
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:
Start with a 2D dynamic programming table
dp[i][j]
wherei
indicates up to thei-th
number andj
counts the numbers of valid inverse pairs. Initialisedp[0][0] = 1
since an array with no numbers has exactly one way to have zero inverse pairs: it's an empty array.Iterate from number
1
ton
. For each number, iterate through possible inverse pairs count from0
tok
, updatingdp[i][j]
based on the preceding values. The key here is determining how placing thei-th
number in the array impacts the count of inverse pairs.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.Updating
dp[i][j]
involves summing possible configurations from previous steps considering valid inverse pairs generated by inserting thei-th
number in possible previous configurations.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.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]
has0
inverse pairs, leading to an answer of1
. - For
n = 3, k = 1
, configurations like[2, 1, 3]
and[1, 3, 2]
provide exactly1
inverse pair, explained by their placements leading to1
cross on the sorted array, leading to an answer of2
.
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
- Java
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);
}
}
This solution involves calculating the number of ways to generate an array where the number of inverse pairs equals a specified target number. The provided Java function, findInversePairs
, uses a dynamic programming approach to solve this problem for arrays of a given size count
.
- Initialize an array
currentArray
, which will track the number of ways to achieve every possible number of inverse pairs up totarget
. - Use a modular arithmetic with
modulo = 1000000007
to ensure that the results fit within standard integer limitations and prevent overflow. - Loop through each possible element count from 1 up to
count
:- Create
nextArray
to temporarily hold the new counts for inverse pairs as you increase the sequence length by one. - Set
nextArray[0] = 1
to initialize the base case where zero inverse pairs can be formed trivially by an array with elements in increasing order. - For each number of inverse pairs from 1 up to
target
, compute the number of ways to form this by adding ways to formj
inverse pairs and differing values up toi
, the current index in the outer loop. - Update
nextArray[j]
using the previously computedcurrentArray
values to track new sums for this particular step. - At the end of inner loop, assign
nextArray
tocurrentArray
to update for the next increasing sequence length iteration.
- Create
- Finally, return the result for the exact
target
of inverse pairs, adjusted with themodulo
to handle any subtractions from previous computations correctly.
In summary, this method uses a tabulation technique in dynamic programming that efficiently calculates possible arrays up to a certain length and specific number of desired inverse pairs by iteratively updating potential counts and making use of previously computed results. Thus, empowering optimal calculation of results for larger inputs.
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