#629

Hard

auto_awesomeState transition dynamic programming

LeetCode Problem Workspace

K Inverse Pairs Array

The K Inverse Pairs Array problem focuses on counting arrays with exactly k inverse pairs using dynamic programming.

Problem Statement

You are given two integers, n and k, and tasked with finding the number of different arrays made up of integers from 1 to n that contain exactly k inverse pairs. An inverse pair is defined as a pair of integers [i, j] where 0 <= i < j < n and nums[i] > nums[j]. Since the result may be large, return it modulo 10^9 + 7.

The problem can be solved using dynamic programming. A DP table can help in tracking the count of inverse pairs for different array configurations, and transitions can be formulated based on the properties of inversions between array elements.

Examples

Example 1

Input: n = 3, k = 0

Output: 1

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

The array [1,3,2] and [2,1,3] have exactly 1 inverse pair.

Constraints

1 <= n <= 1000
0 <= k <= 1000

Solution Approach

Dynamic Programming Approach

The solution involves using dynamic programming to track the number of arrays that form exactly k inverse pairs. We construct a DP table where dp[i][j] represents the number of ways to form an array of size i with exactly j inverse pairs. Transitions are based on adding elements to the array, with each addition affecting the number of inverse pairs.

State Transitions

State transitions occur when new elements are added to the array, where the number of inverse pairs increases depending on the position of the new element. The key is to understand how adding an element contributes to existing inverse pairs and how to calculate the new number of inverse pairs efficiently using cumulative sums.

Modulo Arithmetic

Since the result can be large, all calculations are performed modulo 10^9 + 7. This ensures that the number of valid arrays remains manageable and prevents overflow while keeping the solution correct.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity of this solution is O(n * k), where n is the size of the array and k is the number of inverse pairs. This is due to the dynamic programming table filling. The space complexity is also O(n * k) because of the storage required for the DP table.

What Interviewers Usually Probe

Ensure the candidate can clearly explain dynamic programming table construction.
Ask the candidate to discuss the role of modulo arithmetic in maintaining large results.
Test the candidate's understanding of how state transitions occur with array element additions.

Common Pitfalls or Variants

Common pitfalls

Failing to handle state transitions properly, which leads to incorrect calculations of inverse pairs.
Incorrectly applying the modulo operation, causing overflow or incorrect results.
Not understanding the time and space complexity trade-offs of the dynamic programming solution.

Follow-up variants

Optimizing space complexity by reducing the size of the DP table.
Using a more advanced approach to optimize the time complexity further.
Solving the problem with a non-DP approach, such as a divide-and-conquer algorithm.

FAQ

What is the K Inverse Pairs Array problem?

The K Inverse Pairs Array problem asks for the number of arrays of size n, formed by numbers from 1 to n, that contain exactly k inverse pairs.

How do you approach solving K Inverse Pairs Array?

The problem is solved using dynamic programming. A table is created to track the number of ways to form arrays with exact inverse pairs, and transitions are made based on adding elements.

What is the time complexity of solving K Inverse Pairs Array?

The time complexity is O(n * k) where n is the size of the array and k is the number of inverse pairs.

Why do we use modulo 10^9 + 7 in this problem?

We use modulo 10^9 + 7 to prevent overflow and ensure the results fit within manageable limits.

Can this problem be solved without dynamic programming?

While dynamic programming is the most efficient approach, alternative methods such as divide-and-conquer or brute force could be used but are not optimal.

terminal

Solution

Solution 1: Dynamic Programming + Prefix Sum

We define $f[i][j]$ as the number of arrays of length $i$ with $j$ inverse pairs. Initially, $f[0][0] = 1$, and the rest $f[i][j] = 0$.

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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]

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