K Inverse Pairs Array

Problem Statement

In combinatorial mathematics, an inverse pair in an array of integers is defined as a pair [i, j] such that i < j and the element at the i-th index is greater than the element at the j-th index. This characteristic is vital in understanding the degree of sortedness of an array, with inverse pairs indicating a deviation from an ascending order.

The problem presents a challenge where for given integers n and k, we need to determine the number of distinct arrays that can be formed using all the integers from 1 to n in which there are exactly k inverse pairs. The result is expected to be large hence it should be returned under modulo 10^9 + 7 to keep the number manageable.

This brings about a computational complexity given the constraints that both n and k can go up to 1000. The challenge not only lies in generating the possible arrays but also efficiently calculating the inverse pairs to match exactly k.

Examples

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

Approach and Intuition

The problem is directly relatable to dynamic programming where the subproblems break down as calculating the number of arrays possible for smaller values of n and k. Here’s the basic idea and intuition derived from the provided examples:

1. Start with a 2D dynamic programming table dp[i][j] where i indicates up to the i-th number and j counts the numbers of valid inverse pairs. Initialise dp[0][0] = 1 since an array with no numbers has exactly one way to have zero inverse pairs: it's an empty array.

2. Iterate from number 1 to n. For each number, iterate through possible inverse pairs count from 0 to k, updating dp[i][j] based on the preceding values. The key here is determining how placing the i-th number in the array impacts the count of inverse pairs.

3. Since every new number can be inserted in any position of already formed arrays without new inverse pairs up to its position index, this helps update current dp[i][j] considering the sum from previous values up to the allowed positions.

4. Updating dp[i][j] involves summing possible configurations from previous steps considering valid inverse pairs generated by inserting the i-th number in possible previous configurations.

5. Due to the constraints, efficiencies in calculating dp[i][j] are crucial. This can be facilitated using prefix sums to avoid recalculating sums over overlapping subarrays.

6. Finally, because any configuration can produce a result exceeding standard handling, the answer for every update is taken modulo 10^9 + 7.

From the examples:

• For n = 3, k = 0, the only configuration [1, 2, 3] has 0 inverse pairs, leading to an answer of 1.
• For n = 3, k = 1, configurations like [2, 1, 3] and [1, 3, 2] provide exactly 1 inverse pair, explained by their placements leading to 1 cross on the sorted array, leading to an answer of 2.

The constraints hint that naive generation of all possible arrays and counting inverse pairs is computationally expensive and impractical, hence the reliance on dynamic programming. This approach consolidates understanding of both combinatorial generation and use of dynamic structures to manage complex counting in feasible time frames.

Solutions

public class Solution {
    public int findInversePairs(int count, int target) {
        int[] currentArray = new int[target + 1];
        int modulo = 1000000007;
        for (int i = 1; i <= count; i++) {
            int[] nextArray = new int[target + 1];
            nextArray[0] = 1;
            for (int j = 1; j <= target; j++) {
                int value = (currentArray[j] + modulo - ((j - i) >= 0 ? currentArray[j - i] : 0)) % modulo;
                nextArray[j] = (nextArray[j - 1] + value) % modulo;
            }
            currentArray = nextArray;
        }
        return ((currentArray[target] + modulo - (target > 0 ? currentArray[target - 1] : 0)) % modulo);
    }
}