Rotate function

Given an array of integers A and let n to be its length.

Assume Bk to be an array obtained by rotating the array A k positions clock-wise, we define a "rotation function" F on A as follow:

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

Calculate the maximum value of F(0), F(1), ..., F(n-1).

Note:
n is guaranteed to be less than 105.

Example:

A = [4, 3, 2, 6]

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.

This is a math problem. :(

Do this:

f = 0 *A[0] + 1 * A[1] + ... + (n - 1) * A[n - 1]

sum = A[0] + A[1] + ... + A[n - 1]

f + sum = 1 * A[0] + 2 * A[1] + ... + n * A[n - 1]

k = 1 f = f + sum - n * A[n - k] = 1 * A[0] + 2 * A[1] + ... + n * A[n - 1] - n * A[n - 1]
                                             =  1 * A[0] + 2 * A[1] + ... + 0 * A[n - 1] = f(n - 1)
k = 2 f = 2 * A[0] + 3 * A[1] + ... + n * A[n - 2] + 1 * A[n - 1] - n * A[n - 2] = f(n - 2)
.
.
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public int maxRotateFunction(int[] A) {
        int len = A.length;
        
        int f = 0;
        int sum = 0;
        for (int i = 0; i < len; i++) {
            f += i * A[i];
            sum += A[i];
        }
        
        int max = f;
        
        for (int k = 1; k < len; k++) {
            f = f + sum - len * A[len - k];
            max = Math.max(max, f);
        }
        return max;
    }