What is the time complexity of the Maximum Width Ramp problem?

The time complexity is O(n) due to a single pass through the array using the two-pointer technique, with a stack operation for each element.

How does the two-pointer technique work in the Maximum Width Ramp problem?

The two-pointer technique allows you to efficiently track potential ramps by iterating through the array, using a stack to track indices and calculate the maximum width.

What is a monotonic stack, and how does it help solve the Maximum Width Ramp problem?

A monotonic stack maintains a sequence of indices in non-decreasing order, helping efficiently track potential ramp widths and ensuring the correct pair of indices is selected for maximizing the width.

How does GhostInterview assist with solving the Maximum Width Ramp problem?

GhostInterview guides you step-by-step through the problem, offering clear explanations of the two-pointer and stack techniques, helping you avoid common mistakes.

Can the solution for the Maximum Width Ramp problem be optimized further?

The current solution is already optimal with O(n) time complexity, but variations of the problem with different constraints may require alternative strategies.

#962

Medium

auto_awesome双·指针·invariant

LeetCode 题解工作台

最大宽度坡

给定一个整数数组 A ，坡是元组 (i, j) ，其中 i 且 A[i] 。这样的坡的宽度为 j - i 。找出 A 中的坡的最大宽度，如果不存在，返回 0 。示例 1：输入： [6,0,8,2,1,5] 输出： 4 解释：最大宽度的坡为 (i, j) = (1, 5): A[1] = …

数组双指针栈单调栈

题目描述

给定一个整数数组 A，坡是元组 (i, j)，其中 i < j 且 A[i] <= A[j]。这样的坡的宽度为 j - i。

找出 A 中的坡的最大宽度，如果不存在，返回 0 。

示例 1：

输入：[6,0,8,2,1,5]
输出：4
解释：
最大宽度的坡为 (i, j) = (1, 5): A[1] = 0 且 A[5] = 5.

示例 2：

输入：[9,8,1,0,1,9,4,0,4,1]
输出：7
解释：
最大宽度的坡为 (i, j) = (2, 9): A[2] = 1 且 A[9] = 1.

提示：

2 <= A.length <= 50000
0 <= A[i] <= 50000

lightbulb

解题思路

方法一：单调栈

根据题意，我们可以发现，所有可能的 $\textit{nums}[i]$ 所构成的子序列一定是单调递减的。为什么呢？我们不妨用反证法证明一下。

假设存在 $i_1<i_2$ ，并且 $\textit{nums}[i_1]\leq\textit{nums}[i_2]$ ，那么实际上 $\textit{nums}[i_2]$ 一定不可能是一个候选值，因为 $\textit{nums}[i_1]$ 更靠左，会是一个更优的值。因此 $\textit{nums}[i]$ 所构成的子序列一定单调递减，并且 $i$ 一定是从 0 开始。

我们用一个从栈底到栈顶单调递减的栈 $\textit{stk}$ 来存储所有可能的 $\textit{nums}[i]$ ，然后我们从右边界开始遍历 $j$ ，若遇到 $\textit{nums}[\textit{stk.top()}]\leq\textit{nums}[j]$ ，说明此时构成一个坡，循环弹出栈顶元素，更新 $\textit{ans}$ 。

时间复杂度 $O(n)$ ，空间复杂度 $O(n)$ ，其中 $n$ 表示 $\textit{nums}$ 的长度。

1

2

3

4

5

6

7

8

9

10

11

12

13

14

class Solution:
    def maxWidthRamp(self, nums: List[int]) -> int:
        stk = []
        for i, v in enumerate(nums):
            if not stk or nums[stk[-1]] > v:
                stk.append(i)
        ans = 0
        for i in range(len(nums) - 1, -1, -1):
            while stk and nums[stk[-1]] <= nums[i]:
                ans = max(ans, i - stk.pop())
            if not stk:
                break
        return ans

speed

复杂度分析

指标	值
时间	O(n)
空间	O(n)

psychology

面试官常问的追问

外企场景

question_mark
Candidate understands the two-pointer technique for optimizing problems like this one.
question_mark
Look for clarity in the candidate's explanation of the monotonic stack's role in solving the problem.
question_mark
Assess how well the candidate relates the problem to patterns like invariant tracking and ramp-width maximization.

warning

常见陷阱

外企场景

error
Failing to use the monotonic stack efficiently, leading to incorrect ramp width calculations.
error
Not recognizing that the two-pointer technique must be combined with a stack for optimal performance.
error
Overcomplicating the problem by using a brute-force approach that results in O(n^2) time complexity.

swap_horiz

进阶变体

外企场景

arrow_right_alt
Consider variations where the ramp definition might change, such as requiring nums[i] > nums[j] or other constraints on the array.
arrow_right_alt
Explore edge cases with minimum and maximum array sizes to ensure the algorithm handles these scenarios effectively.
arrow_right_alt
Consider extending the problem to allow ramps in multidimensional arrays or with additional constraints.

help

常见问题

外企场景

继续练习

#581 最短无序连续子数组 #1574 删除最短的子数组使剩余数组有序 #321 拼接最大数