Inversion count in Array using Merge Sort

Do you need a fast, efficient way to keep track of an inversion count for an array of numbers? 

The array's inversion count indicates users how near or how far away the array is from getting sorted. The inversion count is zero if the array is already sorted; nevertheless, it is highest if there is a reverse an array. 

Leveraging the power of Merge Sort algorithms can be one great solution. Using the merge sort method, the left and right sides of the provided array are divided, and both portions are sorted using recursion. 

Then, in order to create our final sorted array, we join the two sorted parts.

This guide will give you complete information on how you can utilise merge sort to solve the inversion count problem. 

Read till the end to become efficient in solving such problems.

But first, let's get a detailed overview of what an inversion count is.

What is Inversion Count?

The quantity of element pairs in an array that are not in chronological order with regard to one another is referred to as the inversion count. For instance, since the pairings (2, 1), (4, 1), and (4, 3) are not in order, the array [2, 4, 1, 3, 5] would have an inversion count of 3. 

Solving Inversion count in Array using Merge Sort

There are numerous methods for counting the count inversions in an array. One approach that is widely employed is merge sort. To use merge sort for solving inversion count, you will first have to divide the arrays in equal halves and then start arranging the arrays in a proper pairing recursively. 

Then, as you start pairing up the combination, then the total of count inversions will be counted at the end.

Let’s understand in detail with an example.

The Problem Statement

In the case of an integer array arr[,] the elements at indexes i and j constitute an inversion, and also the pair i and j is referred to as the inversion of such an array if i and j when arr[i] > arr[j]. To determine the overall inversion counts in an array arr[, you must develop a programme.

Input 1: int arr[] = {3, 2, 5}

Output 1: 3

Explanation: 

Inversion pairs in the given array are 

(3,2), (3,5) and (2,5). Thus, the count of inversion is 3.

Input 2: int arr[] = {2, 3}

Output 2: 1

Explanation: 

Inversion pairs in the given array is (2,1).

Input 3: int arr[] = {3, 3, 3, 2, 2}

Output 3: 0

Explanation: 

Given array is already sorted, so there are no inversions.

Using the Merge Sort, to solve the count inversion problem:

Within the merge sort method, the left and right sides of the provided array are divided, and both portions are sorted using recursion. Then, in order to create our final sorted array, we join the two sorted parts. Depending on the Conquer and divide method, we are aware of merge sort. 

Therefore, the Conquer and divide method will also be used to support this solution. This algorithm repeatedly divides an issue into two or even more subproblems each of the same or similar type, until these are sufficiently straightforward to be solved by itself.

Therefore, we shall split the input array we have been given into two halves. Recursion will be used to obtain the inversion counts for each half. Assume that the array's left and right halves, respectively, have cnt1 and cnt2 inversions each. The elements from both halves are compared to determine the inversion count (pass inversion) during the merge phase. Assume that cnt3 is the inversion count after merging.

Steps to solve the inversion count problem using merge sort

Here are the steps to solve the inversion count problem using merge sort:

• The input array should be split in half.

• Use merge sort to iteratively sort the two parts.

• Combining the two ordered halves while counting the inversions. Use a counter element to keep track of the inversions in order to accomplish this. Set the counter's initial value to 0.

• Compare the components from the two ordered halves as you combine them. An inversion occurs when a component from the initial half is larger than a component from the second half. The counter is increased by how many elements are still present in the initial half.

• As a result, give back the counter.

You can run merge sort on the input array and supply a counter variable that is initialised to zero to use this function to address the inversion count problem. The array will be sorted as a result of merge sort, but the counter parameter will store the inversion count.

For example:

array = [2, 3, 7, 6, 1]

counter = 0

sorted_array = merge_sort(array, counter)

print(counter)  # Output: 5

Some other ways to solve the problem of finding the number of count inversions

Determining the amount of inversions in an array can be done in a variety of different methods. Here are some methods:

Brute Force Method: The array's elements can be compared pair by pair to identify the amount of inversions by counting the instances in which the elements are not in order. For big arrays, this method's temporal complexity of O(n2) makes it ineffective.

Fenwick tree or BIT (Binary Indexed Tree): Another method is to effectively determine the number of inversions by using a Fenwick tree or BIT data structure. This method's time complexity is O. (n log n).

Divide and conquer: The other strategy for resolving the issue is to divide the array into two equal halves, address the issue for each half separately, and then combine the results. This method's time complexity is O. (n log n).

Self-balancing binary search tree: Another approach is to use this binary search tree way to store the elements of the array and find the number of inversions. This approach has a time complexity of O(n log n).

Conclusion

Although there are various other methods for counting inversions, the Merge Sort algorithm provides a fast and efficient solution - especially when confronted with larger arrays. With a little practice, you'll be able to develop a fast and efficient way to keep track of an inversion count - allowing you to keep your arrays sorted or to reverse an array.

Thanks for reading!