629. K Inverse Pairs Array

Explanation

To solve this problem, we can use dynamic programming. We can define a 2D array dp where dp[i][j] represents the number of different arrays of size i with j inverse pairs. We can then iterate through each element in the array and calculate the number of inverse pairs it can form with the previous elements.

Initialize a 2D array dp of size (n+1) x (k+1) with all values set to 0 except dp[0][0] = 1.
Iterate from i = 1 to n and j = 0 to k: a. Calculate dp[i][j] based on the formula: dp[i][j] = dp[i-1][j] + dp[i-1][j-1] + ... + dp[i-1][j-i+1]
Return dp[n][k] modulo 10^9 + 7.

Time Complexity: O(nk^2) Space Complexity: O(nk)

View Leetcode Problem

class Solution {
    public int kInversePairs(int n, int k) {
        int MOD = 1000000007;
        int[][] dp = new int[n+1][k+1];
        dp[0][0] = 1;
        
        for (int i = 1; i <= n; i++) {
            dp[i][0] = 1;
            for (int j = 1; j <= k; j++) {
                for (int p = 0; p <= Math.min(j, i-1); p++) {
                    dp[i][j] = (dp[i][j] + dp[i-1][j-p]) % MOD;
                }
            }
        }
        
        return dp[n][k];
    }
}

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