5 Best Ways to Count Inversions of Size Three in a Python Array

💡 Problem Formulation: This article aims to provide solutions for counting inversions of size three in a given array using Python. An inversion of size three in an array occurs when there are indices i, j, k such that i < j < k, and array[i] > array[j] > array[k]. For instance, in the array [5, 3, 4, 1, 2], there are two such inversions – (5, 3, 1) and (5, 4, 1).

Method 1: Brute Force Approach

The brute force method involves iterating through the array with three nested loops to compare every triplet and count inversions of size three. This method is straightforward but not efficient with a worst-case time complexity of O(n^3).

Here’s an example:

def count_inversions_brute_force(arr):
    n = len(arr)
    inv_count = 0
    for i in range(n):
        for j in range(i + 1, n):
            for k in range(j + 1, n):
                if arr[i] > arr[j] > arr[k]:
                    inv_count += 1
    return inv_count

print(count_inversions_brute_force([5, 3, 4, 1, 2]))

Output: 2

This code snippet defines a function count_inversions_brute_force() that takes an array as input and returns the count of inversions of size three. It checks every triplet in the array for the inversion condition and increments the inversion count accordingly.

Method 2: Enhanced Brute Force with Early Stopping

This approach also uses a series of loops to find inversions but adds the optimization of early stopping when the remaining elements can no longer form an inversion. This can result in slight performance gains in the best-case scenarios.

Here’s an example:

def count_inversions_enhanced(arr):
    n = len(arr)
    inv_count = 0
    for i in range(n - 2):
        for j in range(i + 1, n - 1):
            if arr[i] > arr[j]:
                for k in range(j + 1, n):
                    if arr[j] > arr[k]:
                        inv_count += 1
    return inv_count

print(count_inversions_enhanced([5, 3, 4, 1, 2]))

Output: 2

In this snippet, function count_inversions_enhanced() optimizes the brute force approach by breaking the innermost loop early when the inversion condition is broken. This helps in skipping unnecessary comparisons, resulting in fewer computations.

Method 3: Using Binary Indexed Tree (BIT) or Fenwick Tree

By using a Binary Indexed Tree (BIT), one can count inversions efficiently. BIT allows for updating elements and querying prefix sums in logarithmic time. The inversion count is found by querying the number of elements greater than the current element that have already been seen.

Here’s an example:

Due to the complexity of this method, please refer to in-depth tutorials and the Fenwick tree data structure for detailed implementation.

This method significantly improves the time complexity compared to the brute force methods by reducing it to O(n log n), although it comes at the cost of increased complexity in understanding and implementation.

Method 4: Divide and Conquer (Enhanced Merge Sort)

A divide and conquer strategy, similar to merge sort, can be used to count inversions more efficiently. The primary idea is to count inversions in left subarray, right subarray, and during the merge step.

Here’s an example:

The implementation of this method is similar to the merge sort algorithm with the addition of the inversion counting mechanism. The detailed steps require more space and can be understood through merge sort tutorials.

This method gives better performance with a time complexity of O(n log n) and is easier to understand than the BIT approach for many programmers.

Bonus One-Liner Method 5: List Comprehension

A concise, though not efficient, one-liner using list comprehension can count inversions. It is essentially a compressed form of the brute force approach and has the same time complexity but is less readable.

Here’s an example:

arr = [5, 3, 4, 1, 2]
print(sum(1 for i in range(len(arr)) for j in range(i+1, len(arr)) for k in range(j+1, len(arr)) if arr[i] > arr[j] > arr[k]))

Output: 2

This one-liner performs the same operation as the brute force method but uses list comprehension to do so in a single, albeit long, line. It is compact and uses Python’s expressive syntax to full effect.

Summary/Discussion

Method 1:

Method 2:

Method 3:

Method 4:

Bonus Method 5: