# Solving Leetcode Interviews in Seconds with AI: Rotate Function


	# 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](https://www.chatmagic.app), 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
	```python
	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
	```