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Minimum Inversion Count in Subarrays of Fixed Length - Solution & Explanation

HardArraySegment TreeSliding Window4 min read
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Problem Statement

You are given an integer array nums of length n and an integer k.

An inversion is a pair of indices (i, j) from nums such that i < j and nums[i] > nums[j].

The inversion count of a subarray is the number of inversions within it.

Return the minimum inversion count among all subarrays of nums with length k.

 

Example 1:

Input: nums = [3,1,2,5,4], k = 3

Output: 0

Explanation:

We consider all subarrays of length k = 3 (indices below are relative to each subarray):

  • [3, 1, 2] has 2 inversions: (0, 1) and (0, 2).
  • [1, 2, 5] has 0 inversions.
  • [2, 5, 4] has 1 inversion: (1, 2).

The minimum inversion count among all subarrays of length 3 is 0, achieved by subarray [1, 2, 5].

Example 2:

Input: nums = [5,3,2,1], k = 4

Output: 6

Explanation:

There is only one subarray of length k = 4: [5, 3, 2, 1].
Within this subarray, the inversions are: (0, 1), (0, 2), (0, 3), (1, 2), (1, 3), and (2, 3).
Total inversions is 6, so the minimum inversion count is 6.

Example 3:

Input: nums = [2,1], k = 1

Output: 0

Explanation:

All subarrays of length k = 1 contain only one element, so no inversions are possible.
The minimum inversion count is therefore 0.

 

Constraints:

  • 1 <= n == nums.length <= 105
  • 1 <= nums[i] <= 109
  • 1 <= k <= n

Approach Overview

Problem Overview: Given an array and a fixed window size k, examine every contiguous subarray of length k and compute its inversion count (pairs i < j where arr[i] > arr[j]). The goal is to return the minimum inversion count among all such windows.

Approach 1: Brute Force Window Enumeration (O(n * k log k) time, O(k) space)

Iterate through every subarray of size k. For each window, compute the inversion count independently. A simple method copies the window, then counts inversions using merge sort or nested loops. Merge-sort-based counting takes O(k log k) per window, while naive pair comparison takes O(k²). Since there are n - k + 1 windows, total time becomes O(n * k log k). This approach is straightforward but recomputes inversion counts from scratch for overlapping windows, which wastes work.

Approach 2: Sliding Window + Fenwick Tree / Segment Tree (O(n log n) time, O(n) space)

The key observation: adjacent windows differ by exactly one removed element and one inserted element. Instead of recomputing the entire inversion count, update it incrementally. Maintain element frequencies in a Fenwick Tree (or a segment tree) after applying coordinate compression to the array values.

First build the inversion count for the initial window of size k. For each element inserted, query how many existing elements are greater than it to determine new inversions. Then slide the window. When removing the leftmost element x, subtract the inversions it formed with elements to its right by querying how many values in the structure are smaller than x. After removing it from the tree, insert the new element entering the window and add the inversions it forms with current elements (values greater than it). Each update requires O(log n) operations.

This incremental maintenance keeps the inversion count updated while scanning the array once. The technique combines a sliding window with efficient order statistics queries on a dynamic frequency structure. Total complexity becomes O(n log n), which handles large arrays comfortably.

Recommended for interviews: The sliding window with Fenwick Tree or Segment Tree is the expected solution. Interviewers often look for the insight that overlapping windows allow incremental updates. Showing the brute-force idea demonstrates understanding of inversion counting, but optimizing it with a logarithmic data structure shows strong algorithmic skill.

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Detailed Complexity Analysis

ApproachTimeSpaceWhen to Use
Brute Force with Merge Sort Inversion CountO(n * k log k)O(k)Small arrays or when demonstrating the classic inversion counting technique
Sliding Window + Fenwick TreeO(n log n)O(n)General case with large inputs where windows overlap and updates must be efficient
Sliding Window + Segment TreeO(n log n)O(n)Alternative to Fenwick Tree when range queries and updates need more flexibility

Video Solution

3768. Minimum Inversion Count in Subarrays of Fixed Length (Leetcode Hard) • Programming Live with Larry • 183 views views

Frequently Asked Questions

Is Minimum Inversion Count in Subarrays of Fixed Length easy or hard?
Minimum Inversion Count in Subarrays of Fixed Length is typically classified as a hard problem. The challenge comes from maintaining inversion counts dynamically as the window moves. Solving it efficiently requires understanding sliding window mechanics together with advanced data structures like Fenwick Trees or Segment Trees.
Minimum Inversion Count in Subarrays of Fixed Length Python/Java solution
Python, Java, C++, and Go implementations usually follow the same structure: compress the values, maintain a Fenwick Tree for frequencies, compute the initial window inversion count, and update it as the window slides. Each step performs logarithmic queries and updates, keeping the overall complexity at O(n log n).
How to solve Minimum Inversion Count in Subarrays of Fixed Length in O(n log n)?
Maintain a Fenwick Tree that stores the frequency of elements inside the current window. Compute the inversion count for the first window, then slide the window one position at a time. Remove the leftmost element and subtract its inversion contribution, then insert the new element and add its inversion contribution using range queries. Each update costs O(log n), leading to O(n log n) overall.
What is the best approach for Minimum Inversion Count in Subarrays of Fixed Length?
The most efficient approach uses a sliding window combined with a Fenwick Tree or Segment Tree. Build the inversion count for the first window, then update it as the window slides by removing one element and inserting another. Each update requires logarithmic queries for counting smaller or greater elements. The total time complexity becomes O(n log n).
Is Minimum Inversion Count in Subarrays of Fixed Length asked at Google/Amazon/Meta?
Problems involving inversion counting with sliding windows and Fenwick Trees appear in interviews at companies like Google, Amazon, and Meta. Variants test knowledge of order statistics structures, dynamic inversion updates, and efficient window maintenance. The combination of data structures and window techniques makes it a common hard-level interview problem.
What data structure is used in Minimum Inversion Count in Subarrays of Fixed Length?
The typical data structures are Fenwick Trees (Binary Indexed Trees) or Segment Trees. They support fast prefix queries and updates, which allows counting how many elements are smaller or larger than a given value inside the window. Coordinate compression is often applied so the structure can operate on value ranks.
What is the time complexity of Minimum Inversion Count in Subarrays of Fixed Length?
The optimal solution runs in O(n log n) time using a sliding window with a Fenwick Tree or Segment Tree. Each insertion, deletion, or inversion query takes O(log n), and the window slides across the array once. The brute force approach can take O(n * k log k) or even O(n * k^2) depending on how inversions are counted.

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