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 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.

#493

Hard

auto_awesome二分·搜索·答案·空间

LeetCode 题解工作台

翻转对

给定一个数组 nums ，如果 i 且 nums[i] > 2*nums[j] 我们就将 (i, j) 称作一个重要翻转对。你需要返回给定数组中的重要翻转对的数量。示例 1: 输入 : [1,3,2,3,1] 输出 : 2 示例 2: 输入 : [2,4,3,5,1] 输出 : 3 注意: …

数组二分查找分治树状数组线段树

题目描述

给定一个数组 nums ，如果 i < j 且 nums[i] > 2*nums[j] 我们就将 (i, j) 称作一个重要翻转对。

你需要返回给定数组中的重要翻转对的数量。

示例 1:

输入: [1,3,2,3,1]
输出: 2

示例 2:

输入: [2,4,3,5,1]
输出: 3

注意:

给定数组的长度不会超过50000。
输入数组中的所有数字都在32位整数的表示范围内。

lightbulb

解题思路

方法一：归并排序

归并排序的过程中，如果左边的数大于右边的数，则右边的数与左边的数之后的数都构成逆序对。

时间复杂度 $O(n \times \log n)$ ，空间复杂度 $O(n)$ 。其中 $n$ 为数组长度。

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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)

speed

复杂度分析

指标	值
时间	Depends on the final approach
空间	Depends on the final approach

psychology

面试官常问的追问

外企场景

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

warning

常见陷阱

外企场景

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

swap_horiz

进阶变体

外企场景

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

help

常见问题

外企场景

继续练习

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