Summary of "Count Inversions in an Array | Brute and Optimal"

Summary of “Count Inversions in an Array | Brute and Optimal”

This video explains the problem of counting inversions in an array, where an inversion is defined as a pair of elements (a[i], a[j]) such that i < j and a[i] > a[j]. It covers both the brute force and optimal solutions, focusing on how the optimal approach leverages the merge sort algorithm.

Main Ideas and Concepts

Problem Definition

Given an array of integers, count the number of inversion pairs (a[i], a[j]) where the left element is greater than the right element and the left element appears before the right element in the array.

Example: For the array [5, 3, 2, 4, 1], the inversion pairs are: (5,3), (5,2), (5,4), (5,1), (3,2), (3,1), (2,1), (4,1) Total inversions = 8

Brute Force Approach

Iterate over every element and for each element, check all elements to its right.
Increment count if the left element is greater than the right element.
Time complexity: O(n²)
Space complexity: O(1)
Simple but inefficient for large arrays and generally not acceptable in interviews.

Optimal Approach (Using Merge Sort)

Optimizes the problem to O(n log n) using a modified merge sort.
Key Insight: During the merge step of merge sort, when merging two sorted subarrays, you can count the number of inversions efficiently.
Method:
- Recursively split the array into halves until single elements remain.
- While merging two sorted halves, count how many elements from the left half are greater than elements from the right half.
- Use two pointers to traverse the left and right subarrays.
- If left[i] > right[j], then all remaining elements in the left subarray from index i onwards form inversions with right[j] because the subarrays are sorted.
- Add the count of these inversions (mid - i + 1) to the total count.
- Continue merging until the entire array is sorted.
Time complexity: O(n log n)
Space complexity: O(n) due to temporary arrays used during merging.

Detailed Explanation of the Merge Step

Given two sorted arrays, count pairs where elements from the left array are greater than elements from the right array.
Use two pointers starting at the beginning of each array.
Move pointers accordingly:
- If left[i] <= right[j], move left pointer.
- If left[i] > right[j], add (mid - i + 1) to count because all elements from i to mid in the left array are greater than right[j].
- Move right pointer.
This logic efficiently counts inversions across the two halves.

Implementation Notes

Keep a count variable to store the number of inversions.
Avoid using global variables in interviews; instead, pass count through recursive calls or return counts from merge functions.
The merge sort algorithm sorts the array in place, so mention to the interviewer if the original array is modified or if a copy is used.
This approach is standard in software engineering interviews and shows mastery of divide-and-conquer techniques.

Additional Tips

Before implementing the optimal solution, understand merge sort thoroughly.
Visualize the problem by breaking the array into halves and counting inversions during the merge step.
Explain the space-time tradeoff and mention edge cases such as empty arrays or arrays with duplicates.
Provide clear communication during interviews about assumptions and modifications to the input.

Step-by-Step Methodology for Optimal Solution (Merge Sort Based)

Divide: Recursively split the array into two halves until each subarray has one element.
Conquer: Recursively count inversions in the left half. Recursively count inversions in the right half.
Combine (Merge and Count):
- Merge the two sorted halves.
- While merging, count the number of inversions:
  - Initialize two pointers i and j for left and right halves.
  - If left[i] <= right[j], copy left[i] to temporary array and increment i.
  - Else (left[i] > right[j]), copy right[j] to temporary array, increment j, and add (mid - i + 1) to inversion count.
- Copy the merged array back to the original array.
Return: Sum of inversions from left half, right half, and merge step is the total count.

Speakers / Sources Featured

Primary Speaker: The video host/narrator who explains the problem, walks through examples, and codes the solution.
No other distinct speakers or external sources are mentioned.

Summary Conclusion

The video teaches how to count inversions in an array using both brute force and an optimal merge sort based approach. It emphasizes understanding the merge sort algorithm deeply to leverage its merge step for counting inversions efficiently in O(n log n) time. The video also provides implementation tips and interview advice for handling this classic problem.