What is the primary pattern for solving the Arithmetic Slices problem?

The primary pattern for solving this problem is state transition dynamic programming, where we keep track of valid arithmetic subarrays and update the count efficiently.

How can I optimize the solution for large input sizes?

By using dynamic programming and a sliding window approach, we can optimize both the time and space complexity to handle larger arrays.

Are there any edge cases to consider in the Arithmetic Slices problem?

Yes, edge cases include arrays with fewer than three elements, as they cannot form arithmetic subarrays.

How do I prevent over-counting in this problem?

By carefully updating the count of subarrays only when the sequence continues the arithmetic pattern, you can avoid over-counting subarrays.

What is the time complexity of the Arithmetic Slices solution?

The time complexity is O(n), where n is the length of the input array, since each element is processed once in constant time.

#413

Medium

auto_awesome状态·转移·动态规划

LeetCode 题解工作台

等差数列划分

如果一个数列至少有三个元素，并且任意两个相邻元素之差相同，则称该数列为等差数列。例如， [1,3,5,7,9] 、 [7,7,7,7] 和 [3,-1,-5,-9] 都是等差数列。给你一个整数数组 nums ，返回数组 nums 中所有为等差数组的子数组个数。子数组是数组中的一个连续…

数组动态规划滑动窗口

题目描述

如果一个数列 至少有三个元素 ，并且任意两个相邻元素之差相同，则称该数列为等差数列。

例如，[1,3,5,7,9]、[7,7,7,7] 和 [3,-1,-5,-9] 都是等差数列。

给你一个整数数组 nums ，返回数组 nums 中所有为等差数组的 子数组 个数。

子数组 是数组中的一个连续序列。

示例 1：

输入：nums = [1,2,3,4]
输出：3
解释：nums 中有三个子等差数组：[1, 2, 3]、[2, 3, 4] 和 [1,2,3,4] 自身。

示例 2：

输入：nums = [1]
输出：0

提示：

1 <= nums.length <= 5000
-1000 <= nums[i] <= 1000

lightbulb

解题思路

方法一：遍历计数

我们用 $d$ 表示当前相邻两个元素的差值，用 $cnt$ 表示当前等差数列的长度，初始时 $d = 3000$ , $cnt = 2$ 。

遍历数组 nums，对于相邻的两个元素 $a$ 和 $b$ ，如果 $b - a = d$ ，则说明当前元素 $b$ 也属于当前等差数列，此时 $cnt$ 自增 1；否则说明当前元素 $b$ 不属于当前等差数列，此时更新 $d = b - a$ ，且 $cnt = 2$ 。如果 $cnt \ge 3$ ，则说明当前等差数列的长度至少为 3，此时等差数列的个数为 $cnt - 2$ ，将其加到答案中。

遍历结束后，即可得到答案。

在代码实现上，我们也可以将 $cnt$ 初始化为 $0$ ，重置 $cnt$ 时，直接将 $cnt$ 置为 $0$ ；累加答案时，直接累加 $cnt$ 即可。

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

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class Solution:
    def numberOfArithmeticSlices(self, nums: List[int]) -> int:
        ans = cnt = 0
        d = 3000
        for a, b in pairwise(nums):
            if b - a == d:
                cnt += 1
            else:
                d = b - a
                cnt = 0
            ans += cnt
        return ans

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
Check if the candidate can optimize the solution using dynamic programming and state transition techniques.
question_mark
Assess the candidate's ability to optimize space complexity, particularly when handling large input sizes.
question_mark
Look for a clear understanding of sliding window techniques and how they can apply to sequence-based problems.

warning

常见陷阱

外企场景

error
Failing to account for subarrays with fewer than three elements, which are not considered arithmetic subarrays.
error
Incorrectly updating the count of arithmetic subarrays, leading to over-counting or under-counting.
error
Using brute-force methods to check each possible subarray, resulting in an inefficient solution.

swap_horiz

进阶变体

外企场景

arrow_right_alt
What if the array contains only two elements? How would the solution change?
arrow_right_alt
Can we solve this problem with a more memory-efficient approach while maintaining optimal time complexity?
arrow_right_alt
What happens if the array contains large negative or positive numbers? How would the approach handle extreme values?

help

常见问题

外企场景

继续练习

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