# 629. K Inverse Pairs Array

## 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 consist 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

• 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];
}
};

• 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]

• 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];
}