# 1411. Number of Ways to Paint N × 3 Grid

## Description

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:

$f[i][j] = \sum_{k \in \text{valid}(j)} f[i - 1][k]$

where $\text{valid}(j)$ 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.

• class Solution {
public int numOfWays(int n) {
int mod = (int) 1e9 + 7;
long f0 = 6, f1 = 6;
for (int i = 0; i < n - 1; ++i) {
long g0 = (3 * f0 + 2 * f1) % mod;
long g1 = (2 * f0 + 2 * f1) % mod;
f0 = g0;
f1 = g1;
}
return (int) (f0 + f1) % mod;
}
}

• using ll = long long;

class Solution {
public:
int numOfWays(int n) {
int mod = 1e9 + 7;
ll f0 = 6, f1 = 6;
while (--n) {
ll g0 = (f0 * 3 + f1 * 2) % mod;
ll g1 = (f0 * 2 + f1 * 2) % mod;
f0 = g0;
f1 = g1;
}
return (int) (f0 + f1) % mod;
}
};

• class Solution:
def numOfWays(self, n: int) -> int:
mod = 10**9 + 7
f0 = f1 = 6
for _ in range(n - 1):
g0 = (3 * f0 + 2 * f1) % mod
g1 = (2 * f0 + 2 * f1) % mod
f0, f1 = g0, g1
return (f0 + f1) % mod


• func numOfWays(n int) int {
mod := int(1e9) + 7
f0, f1 := 6, 6
for n > 1 {
n--
g0 := (f0*3 + f1*2) % mod
g1 := (f0*2 + f1*2) % mod
f0, f1 = g0, g1
}
return (f0 + f1) % mod
}

• function numOfWays(n: number): number {
const mod: number = 10 ** 9 + 7;
let f0: number = 6;
let f1: number = 6;

for (let i = 1; i < n; i++) {
const g0: number = (3 * f0 + 2 * f1) % mod;
const g1: number = (2 * f0 + 2 * f1) % mod;
f0 = g0;
f1 = g1;
}

return (f0 + f1) % mod;
}