How do you solve Reverse Pairs on LeetCode?

Reverse Pairs (LeetCode #493) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n) time complexity and O(n) space complexity. Key insight: Understanding how to count pairs efficiently during sorting is crucial.

What is the brute force solution for Reverse Pairs?

The brute force approach for Reverse Pairs is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking every possible pair of indices (i, j) in the array to see if they satisfy the reverse pair condition. This is straightforward but inefficient for larger arrays. The optimal Optimal Solution approach improves this to O(n log n).

What are the interview tips for Reverse Pairs?

Always start with a brute-force solution to understand the problem better. Explain your thought process clearly when transitioning to an optimal solution. Practice writing code for both brute-force and optimal solutions to reinforce your understanding.

What are common mistakes when solving Reverse Pairs?

Not considering the constraints and edge cases, such as very large or very small numbers. Failing to optimize the solution when a brute-force approach is too slow.

#493

Reverse Pairs

Hard

ArrayBinary SearchDivide and ConquerBinary Indexed TreeSegment TreeMerge SortOrdered SetDivide and ConquerMerge Sort

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n log n)
Space	O(1)	O(n)

💡

Intuition

Time O(n log n)Space O(n)

Using a modified merge sort allows us to count reverse pairs efficiently during the sorting process. This takes advantage of the divide-and-conquer strategy to reduce the time complexity significantly.

⚙️

Algorithm

4 steps

1Step 1: Define a recursive function that sorts the array and counts reverse pairs.
2Step 2: Split the array into two halves and recursively sort each half.
3Step 3: While merging the two halves, count how many elements in the left half are greater than twice the elements in the right half.
4Step 4: Merge the two halves back together in sorted order.

solution.py40 lines

1def merge_and_count(nums, temp, left, mid, right):
2    count = 0
3    j = mid + 1
4    for i in range(left, mid + 1):
5        while j <= right and nums[i] > 2 * nums[j]:
6            j += 1
7        count += (j - (mid + 1))
8    i, j, k = left, mid + 1, left
9    while i <= mid and j <= right:
10        if nums[i] <= nums[j]:
11            temp[k] = nums[i]
12            i += 1
13        else:
14            temp[k] = nums[j]
15            j += 1
16        k += 1
17    while i <= mid:
18        temp[k] = nums[i]
19        i += 1
20        k += 1
21    while j <= right:
22        temp[k] = nums[j]
23        j += 1
24        k += 1
25    for i in range(left, right + 1):
26        nums[i] = temp[i]
27    return count
28
29def reversePairs(nums):
30    temp = [0] * len(nums)
31    return merge_sort_and_count(nums, temp, 0, len(nums) - 1)
32
33def merge_sort_and_count(nums, temp, left, right):
34    count = 0
35    if left < right:
36        mid = (left + right) // 2
37        count += merge_sort_and_count(nums, temp, left, mid)
38        count += merge_sort_and_count(nums, temp, mid + 1, right)
39        count += merge_and_count(nums, temp, left, mid, right)
40    return count

ℹ

Complexity note: The time complexity is O(n log n) due to the merge sort process, which divides the array and merges it back while counting reverse pairs. The space complexity is O(n) because we need an additional array for merging.

1Understanding how to count pairs efficiently during sorting is crucial.
2Recognizing the need for a divide-and-conquer strategy can significantly reduce time complexity.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.