Introduction

In this blog post, we will explore how to solve the LeetCode problem "396" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

You are given an integer array nums of length n. Assume arrk to be an array obtained by rotating nums by k positions clock-wise. We define the rotation function F on nums as follow: F(k) = 0 arrk[0] + 1 arrk[1] + ... + (n - 1) arrk[n - 1]. Return the maximum value of F(0), F(1), ..., F(n-1). The test cases are generated so that the answer fits in a 32-bit integer. Example 1: Input: nums = [4,3,2,6] Output: 26 Explanation: 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 So the maximum value of F(0), F(1), F(2), F(3) is F(3) = 26. Example 2: Input: nums = [100] Output: 0 Constraints: n == nums.length 1 <= n <= 105 -100 <= nums[i] <= 100

Explanation

Here's a breakdown of the solution and the corresponding Python code:

• Key Idea: The core idea is to avoid recomputing the entire F(k) for each rotation. Instead, we leverage the relationship between F(k) and F(k-1) to calculate F(k) efficiently.
• Incremental Update: The relationship is: F(k) = F(k-1) + sum(nums) - n * nums[n-k] (or a mathematically equivalent expression for readability). We calculate F(0) initially and then iteratively update it to find F(1), F(2), and so on.

• Optimization: Precompute the sum of the array elements to avoid repeated calculations within the loop.

• Complexity: O(n) time complexity, O(1) space complexity.

Code

    def maxRotateFunction(nums):
    n = len(nums)
    if n == 0:
        return 0

    num_sum = sum(nums)
    f_0 = 0
    for i in range(n):
        f_0 += i * nums[i]

    max_f = f_0
    f_prev = f_0

    for k in range(1, n):
        f_k = f_prev + num_sum - n * nums[n - k]
        max_f = max(max_f, f_k)
        f_prev = f_k

    return max_f