Question

1621. Number of Sets of K Non-Overlapping Line Segments

中文文档

Description

Given n points on a 1-D plane, where the ith point (from 0 to n-1) is at x = i, find the number of ways we can draw exactly k non-overlapping line segments such that each segment covers two or more points. The endpoints of each segment must have integral coordinates. The k line segments do not have to cover all n points, and they are allowed to share endpoints.

Return _the number of ways we can draw _k_ non-overlapping line segments__._ Since this number can be huge, return it modulo 109 + 7.

 

Example 1:

Input: n = 4, k = 2
Output: 5
Explanation: The two line segments are shown in red and blue.
The image above shows the 5 different ways {(0,2),(2,3)}, {(0,1),(1,3)}, {(0,1),(2,3)}, {(1,2),(2,3)}, {(0,1),(1,2)}.

Example 2:

Input: n = 3, k = 1
Output: 3
Explanation: The 3 ways are {(0,1)}, {(0,2)}, {(1,2)}.

Example 3:

Input: n = 30, k = 7
Output: 796297179
Explanation: The total number of possible ways to draw 7 line segments is 3796297200. Taking this number modulo 109 + 7 gives us 796297179.

 

Constraints:

• 2 <= n <= 1000
• 1 <= k <= n-1

Solutions

Solution 1

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

class Solution {
  private static final int MOD = (int) 1e9 + 7;

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

class Solution {
public:
  int f[1010][1010];
  int g[1010][1010];
  const int mod = 1e9 + 7;

  int numberOfSets(int n, int k) {
    memset(f, 0, sizeof(f));
    memset(g, 0, sizeof(g));
    f[1][0] = 1;
    for (int i = 2; i <= n; ++i) {
      for (int j = 0; j <= k; ++j) {
        f[i][j] = (f[i - 1][j] + g[i - 1][j]) % mod;
        g[i][j] = g[i - 1][j];
        if (j > 0) {
          g[i][j] += f[i - 1][j - 1];
          g[i][j] %= mod;
          g[i][j] += g[i - 1][j - 1];
          g[i][j] %= mod;
        }
      }
    }
    return (f[n][k] + g[n][k]) % mod;
  }
};

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

function numberOfSets(n: number, k: number): number {
  const f = Array.from({ length: n + 1 }, _ => new Array(k + 1).fill(0));
  const g = Array.from({ length: n + 1 }, _ => new Array(k + 1).fill(0));
  f[1][0] = 1;
  const mod = 10 ** 9 + 7;
  for (let i = 2; i <= n; ++i) {
    for (let j = 0; j <= k; ++j) {
      f[i][j] = (f[i - 1][j] + g[i - 1][j]) % mod;
      g[i][j] = g[i - 1][j];
      if (j) {
        g[i][j] += f[i - 1][j - 1];
        g[i][j] %= mod;
        g[i][j] += g[i - 1][j - 1];
        g[i][j] %= mod;
      }
    }
  }
  return (f[n][k] + g[n][k]) % mod;
}