Reverse Pairs | Leetcode 493 | Optimal Merge Sort

The “Reverse Pairs” problem is a popular and challenging interview question. In this blog, we’ll explain the problem, walk you through both brute force and optimal solutions, and make everything easy to understand with code comments, dry runs, and clear explanations.

Given an integer array nums, return the number of reverse pairs in the array.

Here’s the [Problem Link] to begin with.

A reverse pair is a pair (i, j) where:

• 0 <= i < j < nums.length and
• nums[i] > 2 * nums[j].

Example 1:

Input: nums = [1,3,2,3,1]
Output: 2
Explanation: The reverse pairs are:
(1, 4) --> nums[1] = 3, nums[4] = 1, 3 > 2 * 1
(3, 4) --> nums[3] = 3, nums[4] = 1, 3 > 2 * 1

Example 2:

Input: nums = [2,4,3,5,1]
Output: 3
Explanation: The reverse pairs are:
(1, 4) --> nums[1] = 4, nums[4] = 1, 4 > 2 * 1
(2, 4) --> nums[2] = 3, nums[4] = 1, 3 > 2 * 1
(3, 4) --> nums[3] = 5, nums[4] = 1, 5 > 2 * 1

Constraints:

• 1 <= nums.length <= 5 * 104
• -231 <= nums[i] <= 231 - 1

Brute Force Solution (Nested Loops)

Intuition and Approach

Let’s solve the problem in the most straightforward way:

1. Check Every Pair:
   For each element, compare it with every element that comes after it.
2. Count Reverse Pairs:
   If nums[i] > 2 * nums[j], count it as a reverse pair.
3. Why does this work?
   We check all possible pairs, so we count every reverse pair.

This approach is easy to understand but not efficient for large arrays.

Code Implementation

from typing import List

class Solution:
    def reversePairs(self, nums: List[int]) -> int:
        count = 0
        n = len(nums)
        for i in range(0, n):
            for j in range(i + 1, n):
                if nums[i] > 2 * nums[j]:      # check reverse pair condition
                    count += 1
        return count

Code Explanation

• We use two nested loops.
• For each pair (i, j) where i < j, we check if nums[i] > 2 * nums[j].
• If true, we increment the reverse pair count.

Dry Run

Input:
nums = [1][1]

• (1,3): 1 > 6? No
• (1,2): 1 > 4? No
• (1,3): 1 > 6? No
• (1,1): 1 > 2? No
• (3,2): 3 > 4? No
• (3,3): 3 > 6? No
• (3,1): 3 > 2? Yes (count = 1)
• (2,3): 2 > 6? No
• (2,1): 2 > 2? No
• (3,1): 3 > 2? Yes (count = 2)

Final count:
2

Time and Space Complexity

• Time Complexity: O(n^2)
  Two nested loops for all pairs.
• Space Complexity: O(1)
  Only a counter variable is used.

Conclusion

The brute force approach is simple and good for understanding, but it’s too slow for large arrays.

Optimal Solution (Merge Sort Approach)

Intuition and Approach

We can count reverse pairs much faster using a modified merge sort:

1. Divide and Conquer:
   Use merge sort to divide the array into smaller parts.
2. Count Reverse Pairs During Merge:
   While merging two sorted halves, for each element in the left half, count how many elements in the right half satisfy the reverse pair condition.
3. Why does this work?
   Merge sort naturally compares and sorts elements, so it’s efficient to count reverse pairs during the merge step.

This approach is efficient and works well for large arrays.

Code Implementation

from typing import List, Tuple

class Solution:
    def mergeList(self, arr1: List[int], arr2: List[int]) -> Tuple[List[int], int]:
        n = len(arr1)
        m = len(arr2)
        count = 0
        result = []
        j = 0
        # Count reverse pairs
        for i in range(0, n):
            while j < m and arr1[i] > 2 * arr2[j]:
                j += 1
            count += j

        i, j = 0, 0
        # Merge two sorted arrays
        while i < n and j < m:
            if arr1[i] <= arr2[j]:
                result.append(arr1[i])
                i += 1
            else:
                result.append(arr2[j])
                j += 1

        while i < n:
            result.append(arr1[i])
            i += 1
        while j < m:
            result.append(arr2[j])
            j += 1

        return result, count

    def mergeSort(self, lst: List[int]) -> Tuple[List[int], int]:
        if len(lst) <= 1:
            return lst, 0

        mid = len(lst) // 2
        first_half = lst[:mid]
        second_half = lst[mid:]

        fh, cnt1 = self.mergeSort(first_half)
        sh, cnt2 = self.mergeSort(second_half)

        merged, count = self.mergeList(fh, sh)
        return merged, cnt1 + cnt2 + count

    def reversePairs(self, nums: List[int]) -> int:
        m, c = self.mergeSort(nums)
        return c

Code Explanation

• We use a modified merge sort.
• During the merge step, for each element in the left half, we count how many elements in the right half satisfy arr1[i] > 2 * arr2[j].
• We merge the two halves as in standard merge sort.
• We sum up the counts from all recursive calls.

Dry Run

Input:
nums = [1][1]

• Split into [1] and [1]
• [1] → [1], “
• [1] → , `[1]` → , [1]
• Merge “ and [1]: check 3 > 2*1? Yes (count = 1)
• Merge “ and [1]: check 2 > 21? No; 2 > 23? No
• Merge [1] and [1]: count reverse pairs across the two halves
• Total count = 2

Time and Space Complexity

• Time Complexity: O(n log n)
  Merge sort divides and merges arrays efficiently.
• Space Complexity: O(n)
  Extra space is used for temporary arrays during merge.

Conclusion

The optimal solution is efficient and works well for large arrays. It’s the best approach for interviews and practical use.

Final Thoughts

The “Reverse Pairs” problem is a great example of how to optimize from brute force to an efficient divide-and-conquer solution. Start with brute force to understand the basics, then use the merge sort approach for best performance.

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