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feat: add solutions to lc problem: No.0920
No.0920.Number of Music Playlists
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‎solution/0900-0999/0920.Number of Music Playlists/README.md

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<!-- 这里可写通用的实现逻辑 -->
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**方法一：动态规划**
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我们定义 $f[i][j]$ 表示听 $i$ 首歌，且这 $i$ 首歌中有 $j$ 首不同歌曲的播放列表的数量。初始时 $f[0][0]=1$。答案为 $f[goal][n]$。
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对于 $f[i][j]$，我们可以选择没听过的歌，那么上一个状态为 $f[i - 1][j - 1]$，这样的选择有 $n - (j - 1) = n - j + 1$ 种，因此 $f[i][j] += f[i - 1][j - 1] \times (n - j + 1)$。我们也可以选择听过的歌，那么上一个状态为 $f[i - 1][j]$，这样的选择有 $j - k$ 种，因此 $f[i][j] += f[i - 1][j] \times (j - k)$，其中 $j \geq k$。
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综上，我们可以得到状态转移方程：
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$$
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f[i][j] = \begin{cases}
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1 & i = 0, j = 0 \\
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f[i - 1][j - 1] \times (n - j + 1) + f[i - 1][j] \times (j - k) & i \geq 1, j \geq 1
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\end{cases}
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$$
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最终的答案为 $f[goal][n]$。
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时间复杂度 $O(goal \times n)$，空间复杂度 $O(goal \times n)$。其中 $goal$ 和 $n$ 为题目中给定的参数。
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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 numMusicPlaylists(self, n: int, goal: int, k: int) -> int:
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mod = 10**9 + 7
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f = [[0] * (n + 1) for _ in range(goal + 1)]
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f[0][0] = 1
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for i in range(1, goal + 1):
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for j in range(1, n + 1):
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f[i][j] += f[i - 1][j - 1] * (n - j + 1)
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if j >= k:
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f[i][j] += f[i - 1][j] * (j - k)
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f[i][j] %= mod
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return f[goal][n]
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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 numMusicPlaylists(int n, int goal, int k) {
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final int mod = (int) 1e9 + 7;
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long[][] f = new long[goal + 1][n + 1];
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f[0][0] = 1;
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for (int i = 1; i <= goal; ++i) {
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for (int j = 1; j <= n; ++j) {
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f[i][j] += f[i - 1][j - 1] * (n - j + 1);
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if (j >= k) {
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f[i][j] += f[i - 1][j] * (j - k);
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}
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f[i][j] %= mod;
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}
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}
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return (int) f[goal][n];
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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 numMusicPlaylists(int n, int goal, int k) {
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const int mod = 1e9 + 7;
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long long f[goal + 1][n + 1];
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memset(f, 0, sizeof(f));
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f[0][0] = 1;
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for (int i = 1; i <= goal; ++i) {
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for (int j = 1; j <= n; ++j) {
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f[i][j] += f[i - 1][j - 1] * (n - j + 1);
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if (j >= k) {
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f[i][j] += f[i - 1][j] * (j - k);
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}
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f[i][j] %= mod;
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}
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}
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return f[goal][n];
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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 numMusicPlaylists(n int, goal int, k int) int {
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const mod = 1e9 + 7
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f := make([][]int, goal+1)
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for i := range f {
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f[i] = make([]int, n+1)
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}
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f[0][0] = 1
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for i := 1; i <= goal; i++ {
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for j := 1; j <= n; j++ {
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f[i][j] += f[i-1][j-1] * (n - j + 1)
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if j >= k {
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f[i][j] += f[i-1][j] * (j - k)
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}
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f[i][j] %= mod
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}
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}
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return f[goal][n]
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}
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```
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### **...**

‎solution/0900-0999/0920.Number of Music Playlists/README_EN.md

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### **Python3**
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```python
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class Solution:
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def numMusicPlaylists(self, n: int, goal: int, k: int) -> int:
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mod = 10**9 + 7
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f = [[0] * (n + 1) for _ in range(goal + 1)]
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f[0][0] = 1
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for i in range(1, goal + 1):
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for j in range(1, n + 1):
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f[i][j] += f[i - 1][j - 1] * (n - j + 1)
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if j >= k:
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f[i][j] += f[i - 1][j] * (j - k)
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f[i][j] %= mod
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return f[goal][n]
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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 numMusicPlaylists(int n, int goal, int k) {
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final int mod = (int) 1e9 + 7;
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long[][] f = new long[goal + 1][n + 1];
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f[0][0] = 1;
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for (int i = 1; i <= goal; ++i) {
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for (int j = 1; j <= n; ++j) {
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f[i][j] += f[i - 1][j - 1] * (n - j + 1);
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if (j >= k) {
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f[i][j] += f[i - 1][j] * (j - k);
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}
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f[i][j] %= mod;
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}
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}
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return (int) f[goal][n];
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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 numMusicPlaylists(int n, int goal, int k) {
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const int mod = 1e9 + 7;
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long long f[goal + 1][n + 1];
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memset(f, 0, sizeof(f));
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f[0][0] = 1;
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for (int i = 1; i <= goal; ++i) {
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for (int j = 1; j <= n; ++j) {
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f[i][j] += f[i - 1][j - 1] * (n - j + 1);
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if (j >= k) {
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f[i][j] += f[i - 1][j] * (j - k);
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}
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f[i][j] %= mod;
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}
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}
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return f[goal][n];
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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 numMusicPlaylists(n int, goal int, k int) int {
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const mod = 1e9 + 7
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f := make([][]int, goal+1)
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for i := range f {
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f[i] = make([]int, n+1)
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}
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f[0][0] = 1
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for i := 1; i <= goal; i++ {
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for j := 1; j <= n; j++ {
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f[i][j] += f[i-1][j-1] * (n - j + 1)
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if j >= k {
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f[i][j] += f[i-1][j] * (j - k)
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}
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f[i][j] %= mod
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}
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}
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return f[goal][n]
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}
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```
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### **...**

‎solution/0900-0999/0920.Number of Music Playlists/Solution.cpp

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class Solution {
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public:
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int numMusicPlaylists(int n, int goal, int k) {
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const int mod = 1e9 + 7;
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long long f[goal + 1][n + 1];
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memset(f, 0, sizeof(f));
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f[0][0] = 1;
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for (int i = 1; i <= goal; ++i) {
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for (int j = 1; j <= n; ++j) {
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f[i][j] += f[i - 1][j - 1] * (n - j + 1);
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if (j >= k) {
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f[i][j] += f[i - 1][j] * (j - k);
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}
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f[i][j] %= mod;
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}
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}
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return f[goal][n];
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}
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};

‎solution/0900-0999/0920.Number of Music Playlists/Solution.go

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func numMusicPlaylists(n int, goal int, k int) int {
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const mod = 1e9 + 7
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f := make([][]int, goal+1)
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for i := range f {
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f[i] = make([]int, n+1)
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}
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f[0][0] = 1
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for i := 1; i <= goal; i++ {
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for j := 1; j <= n; j++ {
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f[i][j] += f[i-1][j-1] * (n - j + 1)
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if j >= k {
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f[i][j] += f[i-1][j] * (j - k)
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}
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f[i][j] %= mod
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}
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}
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return f[goal][n]
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}

‎solution/0900-0999/0920.Number of Music Playlists/Solution.java

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class Solution {
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public int numMusicPlaylists(int n, int goal, int k) {
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final int mod = (int) 1e9 + 7;
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long[][] f = new long[goal + 1][n + 1];
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f[0][0] = 1;
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for (int i = 1; i <= goal; ++i) {
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for (int j = 1; j <= n; ++j) {
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f[i][j] += f[i - 1][j - 1] * (n - j + 1);
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if (j >= k) {
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f[i][j] += f[i - 1][j] * (j - k);
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}
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f[i][j] %= mod;
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}
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}
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return (int) f[goal][n];
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}
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}

‎solution/0900-0999/0920.Number of Music Playlists/Solution.py

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class Solution:
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def numMusicPlaylists(self, n: int, goal: int, k: int) -> int:
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mod = 10**9 + 7
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f = [[0] * (n + 1) for _ in range(goal + 1)]
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f[0][0] = 1
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for i in range(1, goal + 1):
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for j in range(1, n + 1):
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f[i][j] += f[i - 1][j - 1] * (n - j + 1)
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if j >= k:
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f[i][j] += f[i - 1][j] * (j - k)
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f[i][j] %= mod
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return f[goal][n]

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