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Commit 51a32cf

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feat: add solutions to lcp problem: No.25 (doocs#3649)
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‎lcp/LCP 25. 古董键盘/README.md‎

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@@ -42,4 +42,159 @@ edit_url: https://github.com/doocs/leetcode/edit/main/lcp/LCP%2025.%20%E5%8F%A4%
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<!-- solution:start -->
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### 方法一:动态规划
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我们定义 $f[i][j]$ 表示按了 $i$ 次按键,且使用了前 $j$ 个字母的方案数。初始时 $f[0]$ 全部为 1ドル,ドル表示没有按键时只有一种方案,其余 $f[i][j] = 0$。
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对于 $f[i][j],ドル我们考虑其转移方式。我们可以枚举第 $j$ 个字母一共使用了多少次,设为 $h,ドル其中 0ドル \leq h \leq \min(i, k),ドル那么我们可以从 $f[i - h][j - 1]$ 转移过来,且第 $j$ 个字母可以在 $i$ 次按键中使用 $h$ 次,方案数为 $\binom{i}{h}$。即:
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$$
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f[i][j] = \sum_{h = 0}^{\min(i, k)} f[i - h][j - 1] \cdot \binom{i}{h}
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$$
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最终答案即为 $f[n][26]$。
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时间复杂度 $O(n \times k \times |\Sigma|),ドル空间复杂度 $O(n \times |\Sigma|)$。其中 $|\Sigma|$ 表示字母表大小。
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<!-- tabs:start -->
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#### Python3
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```python
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class Solution:
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def keyboard(self, k: int, n: int) -> int:
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f = [[0] * 27 for _ in range(n + 1)]
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f[0] = [1] * 27
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mod = 10**9 + 7
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for i in range(1, n + 1):
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for j in range(1, 27):
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for h in range(min(k, i) + 1):
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f[i][j] += f[i - h][j - 1] * comb(i, h)
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f[i][j] %= mod
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return f[n][26]
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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 keyboard(int k, int n) {
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int[][] c = new int[n + 1][k + 1];
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for (int i = 0; i <= n; ++i) {
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c[i][0] = 1;
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}
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final int mod = (int) 1e9 + 7;
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for (int i = 1; i <= n; ++i) {
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for (int j = 1; j <= k; ++j) {
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c[i][j] = (c[i - 1][j - 1] + c[i - 1][j]) % mod;
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}
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}
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int[][] f = new int[n + 1][27];
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Arrays.fill(f[0], 1);
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for (int i = 1; i <= n; ++i) {
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for (int j = 1; j < 27; ++j) {
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for (int h = 0; h <= Math.min(i, k); ++h) {
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f[i][j] = (f[i][j] + (int) ((long) f[i - h][j - 1] * c[i][h] % mod)) % mod;
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}
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}
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}
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return f[n][26];
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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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class Solution {
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public:
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int keyboard(int k, int n) {
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int f[n + 1][27];
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memset(f, 0, sizeof(f));
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for (int j = 0; j < 27; ++j) {
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f[0][j] = 1;
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}
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int c[n + 1][k + 1];
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memset(c, 0, sizeof(c));
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c[0][0] = 1;
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const int mod = 1e9 + 7;
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for (int i = 1; i <= n; ++i) {
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c[i][0] = 1;
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for (int j = 1; j <= k; ++j) {
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c[i][j] = (c[i - 1][j] + c[i - 1][j - 1]) % mod;
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}
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}
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for (int i = 1; i <= n; ++i) {
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for (int j = 1; j < 27; ++j) {
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for (int h = 0; h <= min(i, k); ++h) {
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f[i][j] = (f[i][j] + 1LL * f[i - h][j - 1] * c[i][h] % mod) % mod;
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}
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}
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}
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return f[n][26];
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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 keyboard(k int, n int) int {
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c := make([][]int, n+1)
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for i := range c {
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c[i] = make([]int, k+1)
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c[i][0] = 1
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}
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const mod int = 1e9 + 7
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for i := 1; i <= n; i++ {
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for j := 1; j <= k; j++ {
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c[i][j] = (c[i-1][j-1] + c[i-1][j]) % mod
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}
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}
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f := make([][27]int, n+1)
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for j := range f[0] {
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f[0][j] = 1
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}
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for i := 1; i <= n; i++ {
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for j := 1; j < 27; j++ {
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for h := 0; h <= min(i, k); h++ {
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f[i][j] = (f[i][j] + f[i-h][j-1]*c[i][h]%mod) % mod
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}
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}
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}
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return f[n][26]
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}
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```
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#### TypeScript
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```ts
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function keyboard(k: number, n: number): number {
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const c: number[][] = Array.from({ length: n + 1 }, () => Array(k + 1).fill(0));
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c[0][0] = 1;
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const mod = 10 ** 9 + 7;
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for (let i = 1; i <= n; ++i) {
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c[i][0] = 1;
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for (let j = 1; j <= k; ++j) {
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c[i][j] = (c[i - 1][j - 1] + c[i - 1][j]) % mod;
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}
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}
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const f: number[][] = Array.from({ length: n + 1 }, () => Array(27).fill(0));
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f[0].fill(1);
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for (let i = 1; i <= n; ++i) {
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for (let j = 1; j < 27; ++j) {
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for (let h = 0; h <= Math.min(i, k); ++h) {
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const v = Number((BigInt(f[i - h][j - 1]) * BigInt(c[i][h])) % BigInt(mod));
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f[i][j] = (f[i][j] + v) % mod;
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}
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}
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}
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return f[n][26];
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}
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```
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<!--- tabs:end -->
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<!-- solution:end -->
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<!-- problem:end -->
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class Solution {
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public:
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int keyboard(int k, int n) {
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int f[n + 1][27];
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memset(f, 0, sizeof(f));
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for (int j = 0; j < 27; ++j) {
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f[0][j] = 1;
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}
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int c[n + 1][k + 1];
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memset(c, 0, sizeof(c));
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c[0][0] = 1;
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const int mod = 1e9 + 7;
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for (int i = 1; i <= n; ++i) {
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c[i][0] = 1;
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for (int j = 1; j <= k; ++j) {
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c[i][j] = (c[i - 1][j] + c[i - 1][j - 1]) % mod;
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}
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}
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for (int i = 1; i <= n; ++i) {
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for (int j = 1; j < 27; ++j) {
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for (int h = 0; h <= min(i, k); ++h) {
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f[i][j] = (f[i][j] + 1LL * f[i - h][j - 1] * c[i][h] % mod) % mod;
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}
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}
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}
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return f[n][26];
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}
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};

‎lcp/LCP 25. 古董键盘/Solution.go‎

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func keyboard(k int, n int) int {
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c := make([][]int, n+1)
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for i := range c {
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c[i] = make([]int, k+1)
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c[i][0] = 1
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}
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const mod int = 1e9 + 7
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for i := 1; i <= n; i++ {
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for j := 1; j <= k; j++ {
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c[i][j] = (c[i-1][j-1] + c[i-1][j]) % mod
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}
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}
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f := make([][27]int, n+1)
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for j := range f[0] {
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f[0][j] = 1
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}
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for i := 1; i <= n; i++ {
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for j := 1; j < 27; j++ {
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for h := 0; h <= min(i, k); h++ {
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f[i][j] = (f[i][j] + f[i-h][j-1]*c[i][h]%mod) % mod
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}
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}
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}
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return f[n][26]
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}
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class Solution {
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public int keyboard(int k, int n) {
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int[][] c = new int[n + 1][k + 1];
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for (int i = 0; i <= n; ++i) {
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c[i][0] = 1;
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}
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final int mod = (int) 1e9 + 7;
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for (int i = 1; i <= n; ++i) {
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for (int j = 1; j <= k; ++j) {
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c[i][j] = (c[i - 1][j - 1] + c[i - 1][j]) % mod;
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}
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}
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int[][] f = new int[n + 1][27];
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Arrays.fill(f[0], 1);
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for (int i = 1; i <= n; ++i) {
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for (int j = 1; j < 27; ++j) {
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for (int h = 0; h <= Math.min(i, k); ++h) {
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f[i][j] = (f[i][j] + (int) ((long) f[i - h][j - 1] * c[i][h] % mod)) % mod;
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}
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}
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}
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return f[n][26];
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}
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}

‎lcp/LCP 25. 古董键盘/Solution.py‎

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class Solution:
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def keyboard(self, k: int, n: int) -> int:
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f = [[0] * 27 for _ in range(n + 1)]
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f[0] = [1] * 27
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mod = 10**9 + 7
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for i in range(1, n + 1):
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for j in range(1, 27):
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for h in range(min(k, i) + 1):
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f[i][j] += f[i - h][j - 1] * comb(i, h)
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f[i][j] %= mod
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return f[n][26]

‎lcp/LCP 25. 古董键盘/Solution.ts‎

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function keyboard(k: number, n: number): number {
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const c: number[][] = Array.from({ length: n + 1 }, () => Array(k + 1).fill(0));
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c[0][0] = 1;
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const mod = 10 ** 9 + 7;
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for (let i = 1; i <= n; ++i) {
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c[i][0] = 1;
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for (let j = 1; j <= k; ++j) {
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c[i][j] = (c[i - 1][j - 1] + c[i - 1][j]) % mod;
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}
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}
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const f: number[][] = Array.from({ length: n + 1 }, () => Array(27).fill(0));
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f[0].fill(1);
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for (let i = 1; i <= n; ++i) {
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for (let j = 1; j < 27; ++j) {
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for (let h = 0; h <= Math.min(i, k); ++h) {
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const v = Number((BigInt(f[i - h][j - 1]) * BigInt(c[i][h])) % BigInt(mod));
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f[i][j] = (f[i][j] + v) % mod;
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}
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}
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}
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return f[n][26];
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}

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