LeetCode Problem Workspace

Reverse Pairs

Count the number of reverse pairs in a given integer array using efficient algorithms like binary search and merge sort.

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Practice Focus

Hard · Binary search over the valid answer space

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Answer-first summary

Count the number of reverse pairs in a given integer array using efficient algorithms like binary search and merge sort.

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To solve the 'Reverse Pairs' problem, efficiently count pairs (i, j) where nums[i] > 2 * nums[j]. This can be approached using binary search and merge sort, improving upon brute force solutions. The key is balancing efficiency and correctness while minimizing time complexity.

Problem Statement

Given an integer array nums, you are tasked with counting the number of reverse pairs in the array. A reverse pair is defined as a pair of indices (i, j) such that i < j and nums[i] > 2 * nums[j].

For example, in the array nums = [1, 3, 2, 3, 1], there are two reverse pairs: (1, 4) and (3, 4). The first reverse pair is (nums[1] = 3, nums[4] = 1), where 3 > 2 * 1, and the second is (nums[3] = 3, nums[4] = 1), where 3 > 2 * 1.

Examples

Example 1

Input: nums = [1,3,2,3,1]

Output: 2

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

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

Solution Approach

Merge Sort and Binary Search

Using a modified merge sort, count reverse pairs during the merge step. Binary search can be applied to optimize the search for valid pairs during the merge, ensuring the solution runs efficiently even for large arrays.

Divide and Conquer Strategy

The problem can be tackled using the divide-and-conquer technique by recursively dividing the array and counting reverse pairs in subarrays, combining the results as you merge the sorted arrays.

Use of Binary Indexed Tree

A Binary Indexed Tree (BIT) or Segment Tree can be leveraged to efficiently count the valid pairs as elements are processed, improving the time complexity of the solution.

Complexity Analysis

Metric Value
Time Depends on the final approach
Space Depends on the final approach

The time complexity depends on the approach. A brute-force solution would run in O(n^2), but with a merge sort-based approach, it reduces to O(n log n). Space complexity depends on the method used but typically would be O(n) for the merge sort or BIT-based solutions.

What Interviewers Usually Probe

  • Assess the candidate's ability to optimize brute force solutions using merge sort or binary search techniques.
  • Look for familiarity with binary indexed trees or segment trees for optimizing the solution's space and time complexity.
  • Evaluate how well the candidate applies divide and conquer strategies to complex counting problems.

Common Pitfalls or Variants

Common pitfalls

  • Overlooking optimization opportunities and resorting to brute force, which is inefficient for larger inputs.
  • Failure to correctly handle edge cases, such as arrays with very small or very large numbers.
  • Misunderstanding the problem statement and incorrectly counting pairs that do not satisfy the reverse pair condition.

Follow-up variants

  • What if the array contains negative numbers or zeros? Ensure the solution handles such edge cases properly.
  • What if the array length exceeds typical constraints? Test the solution’s efficiency for large arrays (up to 5 * 10^4).
  • How would you approach solving this problem if the input array was sorted? Consider variations where sorting the array first can help with optimizations.

FAQ

How does binary search help in solving the Reverse Pairs problem?

Binary search is used to efficiently count the number of valid pairs during the merge step of the merge sort process, reducing the time complexity compared to brute force methods.

What is the best approach for solving the Reverse Pairs problem?

The most efficient approach combines merge sort with binary search, reducing the problem's time complexity to O(n log n), making it feasible for large arrays.

Can a brute force solution work for Reverse Pairs?

While a brute force solution can work, it has a time complexity of O(n^2), which is not efficient enough for large arrays. Optimizing with merge sort and binary search is recommended.

What is a reverse pair in the Reverse Pairs problem?

A reverse pair is a pair of indices (i, j) where i < j and nums[i] > 2 * nums[j]. These pairs need to be counted in the given array.

How does GhostInterview help with solving problems like Reverse Pairs?

GhostInterview helps by breaking down the problem into digestible steps, offering explanations and providing optimal solutions like merge sort with binary search, ensuring an efficient approach.

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Solution

Solution 1

#### Python3

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class Solution:
    def reversePairs(self, nums: List[int]) -> int:
        def merge_sort(l, r):
            if l >= r:
                return 0
            mid = (l + r) >> 1
            ans = merge_sort(l, mid) + merge_sort(mid + 1, r)
            t = []
            i, j = l, mid + 1
            while i <= mid and j <= r:
                if nums[i] <= 2 * nums[j]:
                    i += 1
                else:
                    ans += mid - i + 1
                    j += 1
            i, j = l, mid + 1
            while i <= mid and j <= r:
                if nums[i] <= nums[j]:
                    t.append(nums[i])
                    i += 1
                else:
                    t.append(nums[j])
                    j += 1
            t.extend(nums[i : mid + 1])
            t.extend(nums[j : r + 1])
            nums[l : r + 1] = t
            return ans

        return merge_sort(0, len(nums) - 1)

Solution 2

#### Python3

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class Solution:
    def reversePairs(self, nums: List[int]) -> int:
        def merge_sort(l, r):
            if l >= r:
                return 0
            mid = (l + r) >> 1
            ans = merge_sort(l, mid) + merge_sort(mid + 1, r)
            t = []
            i, j = l, mid + 1
            while i <= mid and j <= r:
                if nums[i] <= 2 * nums[j]:
                    i += 1
                else:
                    ans += mid - i + 1
                    j += 1
            i, j = l, mid + 1
            while i <= mid and j <= r:
                if nums[i] <= nums[j]:
                    t.append(nums[i])
                    i += 1
                else:
                    t.append(nums[j])
                    j += 1
            t.extend(nums[i : mid + 1])
            t.extend(nums[j : r + 1])
            nums[l : r + 1] = t
            return ans

        return merge_sort(0, len(nums) - 1)

Solution 3

#### Python3

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class Solution:
    def reversePairs(self, nums: List[int]) -> int:
        def merge_sort(l, r):
            if l >= r:
                return 0
            mid = (l + r) >> 1
            ans = merge_sort(l, mid) + merge_sort(mid + 1, r)
            t = []
            i, j = l, mid + 1
            while i <= mid and j <= r:
                if nums[i] <= 2 * nums[j]:
                    i += 1
                else:
                    ans += mid - i + 1
                    j += 1
            i, j = l, mid + 1
            while i <= mid and j <= r:
                if nums[i] <= nums[j]:
                    t.append(nums[i])
                    i += 1
                else:
                    t.append(nums[j])
                    j += 1
            t.extend(nums[i : mid + 1])
            t.extend(nums[j : r + 1])
            nums[l : r + 1] = t
            return ans

        return merge_sort(0, len(nums) - 1)
Reverse Pairs Solution: Binary search over the valid answer s… | LeetCode #493 Hard