Reverse Pairs
Given an array `nums`, return the number of 'reverse pairs' where $i < j$ and $nums[i] > 2 \cdot nums[j]$. This requires a modified Merge Sort approach.
Why Interviewers Ask This
LinkedIn asks this to evaluate a candidate's mastery of divide-and-conquer strategies beyond standard sorting. They specifically test the ability to modify established algorithms like Merge Sort to handle complex constraints, such as the factor of two in reverse pairs. This assesses whether you can optimize time complexity from O(n^2) to O(n log n) while maintaining logical rigor under pressure.
How to Answer This Question
1. Clarify Constraints: Immediately confirm input size and data types to address potential integer overflow when calculating 2 * nums[j].
2. Propose Naive Solution: Briefly mention the brute-force O(n^2) approach to establish a baseline, then explain why it fails for large datasets typical at LinkedIn.
3. Define the Strategy: Articulate the plan to use Merge Sort. Explain that during the merge step, since both subarrays are sorted, you can efficiently count valid pairs using a two-pointer technique rather than nested loops.
4. Detail the Logic: Describe how to iterate through the left subarray and find the first index in the right subarray where nums[i] > 2 * nums[j], adding the remaining count to your total before merging.
5. Implement and Trace: Write clean code with clear variable names, then walk through a small example manually to verify the counting logic handles edge cases correctly.
Key Points to Cover
- Demonstrating the transition from O(n^2) brute force to O(n log n) optimized solution
- Explaining the specific two-pointer logic used within the merge step
- Addressing integer overflow risks when multiplying by two
- Clarifying that counting happens before the actual merge operation
- Maintaining the sorted invariant of subarrays throughout the recursion
Sample Answer
To solve the Reverse Pairs problem efficiently, I would start by acknowledging that a brute-force approach checking every pair results in O(n^2) time complexity, which is insufficient for large-scale data processing often seen at LinkedIn. Instead, I propose a modified Merge Sort algorithm to achieve O(n log n) performance.
The core insight is that during the merge phase of Merge Sort, we have two sorted subarrays: left and right. Since they are sorted, we can leverage binary search or a two-pointer technique to count pairs where an element in the left subarray is more than twice an element in the right subarray. Specifically, for each element in the left subarray, we find the first position in the right subarray where the condition holds. All elements from that position to the end of the right subarray form valid reverse pairs with the current left element.
Crucially, we must perform this counting step before merging the arrays to maintain the sorted property required for subsequent levels of recursion. We also need to be vigilant about integer overflow when computing 2 * nums[j], potentially casting to long integers if inputs are large. By integrating this counting logic into the standard merge sort framework, we ensure we process the array in logarithmic passes while linearly counting pairs at each level, resulting in an optimal solution that balances time efficiency with code clarity.
Common Mistakes to Avoid
- Attempting to count pairs after merging, which destroys the sorted order needed for efficient lookup
- Ignoring integer overflow issues when calculating 2 * nums[j] for large numbers
- Using binary search incorrectly inside the merge loop without adjusting pointers based on previous findings
- Failing to reset the pointer for the right subarray for each new element in the left subarray
Practice This Question with AI
Answer this question orally or via text and get instant AI-powered feedback on your response quality, structure, and delivery.