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feat: add solutions to lc problem: No.1621
No.1621.Number of Sets of K Non-Overlapping Line Segments
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‎solution/1600-1699/1621.Number of Sets of K Non-Overlapping Line Segments/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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记 $f[i][j]$ 表示使用前 $i$ 个点构造了 $j$ 条线段，且最后一条线段的右端点不为 $i$ 的方案数；记 $g[i][j]$ 表示使用了前 $i$ 个点构造了 $j$ 条线段，且最后一条线段的右端点为 $i$ 的方案数。初始时 $f[1][0]=1$。
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考虑 $f[i][j]$，由于第 $j$ 条线段的右端点不为 $i$，因此前 $i-1$ 个点构造了 $j$ 条线段，因此有：
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$$
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f[i][j] = f[i-1][j] + g[i - 1][j]
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$$
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考虑 $g[i][j]$，第 $j$ 条线段的右端点为 $i$，如果第 $j$ 条线段的长度超过 $1$，则前 $i-1$ 个点构造了 $j$ 条线段，且第 $j$ 条线段的右端点一定覆盖了 $i-1$，因此有：
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$$
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g[i][j] = g[i - 1][j]
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$$
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如果第 $j$ 条线段的长度为 $1$，则前 $i-1$ 个点构造了 $j-1$ 条线段，有：
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$$
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g[i][j] = f[i - 1][j - 1] + g[i - 1][j - 1]
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$$
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答案为 $f[n][k]+g[n][k]$。
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时间复杂度 $O(n\times k)$，空间复杂度 $O(n\times k)$。
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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 numberOfSets(self, n: int, k: int) -> int:
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mod = 10**9 + 7
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f = [[0] * (k + 1) for _ in range(n + 1)]
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g = [[0] * (k + 1) for _ in range(n + 1)]
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f[1][0] = 1
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for i in range(2, n + 1):
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for j in range(k + 1):
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f[i][j] = (f[i - 1][j] + g[i - 1][j]) % mod
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g[i][j] = g[i - 1][j]
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if j:
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g[i][j] += f[i - 1][j - 1]
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g[i][j] %= mod
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g[i][j] += g[i - 1][j - 1]
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g[i][j] %= mod
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return (f[-1][-1] + g[-1][-1]) % 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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private static final int MOD = (int) 1e9 + 7;
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public int numberOfSets(int n, int k) {
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int[][] f = new int[n + 1][k + 1];
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int[][] g = new int[n + 1][k + 1];
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f[1][0] = 1;
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for (int i = 2; i <= n; ++i) {
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for (int j = 0; j <= k; ++j) {
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f[i][j] = (f[i - 1][j] + g[i - 1][j]) % MOD;
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g[i][j] = g[i - 1][j];
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if (j > 0) {
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g[i][j] += f[i - 1][j - 1];
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g[i][j] %= MOD;
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g[i][j] += g[i - 1][j - 1];
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g[i][j] %= MOD;
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}
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}
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}
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return (f[n][k] + g[n][k]) % 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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class Solution {
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public:
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int f[1010][1010];
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int g[1010][1010];
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const int mod = 1e9 + 7;
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int numberOfSets(int n, int k) {
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memset(f, 0, sizeof(f));
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memset(g, 0, sizeof(g));
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f[1][0] = 1;
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for (int i = 2; i <= n; ++i) {
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for (int j = 0; j <= k; ++j) {
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f[i][j] = (f[i - 1][j] + g[i - 1][j]) % mod;
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g[i][j] = g[i - 1][j];
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if (j > 0) {
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g[i][j] += f[i - 1][j - 1];
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g[i][j] %= mod;
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g[i][j] += g[i - 1][j - 1];
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g[i][j] %= mod;
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}
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}
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}
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return (f[n][k] + g[n][k]) % 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 numberOfSets(n int, k int) int {
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f := make([][]int, n+1)
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g := make([][]int, n+1)
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for i := range f {
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f[i] = make([]int, k+1)
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g[i] = make([]int, k+1)
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}
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f[1][0] = 1
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var mod int = 1e9 + 7
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for i := 2; i <= n; i++ {
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for j := 0; j <= k; j++ {
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f[i][j] = (f[i-1][j] + g[i-1][j]) % mod
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g[i][j] = g[i-1][j]
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if j > 0 {
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g[i][j] += f[i-1][j-1]
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g[i][j] %= mod
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g[i][j] += g[i-1][j-1]
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g[i][j] %= mod
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}
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}
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}
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return (f[n][k] + g[n][k]) % mod
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}
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```
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### **...**

‎solution/1600-1699/1621.Number of Sets of K Non-Overlapping Line Segments/README_EN.md

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### **Python3**
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```python
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class Solution:
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def numberOfSets(self, n: int, k: int) -> int:
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mod = 10**9 + 7
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f = [[0] * (k + 1) for _ in range(n + 1)]
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g = [[0] * (k + 1) for _ in range(n + 1)]
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f[1][0] = 1
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for i in range(2, n + 1):
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for j in range(k + 1):
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f[i][j] = (f[i - 1][j] + g[i - 1][j]) % mod
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g[i][j] = g[i - 1][j]
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if j:
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g[i][j] += f[i - 1][j - 1]
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g[i][j] %= mod
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g[i][j] += g[i - 1][j - 1]
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g[i][j] %= mod
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return (f[-1][-1] + g[-1][-1]) % 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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private static final int MOD = (int) 1e9 + 7;
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public int numberOfSets(int n, int k) {
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int[][] f = new int[n + 1][k + 1];
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int[][] g = new int[n + 1][k + 1];
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f[1][0] = 1;
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for (int i = 2; i <= n; ++i) {
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for (int j = 0; j <= k; ++j) {
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f[i][j] = (f[i - 1][j] + g[i - 1][j]) % MOD;
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g[i][j] = g[i - 1][j];
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if (j > 0) {
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g[i][j] += f[i - 1][j - 1];
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g[i][j] %= MOD;
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g[i][j] += g[i - 1][j - 1];
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g[i][j] %= MOD;
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}
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}
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}
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return (f[n][k] + g[n][k]) % 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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class Solution {
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public:
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int f[1010][1010];
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int g[1010][1010];
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const int mod = 1e9 + 7;
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int numberOfSets(int n, int k) {
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memset(f, 0, sizeof(f));
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memset(g, 0, sizeof(g));
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f[1][0] = 1;
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for (int i = 2; i <= n; ++i) {
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for (int j = 0; j <= k; ++j) {
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f[i][j] = (f[i - 1][j] + g[i - 1][j]) % mod;
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g[i][j] = g[i - 1][j];
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if (j > 0) {
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g[i][j] += f[i - 1][j - 1];
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g[i][j] %= mod;
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g[i][j] += g[i - 1][j - 1];
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g[i][j] %= mod;
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}
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}
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}
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return (f[n][k] + g[n][k]) % 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 numberOfSets(n int, k int) int {
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f := make([][]int, n+1)
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g := make([][]int, n+1)
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for i := range f {
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f[i] = make([]int, k+1)
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g[i] = make([]int, k+1)
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}
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f[1][0] = 1
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var mod int = 1e9 + 7
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for i := 2; i <= n; i++ {
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for j := 0; j <= k; j++ {
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f[i][j] = (f[i-1][j] + g[i-1][j]) % mod
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g[i][j] = g[i-1][j]
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if j > 0 {
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g[i][j] += f[i-1][j-1]
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g[i][j] %= mod
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g[i][j] += g[i-1][j-1]
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g[i][j] %= mod
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}
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}
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}
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return (f[n][k] + g[n][k]) % mod
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}
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```
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### **...**

‎solution/1600-1699/1621.Number of Sets of K Non-Overlapping Line Segments/Solution.cpp

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class Solution {
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public:
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int f[1010][1010];
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int g[1010][1010];
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const int mod = 1e9 + 7;
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int numberOfSets(int n, int k) {
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memset(f, 0, sizeof(f));
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memset(g, 0, sizeof(g));
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f[1][0] = 1;
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for (int i = 2; i <= n; ++i) {
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for (int j = 0; j <= k; ++j) {
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f[i][j] = (f[i - 1][j] + g[i - 1][j]) % mod;
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g[i][j] = g[i - 1][j];
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if (j > 0) {
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g[i][j] += f[i - 1][j - 1];
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g[i][j] %= mod;
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g[i][j] += g[i - 1][j - 1];
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g[i][j] %= mod;
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}
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}
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}
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return (f[n][k] + g[n][k]) % mod;
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}
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};

‎solution/1600-1699/1621.Number of Sets of K Non-Overlapping Line Segments/Solution.go

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func numberOfSets(n int, k int) int {
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f := make([][]int, n+1)
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g := make([][]int, n+1)
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for i := range f {
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f[i] = make([]int, k+1)
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g[i] = make([]int, k+1)
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}
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f[1][0] = 1
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var mod int = 1e9 + 7
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for i := 2; i <= n; i++ {
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for j := 0; j <= k; j++ {
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f[i][j] = (f[i-1][j] + g[i-1][j]) % mod
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g[i][j] = g[i-1][j]
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if j > 0 {
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g[i][j] += f[i-1][j-1]
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g[i][j] %= mod
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g[i][j] += g[i-1][j-1]
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g[i][j] %= mod
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}
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}
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}
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return (f[n][k] + g[n][k]) % mod
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}

‎solution/1600-1699/1621.Number of Sets of K Non-Overlapping Line Segments/Solution.java

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class Solution {
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private static final int MOD = (int) 1e9 + 7;
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public int numberOfSets(int n, int k) {
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int[][] f = new int[n + 1][k + 1];
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int[][] g = new int[n + 1][k + 1];
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f[1][0] = 1;
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for (int i = 2; i <= n; ++i) {
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for (int j = 0; j <= k; ++j) {
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f[i][j] = (f[i - 1][j] + g[i - 1][j]) % MOD;
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g[i][j] = g[i - 1][j];
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if (j > 0) {
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g[i][j] += f[i - 1][j - 1];
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g[i][j] %= MOD;
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g[i][j] += g[i - 1][j - 1];
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g[i][j] %= MOD;
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}
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}
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}
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return (f[n][k] + g[n][k]) % MOD;
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}
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}

‎solution/1600-1699/1621.Number of Sets of K Non-Overlapping Line Segments/Solution.py

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class Solution:
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def numberOfSets(self, n: int, k: int) -> int:
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mod = 10**9 + 7
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f = [[0] * (k + 1) for _ in range(n + 1)]
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g = [[0] * (k + 1) for _ in range(n + 1)]
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f[1][0] = 1
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for i in range(2, n + 1):
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for j in range(k + 1):
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f[i][j] = (f[i - 1][j] + g[i - 1][j]) % mod
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g[i][j] = g[i - 1][j]
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if j:
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g[i][j] += f[i - 1][j - 1]
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g[i][j] %= mod
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g[i][j] += g[i - 1][j - 1]
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g[i][j] %= mod
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return (f[-1][-1] + g[-1][-1]) % mod

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