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feat: add solutions to lc problem: No.1411
No.1411.Number of Ways to Paint N \303\227 3 Grid
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‎solution/1400-1499/1411.Number of Ways to Paint N × 3 Grid/README.md

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<!-- 这里可写通用的实现逻辑 -->
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**方法一：递推**
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把每一行所有可能的状态进行分类。根据对称性原理，当一行只有 $3$ 个元素时，所有合法状态分类为：$010$ 型, $012$ 型。
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- 当状态为 $010$ 型时：下一行可能的状态为：$101$, $102$, $121$, $201$, $202$。这 $5$ 个状态可归纳为 $3$ 个 $010$ 型，$2$ 个 $012$ 型。
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- 当状态为 $012$ 型时：下一行可能的状态为：$101$, $120$, $121$, $201$。这 $4$ 个状态可归纳为 $2$ 个 $010$ 型，$2$ 个 $012$ 型。
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综上所述，可以得到：$newf0 = 3 * f0 + 2 * f1$，$newf1 = 2 * f0 + 2 * f1$。
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时间复杂度 $O(n)$。
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<!-- tabs:start -->
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### **Python3**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```python
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class Solution:
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def numOfWays(self, n: int) -> int:
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mod = 10**9 + 7
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f0 = f1 = 6
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for _ in range(n - 1):
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g0 = (3 * f0 + 2 * f1) % mod
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g1 = (2 * f0 + 2 * f1) % mod
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f0, f1 = g0, g1
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return (f0 + f1) % mod
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```
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### **Java**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```java
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class Solution {
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public int numOfWays(int n) {
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int mod = (int) 1e9 + 7;
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long f0 = 6, f1 = 6;
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for (int i = 0; i < n - 1; ++i) {
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long g0 = (3 * f0 + 2 * f1) % mod;
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long g1 = (2 * f0 + 2 * f1) % mod;
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f0 = g0;
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f1 = g1;
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}
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return (int) (f0 + f1) % mod;
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}
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}
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```
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### **C++**
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```cpp
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using ll = long long;
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class Solution {
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public:
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int numOfWays(int n) {
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int mod = 1e9 + 7;
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ll f0 = 6, f1 = 6;
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while (--n)
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{
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ll g0 = (f0 * 3 + f1 * 2) % mod;
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ll g1 = (f0 * 2 + f1 * 2) % mod;
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f0 = g0;
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f1 = g1;
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}
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return (int) (f0 + f1) % mod;
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}
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};
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```
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### **Go**
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```go
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func numOfWays(n int) int {
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mod := int(1e9) + 7
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f0, f1 := 6, 6
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for n > 1 {
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n--
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g0 := (f0*3 + f1*2) % mod
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g1 := (f0*2 + f1*2) % mod
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f0, f1 = g0, g1
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}
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return (f0 + f1) % mod
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}
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```
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### **...**

‎solution/1400-1499/1411.Number of Ways to Paint N × 3 Grid/README_EN.md

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### **Python3**
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```python
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class Solution:
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def numOfWays(self, n: int) -> int:
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mod = 10**9 + 7
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f0 = f1 = 6
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for _ in range(n - 1):
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g0 = (3 * f0 + 2 * f1) % mod
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g1 = (2 * f0 + 2 * f1) % mod
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f0, f1 = g0, g1
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return (f0 + f1) % mod
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```
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### **Java**
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```java
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class Solution {
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public int numOfWays(int n) {
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int mod = (int) 1e9 + 7;
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long f0 = 6, f1 = 6;
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for (int i = 0; i < n - 1; ++i) {
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long g0 = (3 * f0 + 2 * f1) % mod;
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long g1 = (2 * f0 + 2 * f1) % mod;
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f0 = g0;
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f1 = g1;
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}
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return (int) (f0 + f1) % mod;
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}
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}
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```
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### **C++**
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```cpp
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using ll = long long;
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class Solution {
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public:
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int numOfWays(int n) {
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int mod = 1e9 + 7;
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ll f0 = 6, f1 = 6;
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while (--n)
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{
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ll g0 = (f0 * 3 + f1 * 2) % mod;
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ll g1 = (f0 * 2 + f1 * 2) % mod;
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f0 = g0;
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f1 = g1;
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}
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return (int) (f0 + f1) % mod;
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}
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};
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```
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### **Go**
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```go
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func numOfWays(n int) int {
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mod := int(1e9) + 7
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f0, f1 := 6, 6
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for n > 1 {
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n--
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g0 := (f0*3 + f1*2) % mod
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g1 := (f0*2 + f1*2) % mod
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f0, f1 = g0, g1
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}
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return (f0 + f1) % mod
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}
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```
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### **...**

‎solution/1400-1499/1411.Number of Ways to Paint N × 3 Grid/Solution.cpp

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

‎solution/1400-1499/1411.Number of Ways to Paint N × 3 Grid/Solution.go

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

‎solution/1400-1499/1411.Number of Ways to Paint N × 3 Grid/Solution.java

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

‎solution/1400-1499/1411.Number of Ways to Paint N × 3 Grid/Solution.py

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

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