How do I efficiently count arithmetic triplets in the given array?

Use a hash set to track the elements as you scan through the array. For each number, check if its 'diff' and '2*diff' counterparts exist in the set.

What is the time complexity of the brute force solution?

The brute force approach has a time complexity of O(n^3) due to three nested loops, which is inefficient for larger input sizes.

Can I solve this problem using only two-pointer technique?

Yes, by sorting the array first, you can apply the two-pointer technique to find valid triplets with the given difference.

What if the array is not strictly increasing?

If the array is not strictly increasing, you need to account for duplicates and might need to modify your approach accordingly.

How does GhostInterview help in solving this problem?

GhostInterview assists by providing structured approaches, identifying potential pitfalls, and offering insights into optimization strategies like hash lookup and two-pointer techniques.

#2367

Easy

auto_awesome数组·哈希·扫描

LeetCode 题解工作台

等差三元组的数目

给你一个下标从 0 开始、严格递增的整数数组 nums 和一个正整数 diff 。如果满足下述全部条件，则三元组 (i, j, k) 就是一个等差三元组： i ， nums[j] - nums[i] == diff 且 nums[k] - nums[j] == diff 返回不同等差三元组…

数组哈希表双指针枚举

题目描述

给你一个下标从 0 开始、严格递增 的整数数组 nums 和一个正整数 diff 。如果满足下述全部条件，则三元组 (i, j, k) 就是一个 等差三元组 ：

i < j < k ，
nums[j] - nums[i] == diff 且
nums[k] - nums[j] == diff

返回不同 等差三元组 的数目。

示例 1：

输入：nums = [0,1,4,6,7,10], diff = 3
输出：2
解释：
(1, 2, 4) 是等差三元组：7 - 4 == 3 且 4 - 1 == 3 。
(2, 4, 5) 是等差三元组：10 - 7 == 3 且 7 - 4 == 3 。

示例 2：

输入：nums = [4,5,6,7,8,9], diff = 2
输出：2
解释：
(0, 2, 4) 是等差三元组：8 - 6 == 2 且 6 - 4 == 2 。
(1, 3, 5) 是等差三元组：9 - 7 == 2 且 7 - 5 == 2 。

提示：

3 <= nums.length <= 200
0 <= nums[i] <= 200
1 <= diff <= 50
nums 严格递增

lightbulb

解题思路

方法一：暴力枚举

我们注意到，数组 $nums$ 的长度只有不超过 $200$ ，因此可以直接暴力枚举 $i$ , $j$ , $k$ ，判断是否满足条件，若满足，累加三元组数目。

时间复杂度 $O(n^3)$ ，其中 $n$ 为数组 $nums$ 的长度。空间复杂度 $O(1)$ 。

1

2

3

4

class Solution:
    def arithmeticTriplets(self, nums: List[int], diff: int) -> int:
        return sum(b - a == diff and c - b == diff for a, b, c in combinations(nums, 3))

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 understands the value of hash lookup for optimizing time complexity.
question_mark
The candidate may propose multiple approaches with an understanding of trade-offs in performance.
question_mark
The candidate mentions array scanning or two-pointer techniques to handle the problem efficiently.

warning

常见陷阱

外企场景

error
Failing to correctly handle boundary conditions, such as ensuring the array is strictly increasing.
error
Overcomplicating the solution when a simple hash lookup or two-pointer technique would suffice.
error
Assuming that brute force is always an acceptable solution without considering time constraints for larger inputs.

swap_horiz

进阶变体

外企场景

arrow_right_alt
What if the array was not strictly increasing? This could require a different approach to ensure the difference condition holds.
arrow_right_alt
What if the diff value is much larger than the array size? This would require careful handling to avoid unnecessary checks.
arrow_right_alt
Can the problem be adapted to find the number of arithmetic triplets in a non-zero-indexed array or with negative integers?

help

常见问题

外企场景

继续练习

#2122 还原原数组 #2306 公司命名 #2441 与对应负数同时存在的最大正整数