What is the prefix sum technique?

The prefix sum technique involves computing the cumulative sum of an array as you iterate through it, allowing you to efficiently calculate values like total altitude at each point in this problem.

How can I optimize the space complexity for this problem?

You can optimize space complexity by tracking only the current and maximum altitude reached, avoiding the need to store all altitudes.

Why does the time complexity of this solution remain O(N)?

The time complexity is O(N) because we traverse the gain array exactly once, updating the altitude and maximum altitude in each step.

What are some common mistakes in solving this problem?

Some common mistakes include unnecessary nested loops or storing all intermediate altitude values instead of just tracking the maximum altitude.

How does GhostInterview help with problems like this?

GhostInterview provides step-by-step guidance, helping you understand the core concepts like prefix sum and space optimization, so you can solve problems efficiently in interviews.

#1732

Easy

auto_awesome前缀和

LeetCode 题解工作台

找到最高海拔

有一个自行车手打算进行一场公路骑行，这条路线总共由 n + 1 个不同海拔的点组成。自行车手从海拔为 0 的点 0 开始骑行。给你一个长度为 n 的整数数组 gain ，其中 gain[i] 是点 i 和点 i + 1 的净海拔高度差（ 0 ）。请你返回最高点的海拔。示例 1：输入： …

数组前缀和

题目描述

有一个自行车手打算进行一场公路骑行，这条路线总共由 n + 1 个不同海拔的点组成。自行车手从海拔为 0 的点 0 开始骑行。

给你一个长度为 n 的整数数组 gain ，其中 gain[i] 是点 i 和点 i + 1 的 净海拔高度差（0 <= i < n）。请你返回 最高点的海拔 。

示例 1：

输入：gain = [-5,1,5,0,-7]
输出：1
解释：海拔高度依次为 [0,-5,-4,1,1,-6] 。最高海拔为 1 。

示例 2：

输入：gain = [-4,-3,-2,-1,4,3,2]
输出：0
解释：海拔高度依次为 [0,-4,-7,-9,-10,-6,-3,-1] 。最高海拔为 0 。

提示：

n == gain.length
1 <= n <= 100
-100 <= gain[i] <= 100

lightbulb

解题思路

方法一：前缀和（差分数组）

我们假设每个点的海拔为 $h_i$ ，由于 $gain[i]$ 表示第 $i$ 个点和第 $i + 1$ 个点的海拔差，因此 $gain[i] = h_{i + 1} - h_i$ 。那么：

\sum_{i = 0}^{n-1} gain[i] = h_1 - h_0 + h_2 - h_1 + \cdots + h_n - h_{n - 1} = h_n - h_0 = h_n

即：

h_{i+1} = \sum_{j = 0}^{i} gain[j]

可以发现，每个点的海拔都可以通过前缀和的方式计算出来。因此，我们只需要遍历一遍数组，求出前缀和的最大值，即为最高点的海拔。

实际上题目中的 $gain$ 数组是一个差分数组，对差分数组求前缀和即可得到原海拔数组。然后求出原海拔数组的最大值即可。

时间复杂度 $O(n)$ ，空间复杂度 $O(1)$ 。其中 $n$ 为数组 $gain$ 的长度。

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class Solution:
    def largestAltitude(self, gain: List[int]) -> int:
        return max(accumulate(gain, initial=0))

speed

复杂度分析

指标	值
时间	O(N)
空间	O(1)

psychology

面试官常问的追问

外企场景

question_mark
Look for the candidate's ability to apply prefix sum in a simple problem setting.
question_mark
Assess whether the candidate can explain the efficiency of the O(N) time complexity and O(1) space complexity.
question_mark
Gauge how well the candidate can identify and optimize unnecessary space usage.

warning

常见陷阱

外企场景

error
Overcomplicating the problem by trying to store the altitudes at every point instead of just tracking the highest altitude.
error
Misunderstanding the prefix sum concept and attempting a more complex solution like nested loops.
error
Failing to recognize that only a few variables are needed to track the current and maximum altitude, leading to unnecessary space usage.

swap_horiz

进阶变体

外企场景

arrow_right_alt
Extend the problem to handle negative altitude changes more effectively by considering additional edge cases.
arrow_right_alt
Consider situations where the biker stays at the starting altitude or only increases or decreases in a simple pattern.
arrow_right_alt
Explore the problem with varying input sizes to assess performance with maximum constraints.

help

常见问题

外企场景

继续练习

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