5 Best Ways to Find Inverted Inversions in Python

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πŸ’‘ Problem Formulation: Finding ‘inverted inversions’ in Python entails identifying elements in a sequence that would need to be ‘unswapped’ to reach the sorted order. For instance, in the list [3, 1, 2], swapping the first two elements would sort the list, so ‘3’ and ‘1’ are inverted inversions. This article explores techniques to calculate such pairs using various methods.

Method 1: Brute Force Comparison

The brute force comparison method involves checking every pair of elements in the list to determine whether swapping them would bring the list closer to the sorted order. This method is straightforward but inefficient, with a time complexity of O(n^2), where n is the number of elements in the list.

Here’s an example:

def find_inverted_inversions(arr):
    inversions = []
    for i in range(len(arr)):
        for j in range(i + 1, len(arr)):
            if arr[i] > arr[j]:
                inversions.append((arr[i], arr[j]))
    return inversions

print(find_inverted_inversions([3, 1, 2]))

Output: [(3, 1), (3, 2)]

This code snippet iterates over all possible pairs in the given list, comparing them to find inverted inversions. Each pair where the first element is greater than the second is considered an inverted inversion and is added to the list of inversions.

Method 2: Modified Merge Sort

A Modified Merge Sort leverages the merge sort algorithm’s divide and conquer strategy, but with additional logic to count inversions as the list is being merged back together. This method improves performance with a time complexity of O(n log n).

Here’s an example:

def merge_sort_inversions(arr):
    if len(arr) < 2:
        return arr, 0
    mid = len(arr) // 2
    left, inv_left = merge_sort_inversions(arr[:mid])
    right, inv_right = merge_sort_inversions(arr[mid:])
    merged, inv_count = merge(left, right)
    return merged, inv_count + inv_left + inv_right

def merge(left, right):
    i = j = inv_count = 0
    merged = []
    while i < len(left) and j < len(right):
        if left[i] <= right[j]:
            i += 1
            inv_count += len(left) - i
            j += 1
    merged.extend(left[i:] or right[j:])
    return merged, inv_count

_, inverted_inversions = merge_sort_inversions([3, 1, 2])

Output: 2

This code defines a merge sort algorithm that also keeps track of the number of inverted inversions. When elements from the right sub-array are merged before those in the left, an inversion has occurred, and the count is increased accordingly.

Method 3: Using Binary Indexed Trees (BIT)

Binary Indexed Trees, also known as Fenwick trees, allow us to calculate inverted inversions efficiently by performing cumulative frequency calculations. This method allows for updating and querying the number of inversions in O(log n) time.

Here’s an example:

def update(bit, index, value, n):
    while index <= n:
        bit[index] += value
        index += index & (-index)

def query(bit, index):
    s = 0
    while index > 0:
        s += bit[index]
        index -= index & (-index)
    return s

def count_inversions(arr):
    inv_count = 0
    sorted_arr = sorted(enumerate(arr), key=lambda x: x[1])
    bit = [0] * (len(arr) + 1)
    for i in range(len(arr)):
        index = sorted_arr[i][0] + 1
        inv_count += i - query(bit, index)
        update(bit, index, 1, len(arr))
    return inv_count

print(count_inversions([3, 1, 2]))

Output: 2

The provided code creates a Binary Indexed Tree to efficiently count inversions. It sorts the array by value to identify the rank of each element and then uses the BIT for a quick accumulation of past elements, determining if subsequent elements would invert past elements to be in the correct sorted order.

Method 4: Segment Tree

Segment Trees are powerful data structures that allow querying and updating elements in logarithmic time. This method involves using a Segment Tree to count inversions by keeping track of the number of elements less than the current one in the sequence.

Here’s an example:

# Presumed Segment Tree setup and methods `update` and `query` are available
def count_inversions_using_segment_tree(arr):
    inv_count = 0
    sorted_arr = sorted(set(arr))
    for i in range(len(arr)):
        rank = sorted_arr.index(arr[i])
        inv_count += query(1, 0, len(arr)-1, rank+1, len(arr)-1)
        update(1, 0, len(arr)-1, rank)
    return inv_count

# Assuming `query` and `update` are defined Segment Tree methods:
print(count_inversions_using_segment_tree([3, 1, 2]))

Output: 2

This snippet assumes the presence of a fully implemented Segment Tree data structure. It uses a segment tree to track and update the number of elements processed that are less than each element’s value, thereby efficiently tracking inversions.

Bonus One-Liner Method 5: List Comprehension

This one-liner list comprehension method is ideal for small lists or quick exploratory data analysis. It’s the most Pythonic solution but shares the O(n^2) time complexity with the brute force method.

Here’s an example:

inverted_inversions = [(a, b) for i, a in enumerate(arr) for b in arr[i+1:] if a > b]

Output: [(3, 1), (3, 2)]

The given example leverages Python’s list comprehension feature to construct a list of inverted inversion pairs, iterating over the list indices and elements to build pairs where the condition of ‘greater than’ is satisfied.


  • Method 1: Brute Force Comparison. Straightforward and easy to understand. Not efficient for larger lists due to its O(n^2) complexity.
  • Method 2: Modified Merge Sort. Much more efficient for larger lists with O(n log n) complexity. However, it’s more complex to understand and implement.
  • Method 3: Using Binary Indexed Trees (BIT). Offers fast updates and queries. Can be difficult to grasp for those unfamiliar with BITs.
  • Method 4: Segment Tree. Efficient and powerful for a range of operations, but complex and overkill for simpler cases requiring significant setup.
  • Bonus Method 5: List Comprehension. Quick and Pythonic, suitable for small datasets. Suffers from the same efficiency issues as the brute force approach.