Solution Explanation

This problem asks to find the maximum value of the rotation function F(k) for an integer array nums, where F(k) is the sum of the products of each element and its index after rotating the array k times. A naive approach would be to calculate F(k) for all k from 0 to n-1, but this would lead to O(n^2) time complexity. The optimized solution uses the relationship between consecutive rotations to achieve linear time complexity.

Core Idea:

The key observation is that the rotation function F(k) and F(k+1) are closely related. When we rotate the array by one position, the value of F changes in a predictable way. Let's analyze this change:

• F(k): 0*a[0] + 1*a[1] + 2*a[2] + ... + (n-1)*a[n-1] (assuming a is the rotated array).
• F(k+1): 0*a[n-1] + 1*a[0] + 2*a[1] + ... + (n-1)*a[n-2]

Notice the difference: The last element a[n-1] is now multiplied by 0 instead of n-1, while other elements shift one position to the left with their multipliers increased by 1. To get F(k+1) from F(k):

1. Subtract (n-1)*a[n-1] (the contribution of the last element in F(k)).
2. Add sum(nums) (because each element’s multiplier increased by 1).
3. Subtract n*a[n-1] (because the last element’s multiplier is now 0 instead of (n-1), we need to subtract the excess).

Therefore, we can iteratively compute F(k+1) from F(k) in O(1) time. This avoids recalculating the entire sum for each rotation.

Algorithm:

1. Initialization: Calculate F(0) directly.
2. Iteration: Iteratively compute F(k) for k = 1, 2, ..., n-1 using the formula derived above: F(k+1) = F(k) + sum(nums) - n * nums[n-k-1].
3. Maximization: Keep track of the maximum value of F(k) encountered during the iteration.
4. Return: Return the maximum value.

Time Complexity: O(n) - We iterate through the array once to calculate F(0) and then n-1 times to compute subsequent F(k) values.

Space Complexity: O(1) - We use only a few variables to store the intermediate results.

Code Explanation (Python)

The Python code efficiently implements this algorithm:

class Solution:
    def maxRotateFunction(self, nums: List[int]) -> int:
        f = sum(i * v for i, v in enumerate(nums)) #calculate F(0)
        n, s = len(nums), sum(nums) #store length and sum
        ans = f # initialize max value
        for i in range(1, n): #iterate to calculate remaining F(k)
            f = f + s - n * nums[n - i] #apply the formula
            ans = max(ans, f) # update max value if needed
        return ans
 

The code in other languages (Java, C++, Go, Typescript, Rust) follows the same logic, adapting the syntax and data structures accordingly. The core algorithmic steps remain consistent across all implementations.