What is the primary pattern used in solving the Rotate Function problem?

The primary pattern used is state transition dynamic programming, where each rotation's result is computed based on the previous one.

How can I optimize the solution to handle large arrays?

Use dynamic programming to reduce the time complexity to O(n) by avoiding recalculating the sum for each rotation.

Why does the solution need to keep track of the maximum rotation function value?

Tracking the maximum ensures you find the optimal rotation value, which is the solution to the problem.

Can the Rotate Function problem be solved without dynamic programming?

While it is possible, a brute force solution would require recalculating the sum for each rotation, leading to a time complexity of O(n^2), which is inefficient for large inputs.

How does the Rotate Function solution apply to other array-based dynamic programming problems?

The problem illustrates how to optimize array operations through dynamic programming, a technique applicable to other problems involving rotating or shifting array elements.

#396

Medium

auto_awesome状态·转移·动态规划

LeetCode 题解工作台

旋转函数

给定一个长度为 n 的整数数组 nums 。假设 arr k 是数组 nums 顺时针旋转 k 个位置后的数组，我们定义 nums 的旋转函数 F 为： F(k) = 0 * arr k [0] + 1 * arr k [1] + ... + (n - 1) * arr k [n - 1] 返回…

数组数学动态规划

题目描述

给定一个长度为 n 的整数数组 nums 。

假设 arr_k 是数组 nums 顺时针旋转 k 个位置后的数组，我们定义 nums 的 旋转函数 F 为：

F(k) = 0 * arr_k[0] + 1 * arr_k[1] + ... + (n - 1) * arr_k[n - 1]

返回 F(0), F(1), ..., F(n-1)中的最大值 。

生成的测试用例让答案符合 32 位 整数。

示例 1:

输入: nums = [4,3,2,6]
输出: 26
解释:
F(0) = (0 * 4) + (1 * 3) + (2 * 2) + (3 * 6) = 0 + 3 + 4 + 18 = 25
F(1) = (0 * 6) + (1 * 4) + (2 * 3) + (3 * 2) = 0 + 4 + 6 + 6 = 16
F(2) = (0 * 2) + (1 * 6) + (2 * 4) + (3 * 3) = 0 + 6 + 8 + 9 = 23
F(3) = (0 * 3) + (1 * 2) + (2 * 6) + (3 * 4) = 0 + 2 + 12 + 12 = 26
所以 F(0), F(1), F(2), F(3) 中的最大值是 F(3) = 26 。

示例 2:

输入: nums = [100]
输出: 0

提示:

n == nums.length
1 <= n <= 10⁵
-100 <= nums[i] <= 100

lightbulb

解题思路

方法一

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class Solution:
    def maxRotateFunction(self, nums: List[int]) -> int:
        f = sum(i * v for i, v in enumerate(nums))
        n, s = len(nums), sum(nums)
        ans = f
        for i in range(1, n):
            f = f + s - n * nums[n - i]
            ans = max(ans, f)
        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
Look for the candidate's understanding of state transition dynamic programming and its application to rotating arrays.
question_mark
Evaluate how well the candidate avoids recalculating the sum for each rotation and instead uses previous results.
question_mark
Assess if the candidate can optimize the algorithm to meet the problem's constraints, especially with a time complexity of O(n).

warning

常见陷阱

外企场景

error
Failing to utilize dynamic programming, leading to redundant calculations for each rotation.
error
Not maintaining the maximum function value during the iterations, which can result in incorrect answers.
error
Overcomplicating the solution by recalculating the weighted sum from scratch for each rotation.

swap_horiz

进阶变体

外企场景

arrow_right_alt
Rotate the array k positions to the left instead of the right.
arrow_right_alt
Consider variations where the array elements are all the same, simplifying the rotation function calculation.
arrow_right_alt
Change the weights used in the rotation function from simple indices to custom values based on a given formula.

help

常见问题

外企场景

继续练习

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