How do I find the longest alternating subarray in the problem "Longest Alternating Subarray"?

Iterate through the array, checking consecutive elements for alternation. Keep track of the longest subarray you find.

What is the time complexity of solving the "Longest Alternating Subarray" problem?

The time complexity is O(n) where n is the length of the array, assuming you efficiently check consecutive elements.

What should I do if no alternating subarray exists?

Return -1 if no alternating subarray is found in the array.

How do I handle edge cases for the "Longest Alternating Subarray" problem?

Consider arrays where no alternations occur, such as arrays with all equal elements. Also handle the minimum array length.

Can the "Longest Alternating Subarray" problem be solved by brute force?

While brute force is possible, it is not efficient. Use enumeration and careful iteration to optimize the solution.

#2765

Easy

auto_awesome数组·结合·enumeration

LeetCode 题解工作台

最长交替子数组

给你一个下标从 0 开始的整数数组 nums 。如果 nums 中长度为 m 的子数组 s 满足以下条件，我们称它是一个交替子数组： m 大于 1 。 s 1 = s 0 + 1 。下标从 0 开始的子数组 s 与数组 [s 0 , s 1 , s 0 , s 1 ,...,s (m-1) %…

数组枚举

题目描述

给你一个下标从 0 开始的整数数组 nums 。如果 nums 中长度为 m 的子数组 s 满足以下条件，我们称它是一个 交替子数组 ：

m 大于 1 。
s₁ = s₀ + 1 。
下标从 0 开始的子数组 s 与数组 [s₀, s₁, s₀, s₁,...,s_{(m-1) % 2}] 一样。也就是说，s₁ - s₀ = 1 ，s₂ - s₁ = -1 ，s₃ - s₂ = 1 ，s₄ - s₃ = -1 ，以此类推，直到 s[m - 1] - s[m - 2] = (-1)^m 。

请你返回 nums 中所有交替子数组中，最长的长度，如果不存在交替子数组，请你返回 -1 。

子数组是一个数组中一段连续非空的元素序列。

示例 1：

输入：nums = [2,3,4,3,4]

输出：4

解释：交替子数组有 [2,3]，[3,4]，[3,4,3] 和 [3,4,3,4]。最长的子数组为 [3,4,3,4]，长度为 4。

示例 2：

输入：nums = [4,5,6]

输出：2

解释：[4,5] 和 [5,6] 是仅有的两个交替子数组。它们长度都为 2 。

提示：

2 <= nums.length <= 100
1 <= nums[i] <= 10⁴

lightbulb

解题思路

方法一：枚举

我们可以枚举子数组的左端点 $i$ ，对于每个 $i$ ，我们需要找到最长的满足条件的子数组。我们可以从 $i$ 开始向右遍历，每次遇到相邻元素差值不满足交替条件时，我们就找到了一个满足条件的子数组。我们可以用一个变量 $k$ 来记录当前元素的差值应该是 $1$ 还是 $-1$ ，如果当前元素的差值应该是 $-k$ ，那么我们就将 $k$ 取反。当我们找到一个满足条件的子数组 $nums[i..j]$ 时，我们更新答案为 $\max(ans, j - i + 1)$ 。

时间复杂度 $O(n^2)$ ，其中 $n$ 是数组的长度。我们需要枚举子数组的左端点 $i$ ，对于每个 $i$ ，我们需要 $O(n)$ 的时间来找到最长的满足条件的子数组。空间复杂度 $O(1)$ 。

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class Solution:
    def alternatingSubarray(self, nums: List[int]) -> int:
        ans, n = -1, len(nums)
        for i in range(n):
            k = 1
            j = i
            while j + 1 < n and nums[j + 1] - nums[j] == k:
                j += 1
                k *= -1
            if j - i + 1 > 1:
                ans = max(ans, j - i + 1)
        return ans

speed

复杂度分析

指标	值
时间	complexity can be O(n) where n is the length of the array, depending on the approach used to detect and count alternating subarrays. Space complexity is O(1) since we are not using extra space aside from counters and temporary variables.
空间	Depends on the final approach

psychology

面试官常问的追问

外企场景

question_mark
Candidate understands the pattern of alternating subarrays.
question_mark
The candidate can efficiently loop through the array without unnecessary complexity.
question_mark
Candidate considers edge cases such as no valid alternating subarrays.

warning

常见陷阱

外企场景

error
Not handling the case where no alternating subarrays exist, leading to an incorrect result.
error
Overcomplicating the solution with extra loops or unnecessary calculations.
error
Failing to properly track the maximum length of alternating subarrays.

swap_horiz

进阶变体

外企场景

arrow_right_alt
Consider optimizing the approach for larger arrays.
arrow_right_alt
Extend the problem to find alternating subarrays with specific length constraints.
arrow_right_alt
Alter the problem to return the longest alternating subarray itself, not just the length.

help

常见问题

外企场景

继续练习

#2768 黑格子的数目 #2761 和等于目标值的质数对 #2778 特殊元素平方和