629. K Inverse Pairs Array

Problem

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 10^9 + 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

Solution (Java)

class Solution {
    public int kInversePairs(int n, int k) {
        k = Math.min(k, n * (n - 1) / 2 - k);
        if (k < 0) {
            return 0;
        }
        int[] dp = new int[k + 1];
        int[] dp1 = new int[k + 1];
        dp[0] = 1;
        dp1[0] = 1;
        int mod = 1_000_000_007;
        for (int i = 1; i <= n; i++) {
            int[] temp = dp;
            dp = dp1;
            dp1 = temp;
            for (int j = 1, m = Math.min(k, i * (i - 1) / 2); j <= m; j++) {
                dp[j] = (dp1[j] + dp[j - 1] - (j >= i ? dp1[j - i] : 0)) % mod;
                if (dp[j] < 0) {
                    dp[j] += mod;
                }
            }
        }
        return dp[k];
    }
}

Solution (Python3)

class Solution:
    def kInversePairs(self, n: int, k: int) -> int:
        kMod = int(1e9) + 7
        # dp[i][j] := # of permutations of numbers 1..i with j inverse pairs
        dp = [[0] * (k + 1) for _ in range(n + 1)]

        # if there's no inverse pair, the permutation is unique '123..i'
        for i in range(n + 1):
            dp[i][0] = 1

        for i in range(1, n + 1):
            for j in range(1, k + 1):
                dp[i][j] = (dp[i][j - 1] + dp[i - 1][j]) % kMod
                if j - i >= 0:
                    dp[i][j] = (dp[i][j] - dp[i - 1][j - i] + kMod) % kMod

        return dp[n][k]

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).