feat: add solutions to lcp problem: No.25

lcp/LCP 25. 古董键盘/README.md

@@ -42,4 +42,159 @@ edit_url: https://github.com/doocs/leetcode/edit/main/lcp/LCP%2025.%20%E5%8F%A4%
 
 <!-- solution:start -->
 
+### 方法一：动态规划
+
+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 $1$，表示没有按键时只有一种方案，其余 $f[i][j] = 0$。
+
+对于 $f[i][j]$，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 $h$，其中 $0 \leq h \leq \min(i, k)$，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：
+
+$$
+f[i][j] = \sum_{h = 0}^{\min(i, k)} f[i - h][j - 1] \cdot \binom{i}{h}
+$$
+
+最终答案即为 $f[n][26]$。
+
+时间复杂度 $O(n \times k \times |\Sigma|)$，空间复杂度 $O(n \times |\Sigma|)$。其中 $|\Sigma|$ 表示字母表大小。
+
+<!-- tabs:start -->
+
+#### Python3
+
+```python
+class Solution:
+    def keyboard(self, k: int, n: int) -> int:
+        f = [[0] * 27 for _ in range(n + 1)]
+        f[0] = [1] * 27
+        mod = 10**9 + 7
+        for i in range(1, n + 1):
+            for j in range(1, 27):
+                for h in range(min(k, i) + 1):
+                    f[i][j] += f[i - h][j - 1] * comb(i, h)
+                    f[i][j] %= mod
+        return f[n][26]
+```
+
+#### Java
+
+```java
+class Solution {
+    public int keyboard(int k, int n) {
+        int[][] c = new int[n + 1][k + 1];
+        for (int i = 0; i <= n; ++i) {
+            c[i][0] = 1;
+        }
+        final int mod = (int) 1e9 + 7;
+        for (int i = 1; i <= n; ++i) {
+            for (int j = 1; j <= k; ++j) {
+                c[i][j] = (c[i - 1][j - 1] + c[i - 1][j]) % mod;
+            }
+        }
+        int[][] f = new int[n + 1][27];
+        Arrays.fill(f[0], 1);
+        for (int i = 1; i <= n; ++i) {
+            for (int j = 1; j < 27; ++j) {
+                for (int h = 0; h <= Math.min(i, k); ++h) {
+                    f[i][j] = (f[i][j] + (int) ((long) f[i - h][j - 1] * c[i][h] % mod)) % mod;
+                }
+            }
+        }
+        return f[n][26];
+    }
+}
+```
+
+#### C++
+
+```cpp
+class Solution {
+public:
+    int keyboard(int k, int n) {
+        int f[n + 1][27];
+        memset(f, 0, sizeof(f));
+        for (int j = 0; j < 27; ++j) {
+            f[0][j] = 1;
+        }
+        int c[n + 1][k + 1];
+        memset(c, 0, sizeof(c));
+        c[0][0] = 1;
+        const int mod = 1e9 + 7;
+        for (int i = 1; i <= n; ++i) {
+            c[i][0] = 1;
+            for (int j = 1; j <= k; ++j) {
+                c[i][j] = (c[i - 1][j] + c[i - 1][j - 1]) % mod;
+            }
+        }
+        for (int i = 1; i <= n; ++i) {
+            for (int j = 1; j < 27; ++j) {
+                for (int h = 0; h <= min(i, k); ++h) {
+                    f[i][j] = (f[i][j] + 1LL * f[i - h][j - 1] * c[i][h] % mod) % mod;
+                }
+            }
+        }
+        return f[n][26];
+    }
+};
+```
+
+#### Go
+
+```go
+func keyboard(k int, n int) int {
+	c := make([][]int, n+1)
+	for i := range c {
+		c[i] = make([]int, k+1)
+		c[i][0] = 1
+	}
+	const mod int = 1e9 + 7
+	for i := 1; i <= n; i++ {
+		for j := 1; j <= k; j++ {
+			c[i][j] = (c[i-1][j-1] + c[i-1][j]) % mod
+		}
+	}
+	f := make([][27]int, n+1)
+	for j := range f[0] {
+		f[0][j] = 1
+	}
+	for i := 1; i <= n; i++ {
+		for j := 1; j < 27; j++ {
+			for h := 0; h <= min(i, k); h++ {
+				f[i][j] = (f[i][j] + f[i-h][j-1]*c[i][h]%mod) % mod
+			}
+		}
+	}
+	return f[n][26]
+}
+```
+
+#### TypeScript
+
+```ts
+function keyboard(k: number, n: number): number {
+    const c: number[][] = Array.from({ length: n + 1 }, () => Array(k + 1).fill(0));
+    c[0][0] = 1;
+    const mod = 10 ** 9 + 7;
+    for (let i = 1; i <= n; ++i) {
+        c[i][0] = 1;
+        for (let j = 1; j <= k; ++j) {
+            c[i][j] = (c[i - 1][j - 1] + c[i - 1][j]) % mod;
+        }
+    }
+    const f: number[][] = Array.from({ length: n + 1 }, () => Array(27).fill(0));
+    f[0].fill(1);
+    for (let i = 1; i <= n; ++i) {
+        for (let j = 1; j < 27; ++j) {
+            for (let h = 0; h <= Math.min(i, k); ++h) {
+                const v = Number((BigInt(f[i - h][j - 1]) * BigInt(c[i][h])) % BigInt(mod));
+                f[i][j] = (f[i][j] + v) % mod;
+            }
+        }
+    }
+    return f[n][26];
+}
+```
+
+<!--- tabs:end -->
+
+<!-- solution:end -->
+
 <!-- problem:end -->

lcp/LCP 25. 古董键盘/Solution.cpp

@@ -0,0 +1,28 @@
+class Solution {
+public:
+    int keyboard(int k, int n) {
+        int f[n + 1][27];
+        memset(f, 0, sizeof(f));
+        for (int j = 0; j < 27; ++j) {
+            f[0][j] = 1;
+        }
+        int c[n + 1][k + 1];
+        memset(c, 0, sizeof(c));
+        c[0][0] = 1;
+        const int mod = 1e9 + 7;
+        for (int i = 1; i <= n; ++i) {
+            c[i][0] = 1;
+            for (int j = 1; j <= k; ++j) {
+                c[i][j] = (c[i - 1][j] + c[i - 1][j - 1]) % mod;
+            }
+        }
+        for (int i = 1; i <= n; ++i) {
+            for (int j = 1; j < 27; ++j) {
+                for (int h = 0; h <= min(i, k); ++h) {
+                    f[i][j] = (f[i][j] + 1LL * f[i - h][j - 1] * c[i][h] % mod) % mod;
+                }
+            }
+        }
+        return f[n][26];
+    }
+};

lcp/LCP 25. 古董键盘/Solution.go

@@ -0,0 +1,25 @@
+func keyboard(k int, n int) int {
+	c := make([][]int, n+1)
+	for i := range c {
+		c[i] = make([]int, k+1)
+		c[i][0] = 1
+	}
+	const mod int = 1e9 + 7
+	for i := 1; i <= n; i++ {
+		for j := 1; j <= k; j++ {
+			c[i][j] = (c[i-1][j-1] + c[i-1][j]) % mod
+		}
+	}
+	f := make([][27]int, n+1)
+	for j := range f[0] {
+		f[0][j] = 1
+	}
+	for i := 1; i <= n; i++ {
+		for j := 1; j < 27; j++ {
+			for h := 0; h <= min(i, k); h++ {
+				f[i][j] = (f[i][j] + f[i-h][j-1]*c[i][h]%mod) % mod
+			}
+		}
+	}
+	return f[n][26]
+}

lcp/LCP 25. 古董键盘/Solution.java

@@ -0,0 +1,24 @@
+class Solution {
+    public int keyboard(int k, int n) {
+        int[][] c = new int[n + 1][k + 1];
+        for (int i = 0; i <= n; ++i) {
+            c[i][0] = 1;
+        }
+        final int mod = (int) 1e9 + 7;
+        for (int i = 1; i <= n; ++i) {
+            for (int j = 1; j <= k; ++j) {
+                c[i][j] = (c[i - 1][j - 1] + c[i - 1][j]) % mod;
+            }
+        }
+        int[][] f = new int[n + 1][27];
+        Arrays.fill(f[0], 1);
+        for (int i = 1; i <= n; ++i) {
+            for (int j = 1; j < 27; ++j) {
+                for (int h = 0; h <= Math.min(i, k); ++h) {
+                    f[i][j] = (f[i][j] + (int) ((long) f[i - h][j - 1] * c[i][h] % mod)) % mod;
+                }
+            }
+        }
+        return f[n][26];
+    }
+}

lcp/LCP 25. 古董键盘/Solution.py

@@ -0,0 +1,11 @@
+class Solution:
+    def keyboard(self, k: int, n: int) -> int:
+        f = [[0] * 27 for _ in range(n + 1)]
+        f[0] = [1] * 27
+        mod = 10**9 + 7
+        for i in range(1, n + 1):
+            for j in range(1, 27):
+                for h in range(min(k, i) + 1):
+                    f[i][j] += f[i - h][j - 1] * comb(i, h)
+                    f[i][j] %= mod
+        return f[n][26]

lcp/LCP 25. 古董键盘/Solution.ts

@@ -0,0 +1,22 @@
+function keyboard(k: number, n: number): number {
+    const c: number[][] = Array.from({ length: n + 1 }, () => Array(k + 1).fill(0));
+    c[0][0] = 1;
+    const mod = 10 ** 9 + 7;
+    for (let i = 1; i <= n; ++i) {
+        c[i][0] = 1;
+        for (let j = 1; j <= k; ++j) {
+            c[i][j] = (c[i - 1][j - 1] + c[i - 1][j]) % mod;
+        }
+    }
+    const f: number[][] = Array.from({ length: n + 1 }, () => Array(27).fill(0));
+    f[0].fill(1);
+    for (let i = 1; i <= n; ++i) {
+        for (let j = 1; j < 27; ++j) {
+            for (let h = 0; h <= Math.min(i, k); ++h) {
+                const v = Number((BigInt(f[i - h][j - 1]) * BigInt(c[i][h])) % BigInt(mod));
+                f[i][j] = (f[i][j] + v) % mod;
+            }
+        }
+    }
+    return f[n][26];
+}