <= 5000
Solutions
Solution 1: Recursion
We classify all possible states for each row. According to the principle of symmetry, when a row only has $3$ elements, all legal states are classified as: $010$ type, $012$ type.
When the state is $010$ type: The possible states for the next row are: $101$, $102$, $121$, $201$, $202$. These $5$ states can be summarized as $3$ $010$ types and $2$ $012$ types.
When the state is $012$ type: The possible states for the next row are: $101$, $120$, $121$, $201$. These $4$ states can be summarized as $2$ $010$ types and $2$ $012$ types.
In summary, we can get: $newf0 = 3 \times f0 + 2 \times f1$, $newf1 = 2 \times f0 + 2 \times f1$.
The time complexity is $O(n)$, where $n$ is the number of rows in the grid. The space complexity is $O(1)$.
Solution 2: State Compression + Dynamic Programming
We notice that the grid only has $3$ columns, so there are at most $3^3=27$ different coloring schemes in a row.
Therefore, we define $f[i][j]$ to represent the number of schemes in the first $i$ rows, where the coloring state of the $i$th row is $j$. The state $f[i][j]$ is transferred from $f[i - 1][k]$, where $k$ is the coloring state of the $i - 1$th row, and $k$ and $j$ meet the requirement of different colors being adjacent. That is:
where $\textit{valid}(j)$ represents all legal predecessor states of state $j$.
The final answer is the sum of $f[n][j]$, where $j$ is any legal state.
We notice that $f[i][j]$ is only related to $f[i - 1][k]$, so we can use a rolling array to optimize the space complexity.
The time complexity is $O((m + n) \times 3^{2m})$, and the space complexity is $O(3^m)$. Here, $m$ and $n$ are the number of rows and columns of the grid, respectively.