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feat: add solutions to lc problem: No.1411 #2003

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Nov 23, 2023
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feat: add solutions to lc problem: No.1411 #2003

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319 changes: 317 additions & 2 deletions solution/1400-1499/1411.Number of Ways to Paint N × 3 Grid/README.md
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Original file line number Diff line number Diff line change
Expand Up @@ -67,9 +67,27 @@
- 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。
- 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。

综上所述，可以得到：$newf0 = 3 * f0 + 2 * f1$，$newf1 = 2 * f0 + 2 * f1$。
综上所述，可以得到：$newf0 = 3 \times f0 + 2 \times f1$, $newf1 = 2 \times f0 + 2 \times f1$。

时间复杂度 $O(n)$。
时间复杂度 $O(n)$，其中 $n$ 是网格的行数。空间复杂度 $O(1)$。

**方法二：状态压缩 + 动态规划**

我们注意到，网格只有 $3$ 列，那么一行中最多有 $3^3=27$ 种不同的涂色方案。

因此，我们定义 $f[i][j]$ 表示前 $i$ 行中，第 $i$ 行的涂色状态为 $j$ 的方案数。状态 $f[i][j]$ 由 $f[i - 1][k]$ 转移而来，其中 $k$ 是第 $i - 1$ 行的涂色状态，且 $k$ 和 $j$ 满足不同颜色相邻的要求。即：

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

其中 $\text{valid}(j)$ 表示状态 $j$ 的所有合法前驱状态。

最终的答案即为 $f[n][j]$ 的总和，其中 $j$ 是任意合法的状态。

我们注意到，$f[i][j]$ 只和 $f[i - 1][k]$ 有关，因此我们可以使用滚动数组优化空间复杂度。

时间复杂度 $O((m + n) \times 3^{2m})$，空间复杂度 $O(3^m)$。其中 $n$ 和 $m$ 分别是网格的行数和列数。

<!-- tabs:start -->

Expand All @@ -89,6 +107,44 @@ class Solution:
return (f0 + f1) % mod
```

```python
class Solution:
def numOfWays(self, n: int) -> int:
def f1(x: int) -> bool:
last = -1
for _ in range(3):
if x % 3 == last:
return False
last = x % 3
x //= 3
return True

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

mod = 10**9 + 7
m = 27
valid = {i for i in range(m) if f1(i)}
d = defaultdict(list)
for i in valid:
for j in valid:
if f2(i, j):
d[i].append(j)
f = [int(i in valid) for i in range(m)]
for _ in range(n - 1):
g = [0] * m
for i in valid:
for j in d[i]:
g[j] = (g[j] + f[i]) % mod
f = g
return sum(f) % mod
```

### **Java**

<!-- 这里可写当前语言的特殊实现逻辑 -->
Expand All @@ -109,6 +165,68 @@ class Solution {
}
```

```java
class Solution {
public int numOfWays(int n) {
final int mod = (int) 1e9 + 7;
int m = 27;
Set<Integer> valid = new HashSet<>();
int[] f = new int[m];
for (int i = 0; i < m; ++i) {
if (f1(i)) {
valid.add(i);
f[i] = 1;
}
}
Map<Integer, List<Integer>> d = new HashMap<>();
for (int i : valid) {
for (int j : valid) {
if (f2(i, j)) {
d.computeIfAbsent(i, k -> new ArrayList<>()).add(j);
}
}
}
for (int k = 1; k < n; ++k) {
int[] g = new int[m];
for (int i : valid) {
for (int j : d.getOrDefault(i, List.of())) {
g[j] = (g[j] + f[i]) % mod;
}
}
f = g;
}
int ans = 0;
for (int x : f) {
ans = (ans + x) % mod;
}
return ans;
}

private boolean f1(int x) {
int last = -1;
for (int i = 0; i < 3; ++i) {
if (x % 3 == last) {
return false;
}
last = x % 3;
x /= 3;
}
return true;
}

private boolean f2(int x, int y) {
for (int i = 0; i < 3; ++i) {
if (x % 3 == y % 3) {
return false;
}
x /= 3;
y /= 3;
}
return true;
}
}
```

### **C++**

```cpp
Expand All @@ -130,6 +248,69 @@ public:
};
```

```cpp
class Solution {
public:
int numOfWays(int n) {
int m = 27;

auto f1 = [&](int x) {
int last = -1;
for (int i = 0; i < 3; ++i) {
if (x % 3 == last) {
return false;
}
last = x % 3;
x /= 3;
}
return true;
};
auto f2 = [&](int x, int y) {
for (int i = 0; i < 3; ++i) {
if (x % 3 == y % 3) {
return false;
}
x /= 3;
y /= 3;
}
return true;
};

const int mod = 1e9 + 7;
unordered_set<int> valid;
vector<int> f(m);
for (int i = 0; i < m; ++i) {
if (f1(i)) {
valid.insert(i);
f[i] = 1;
}
}
unordered_map<int, vector<int>> d;
for (int i : valid) {
for (int j : valid) {
if (f2(i, j)) {
d[i].push_back(j);
}
}
}
for (int k = 1; k < n; ++k) {
vector<int> g(m);
for (int i : valid) {
for (int j : d[i]) {
g[j] = (g[j] + f[i]) % mod;
}
}
f = move(g);
}
int ans = 0;
for (int x : f) {
ans = (ans + x) % mod;
}
return ans;
}
};
```

### **Go**

```go
Expand All @@ -146,6 +327,140 @@ func numOfWays(n int) int {
}
```

```go
func numOfWays(n int) (ans int) {
f1 := func(x int) bool {
last := -1
for i := 0; i < 3; i++ {
if x%3 == last {
return false
}
last = x % 3
x /= 3
}
return true
}
f2 := func(x, y int) bool {
for i := 0; i < 3; i++ {
if x%3 == y%3 {
return false
}
x /= 3
y /= 3
}
return true
}
m := 27
valid := map[int]bool{}
f := make([]int, m)
for i := 0; i < m; i++ {
if f1(i) {
valid[i] = true
f[i] = 1
}
}
d := map[int][]int{}
for i := range valid {
for j := range valid {
if f2(i, j) {
d[i] = append(d[i], j)
}
}
}
const mod int = 1e9 + 7
for k := 1; k < n; k++ {
g := make([]int, m)
for i := range valid {
for _, j := range d[i] {
g[i] = (g[i] + f[j]) % mod
}
}
f = g
}
for _, x := range f {
ans = (ans + x) % mod
}
return
}
```

### **TypeScript**

```ts
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;
}
```

```ts
function numOfWays(n: number): number {
const f1 = (x: number): boolean => {
let last = -1;
for (let i = 0; i < 3; ++i) {
if (x % 3 === last) {
return false;
}
last = x % 3;
x = Math.floor(x / 3);
}
return true;
};
const f2 = (x: number, y: number): boolean => {
for (let i = 0; i < 3; ++i) {
if (x % 3 === y % 3) {
return false;
}
x = Math.floor(x / 3);
y = Math.floor(y / 3);
}
return true;
};
const m = 27;
const valid = new Set<number>();
const f: number[] = Array(m).fill(0);
for (let i = 0; i < m; ++i) {
if (f1(i)) {
valid.add(i);
f[i] = 1;
}
}
const d: Map<number, number[]> = new Map();
for (const i of valid) {
for (const j of valid) {
if (f2(i, j)) {
d.set(i, (d.get(i) || []).concat(j));
}
}
}
const mod = 10 ** 9 + 7;
for (let k = 1; k < n; ++k) {
const g: number[] = Array(m).fill(0);
for (const i of valid) {
for (const j of d.get(i) || []) {
g[i] = (g[i] + f[j]) % mod;
}
}
f.splice(0, f.length, ...g);
}
let ans = 0;
for (const x of f) {
ans = (ans + x) % mod;
}
return ans;
}
```

### **...**

```
Expand Down
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