Question

629. K Inverse Pairs Array

中文文档

Description

For an integer array nums, an inverse pair is a pair of integers [i, j] where 0 <= i < j < nums.length and nums[i] > nums[j].

Given two integers n and k, return the number of different arrays consisting of numbers from 1 to n such that there are exactly k inverse pairs. Since the answer can be huge, return it modulo 109 + 7.

 

Example 1:

Input: n = 3, k = 0
Output: 1
Explanation: Only the array [1,2,3] which consists of numbers from 1 to 3 has exactly 0 inverse pairs.

Example 2:

Input: n = 3, k = 1
Output: 2
Explanation: The array [1,3,2] and [2,1,3] have exactly 1 inverse pair.

 

Constraints:

• 1 <= n <= 1000
• 0 <= k <= 1000

Solutions

Solution 1

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

class Solution {
  public int kInversePairs(int n, int k) {
    final int mod = (int) 1e9 + 7;
    int[] f = new int[k + 1];
    int[] s = new int[k + 2];
    f[0] = 1;
    Arrays.fill(s, 1);
    s[0] = 0;
    for (int i = 1; i <= n; ++i) {
      for (int j = 1; j <= k; ++j) {
        f[j] = (s[j + 1] - s[Math.max(0, j - (i - 1))] + mod) % mod;
      }
      for (int j = 1; j <= k + 1; ++j) {
        s[j] = (s[j - 1] + f[j - 1]) % mod;
      }
    }
    return f[k];
  }
}

class Solution {
public:
  int kInversePairs(int n, int k) {
    int f[k + 1];
    int s[k + 2];
    memset(f, 0, sizeof(f));
    f[0] = 1;
    fill(s, s + k + 2, 1);
    s[0] = 0;
    const int mod = 1e9 + 7;
    for (int i = 1; i <= n; ++i) {
      for (int j = 1; j <= k; ++j) {
        f[j] = (s[j + 1] - s[max(0, j - (i - 1))] + mod) % mod;
      }
      for (int j = 1; j <= k + 1; ++j) {
        s[j] = (s[j - 1] + f[j - 1]) % mod;
      }
    }
    return f[k];
  }
};

func kInversePairs(n int, k int) int {
  f := make([]int, k+1)
  s := make([]int, k+2)
  f[0] = 1
  for i, x := range f {
    s[i+1] = s[i] + x
  }
  const mod = 1e9 + 7
  for i := 1; i <= n; i++ {
    for j := 1; j <= k; j++ {
      f[j] = (s[j+1] - s[max(0, j-(i-1))] + mod) % mod
    }
    for j := 1; j <= k+1; j++ {
      s[j] = (s[j-1] + f[j-1]) % mod
    }
  }
  return f[k]
}

function kInversePairs(n: number, k: number): number {
  const f: number[] = new Array(k + 1).fill(0);
  f[0] = 1;
  const s: number[] = new Array(k + 2).fill(1);
  s[0] = 0;
  const mod: number = 1e9 + 7;
  for (let i = 1; i <= n; ++i) {
    for (let j = 1; j <= k; ++j) {
      f[j] = (s[j + 1] - s[Math.max(0, j - (i - 1))] + mod) % mod;
    }
    for (let j = 1; j <= k + 1; ++j) {
      s[j] = (s[j - 1] + f[j - 1]) % mod;
    }
  }
  return f[k];
}