Mastering the Counting Inversions Problem on LeetCode

Counting inversions is a classic problem in computer science that has applications in sorting algorithms, data analysis, and more. If you’re working through algorithmic challenges on LeetCode, you might encounter this problem, which can help you understand the intricacies of sorting and inversion counting. In this article, we’ll explore the “Count Inversions” problem, discuss its significance, and provide a step-by-step solution.

What Is the Counting Inversions Problem?

Problem Statement: Given an array of integers, an inversion is a pair of indices $(i,j)$ such that $i<j$ and $arr[i]>arr[j]$. The task is to count the total number of inversions in the array.

For example, given the array $[2,3,8,6,1]$, the inversions are:

• $(2,1)$ because 2 > 1 and 2 appears before 1.
• $(3,1)$ because 3 > 1 and 3 appears before 1.
• $(8,6)$ because 8 > 6 and 8 appears before 6.
• $(8,1)$ because 8 > 1 and 8 appears before 1.
• $(6,1)$ because 6 > 1 and 6 appears before 1.

So, the total number of inversions in the array is 5.

Why Is Counting Inversions Important?

Counting inversions is more than a theoretical exercise; it has practical applications in:

• Sorting Algorithms: Inversions can help in analyzing the efficiency of sorting algorithms.
• Data Analysis: Understanding the order and relationships between elements can provide insights into data distributions.
• Computational Complexity: It serves as a benchmark for algorithms and can be used to study algorithmic efficiency and complexity.

Efficient Approach to Counting Inversions

While a naive approach would involve a nested loop, resulting in O(n2)O(n^2)O(n2) time complexity, we can solve this problem more efficiently using a modified merge sort algorithm with $O(n\log n)$ time complexity.

Merge Sort Approach

1. Divide: Recursively split the array into two halves until you reach individual elements.
2. Conquer: Merge the two halves while counting inversions.
3. Combine: Count inversions across the split by counting how many elements from the right half are less than elements in the left half during the merge process.

Python Code Implementation

Here’s how you can implement this approach in Python:

def countInversions(arr):
    def merge_count_split_inv(arr, temp_arr, left, mid, right):
        i = left    # Starting index for left subarray
        j = mid + 1 # Starting index for right subarray
        k = left    # Starting index to be sorted
        inv_count = 0

        # Conditions are checked to ensure that i doesn't exceed mid and j doesn't exceed right
        while i <= mid and j <= right:
            if arr[i] <= arr[j]:
                temp_arr[k] = arr[i]
                i += 1
            else:
                # There are mid - i inversions, because all the remaining elements in the left subarray
                # (arr[i], arr[i+1], ..., arr[mid]) are greater than arr[j]
                temp_arr[k] = arr[j]
                inv_count += (mid - i + 1)
                j += 1
            k += 1

        # Copy the remaining elements of left subarray, if any
        while i <= mid:
            temp_arr[k] = arr[i]
            i += 1
            k += 1

        # Copy the remaining elements of right subarray, if any
        while j <= right:
            temp_arr[k] = arr[j]
            j += 1
            k += 1

        # Copy the sorted subarray into Original array
        for i in range(left, right + 1):
            arr[i] = temp_arr[i]
        
        return inv_count

    def merge_sort_and_count(arr, temp_arr, left, right):
        inv_count = 0
        if left < right:
            mid = (left + right)//2

            inv_count += merge_sort_and_count(arr, temp_arr, left, mid)
            inv_count += merge_sort_and_count(arr, temp_arr, mid + 1, right)
            inv_count += merge_count_split_inv(arr, temp_arr, left, mid, right)

        return inv_count

    temp_arr = [0]*len(arr)
    return merge_sort_and_count(arr, temp_arr, 0, len(arr) - 1)

# Example usage:
arr = [2, 3, 8, 6, 1]
print(countInversions(arr))  # Output: 5

Explanation of the Code

1. merge_count_split_inv: This function merges two sorted halves of the array while counting inversions. It tracks how many times elements from the right half are smaller than elements from the left half.
2. merge_sort_and_count: This function recursively sorts the array and counts inversions using the merge function.
3. countInversions: This is the main function that initializes the temporary array and starts the merge sort process.

Conclusion

The “Count Inversions” problem is a valuable exercise for understanding sorting algorithms and inversion counting. By leveraging a modified merge sort algorithm, you can efficiently count inversions with $O(n\log n)$ time complexity. This problem not only enhances your algorithmic skills but also provides insights into the performance of sorting algorithms and data structures.

Whether you’re preparing for coding interviews or seeking to deepen your understanding of algorithms, mastering inversion counting will be a valuable addition to your problem-solving toolkit. Happy coding!