What is the time complexity of Minimum Inversion Count in Subarrays of Fixed Length?

The optimal solution for Minimum Inversion Count in Subarrays of Fixed Length runs in O(n log n) time and uses O(n) space. The Fenwick Tree operations are O(log n), and we perform them for each of the n elements.

What is the brute force solution for Minimum Inversion Count in Subarrays of Fixed Length?

The brute force approach for Minimum Inversion Count in Subarrays of Fixed Length is Brute Force, with O(n²) time complexity and O(1) space complexity. We can check every subarray of length k and count inversions by comparing each pair of elements. This is straightforward but inefficient for large arrays. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve Minimum Inversion Count in Subarrays of Fixed Length?

Minimum Inversion Count in Subarrays of Fixed Length uses the following concepts: Array, Segment Tree, Sliding Window. The recommended patterns to study are: Fenwick Tree, Sliding Window.

What are the interview tips for Minimum Inversion Count in Subarrays of Fixed Length?

Explain your thought process clearly. Discuss the trade-offs between brute force and optimal solutions. Practice coding data structures like Fenwick Tree beforehand.

What are common mistakes when solving Minimum Inversion Count in Subarrays of Fixed Length?

Not using a data structure to maintain counts efficiently. Forgetting to compress values when they are not in a small range.

#3768

Minimum Inversion Count in Subarrays of Fixed Length

Hard

ArraySegment TreeSliding WindowFenwick TreeSliding Window

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n log n)
Space	O(1)	O(n)

💡

Intuition

Time O(n log n)Space O(n)

Using a Fenwick Tree (BIT) allows us to efficiently count inversions as we slide the window across the array, maintaining counts dynamically.

⚙️

Algorithm

3 steps

1Step 1: Compress the values in nums to a range of 1 to n.
2Step 2: Initialize a Fenwick Tree to count occurrences of elements in the current window.
3Step 3: For each new element added to the window, update the tree and count inversions, then slide the window by removing the oldest element.

solution.py30 lines

1class FenwickTree:
2    def __init__(self, size):
3        self.size = size
4        self.tree = [0] * (size + 1)
5    def update(self, index, delta):
6        while index <= self.size:
7            self.tree[index] += delta
8            index += index & -index
9    def query(self, index):
10        sum = 0
11        while index > 0:
12            sum += self.tree[index]
13            index -= index & -index
14        return sum
15
16def min_inversion_count(nums, k):
17    compressed = {v: i + 1 for i, v in enumerate(sorted(set(nums))) }
18    ft = FenwickTree(len(compressed))
19    inversions = 0
20    for i in range(k):
21        inversions += i - ft.query(compressed[nums[i]] - 1)
22        ft.update(compressed[nums[i]], 1)
23    min_inversions = inversions
24    for i in range(k, len(nums)):
25        inversions += k - ft.query(compressed[nums[i]])
26        inversions -= ft.query(compressed[nums[i - k]])
27        ft.update(compressed[nums[i]], 1)
28        ft.update(compressed[nums[i - k]], -1)
29        min_inversions = min(min_inversions, inversions)
30    return min_inversions

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Complexity note: The Fenwick Tree operations are O(log n), and we perform them for each of the n elements.

1Inversions count can be efficiently tracked using data structures like Fenwick Tree.
2Sliding window technique helps in maintaining the current state without recalculating from scratch.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.