What is the approach for solving the Longest Arithmetic Subsequence of Given Difference problem?

The approach combines dynamic programming and hash tables to track subsequence lengths while scanning the array. The hash table stores the length of each subsequence ending at a specific value.

How do hash tables optimize the solution for the Longest Arithmetic Subsequence of Given Difference?

Hash tables allow for constant-time lookups, enabling quick access to the length of previous subsequences that can extend with the current element. This reduces the time complexity compared to brute-force approaches.

What is the time complexity of the solution?

The time complexity is O(n), where n is the length of the input array. Each element is processed once, and hash lookups and insertions are O(1).

How does the space complexity of the solution scale?

The space complexity is O(n) because we store the length of subsequences for each element in a hash table, which requires linear space.

Can the solution handle negative differences in the Longest Arithmetic Subsequence of Given Difference?

Yes, the solution can handle negative differences by tracking subsequences where elements decrease by the given negative difference.

#1218

Medium

auto_awesome数组·哈希·扫描

LeetCode 题解工作台

最长定差子序列

给你一个整数数组 arr 和一个整数 difference ，请你找出并返回 arr 中最长等差子序列的长度，该子序列中相邻元素之间的差等于 difference 。子序列是指在不改变其余元素顺序的情况下，通过删除一些元素或不删除任何元素而从 arr 派生出来的序列。示例 1：输入： arr…

数组哈希表动态规划

题目描述

给你一个整数数组 arr 和一个整数 difference，请你找出并返回 arr 中最长等差子序列的长度，该子序列中相邻元素之间的差等于 difference 。

子序列 是指在不改变其余元素顺序的情况下，通过删除一些元素或不删除任何元素而从 arr 派生出来的序列。

示例 1：

输入：arr = [1,2,3,4], difference = 1
输出：4
解释：最长的等差子序列是 [1,2,3,4]。

示例 2：

输入：arr = [1,3,5,7], difference = 1
输出：1
解释：最长的等差子序列是任意单个元素。

示例 3：

输入：arr = [1,5,7,8,5,3,4,2,1], difference = -2
输出：4
解释：最长的等差子序列是 [7,5,3,1]。

提示：

1 <= arr.length <= 10⁵
-10⁴ <= arr[i], difference <= 10⁴

lightbulb

解题思路

方法一：动态规划

我们可以使用哈希表 $f$ 来存储以 $x$ 结尾的最长等差子序列的长度。

遍历数组 $\textit{arr}$ ，对于每个元素 $x$ ，我们更新 $f[x]$ 为 $f[x - \textit{difference}] + 1$ 。

遍历结束后，我们返回 $f$ 中的最大值作为答案返回即可。

时间复杂度 $O(n)$ ，空间复杂度 $O(n)$ 。其中 $n$ 为数组 $\textit{arr}$ 的长度。

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class Solution:
    def longestSubsequence(self, arr: List[int], difference: int) -> int:
        f = defaultdict(int)
        for x in arr:
            f[x] = f[x - difference] + 1
        return max(f.values())

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
The candidate should demonstrate familiarity with dynamic programming techniques.
question_mark
They should understand how hash tables are used to optimize subsequence tracking.
question_mark
The candidate must ensure they can explain the trade-offs between time and space complexity in this approach.

warning

常见陷阱

外企场景

error
Forgetting to account for negative differences when scanning the array.
error
Overcomplicating the problem by trying to brute force through all possible subsequences.
error
Not using hash tables efficiently, leading to unnecessary recomputation or slower performance.

swap_horiz

进阶变体

外企场景

arrow_right_alt
Consider variations where the difference is zero, which means the subsequence would consist of identical elements.
arrow_right_alt
Explore scenarios where elements of the array are all unique but can form an arithmetic subsequence with a given difference.
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Test with arrays that have negative and positive numbers mixed, making sure the algorithm handles both directions of differences.

help

常见问题

外企场景

继续练习

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