Summary of "Reverse Pairs | Hard Interview Question"

Summary of “Reverse Pairs | Hard Interview Question”

This video is a detailed tutorial on solving the “Reverse Pairs” problem, a challenging data structures and algorithms (DSA) question commonly asked in technical interviews. The problem involves counting pairs in an integer array where the left element is greater than twice the right element.

Main Ideas and Concepts

Problem Definition

Given an array of integers, count the number of pairs (i, j) such that:

i < j
arr[i] > 2 * arr[j]

Relation to Count Inversions Problem

This problem is similar to the classic “count inversions” problem but with a stricter condition: arr[i] > 2 * arr[j] instead of arr[i] > arr[j].

Naive Brute Force Approach

Iterate over all pairs (i, j) with i < j.
Check if arr[i] > 2 * arr[j].
Increment count if the condition is true.

Complexity: - Time: O(n²) — inefficient for large arrays - Space: O(1)

Need for Optimization

Interviewers expect a better-than-quadratic solution, ideally O(n log n).

Optimized Approach Using Modified Merge Sort

Use the divide-and-conquer strategy of merge sort.
When merging two sorted halves, count valid pairs between the left and right halves before merging.
Since both halves are sorted, use a two-pointer technique to efficiently count pairs:
- For each element in the left half, find how many elements in the right half satisfy left[i] > 2 * right[j].
- Move pointers accordingly without rechecking from the start each time.
After counting, merge the two halves normally to maintain sorted order.
Recursively apply this process to the entire array.

Key Insight

The sorted halves enable efficient counting of pairs by leveraging the sorted property and avoiding redundant comparisons.

Step-by-Step Methodology

Divide: Recursively split the array into halves until subarrays of size one are reached.
Conquer (Count Pairs): Before merging two sorted halves:
- Initialize two pointers: one for the left half (i), one for the right half (j).
- For each i in left half, move j in right half forward while arr[i] > 2 * arr[j] holds.
- Count the number of valid pairs as j - (mid + 1) for each i.
Merge: Merge the two sorted halves into a single sorted array.
Combine Counts: Sum counts from left half, right half, and cross pairs found during merging.

Implementation Details

Maintain a count variable to accumulate the number of valid pairs.
Avoid using global variables by returning counts from recursive calls.
Mention to the interviewer if the input array is modified (distorted) during sorting.
Time complexity is approximately O(n log n).
Space complexity is O(n) due to auxiliary arrays used during merge.

Additional Notes

The algorithm is a modification of the classic merge sort inversion count.
The video stresses the importance of understanding merge sort and the count inversion problem before attempting this.
The problem is hard and typically not invented from scratch in an interview; prior practice is recommended.
The presenter encourages viewers to practice the problem via the provided link.

Detailed Step-by-Step Instructions (Algorithm)

Brute Force (Inefficient)

Initialize count = 0.
For i from 0 to n-2:
- For j from i+1 to n-1:
  - If arr[i] > 2 * arr[j], increment count.
Return count.

Optimized Merge Sort Based Approach

function mergeSort(arr, low, high):
    if low >= high:
        return 0
    mid = (low + high) / 2
    count = mergeSort(arr, low, mid)
    count += mergeSort(arr, mid + 1, high)
    count += countPairs(arr, low, mid, high)
    merge(arr, low, mid, high)
    return count

function countPairs(arr, low, mid, high):
    count = 0
    j = mid + 1
    for i in range(low, mid+1):
        while j <= high and arr[i] > 2 * arr[j]:
            j += 1
        count += (j - (mid + 1))
    return count

function merge(arr, low, mid, high):
    // Standard merge procedure to merge two sorted halves

Usage

Call mergeSort(arr, 0, n-1) to get the total count of reverse pairs.

Speakers / Sources Featured

Primary Speaker: The video is presented by a single instructor from the “Strivers A to Z DSA course” YouTube channel (name not explicitly mentioned in the subtitles).

Summary

The video explains the “Reverse Pairs” problem, starting from the brute force solution and progressing to an optimized O(n log n) solution using a modified merge sort approach. It emphasizes understanding the count inversions problem and merge sort fundamentals before tackling this problem. The key is counting valid pairs before merging sorted halves efficiently using two pointers. The presenter also discusses implementation nuances and complexity analysis, encouraging viewers to practice the problem ahead of interviews.

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Summary of "Reverse Pairs | Hard Interview Question"