Rotate Function

LeetCode Q 396 - 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” Fon 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.

Solution:

F(k) = 0 * Bk[0] + 1 * Bk[1] + ... + (n-1) * Bk[n-1]
F(k+1) = 0 * Bk[n -1] + 1 * Bk[0] + ... + (n-1) * Bk[n-2]
So,
F(k+1) - F(k) = Bk[0] + Bk[1] + ... + Bk[n-2] - (n-1) * Bk[n-1]
F(k+1) - F(k) = sum - n * Bk[n - 1]
F(k+1) = sum - n * Bk[n - 1] + F(k)

Then, what is Bk[n - 1]?
For k = 0, B0[n-1] = A[n - 1]
For k = 1, B1[n-1] = A[n - 2]
…
For k, Bk[n-1] = A[n - 1 - k]

So,
F(k + 1) = sum + F(k) + A[n - 1 - k]. For example,
F(1) = sum + F(0) + A[n - 1]; F(2) = sum + F(1) + A[n - 2];…

Code:

public int maxRotateFunction(int[] A) {
  if (A == null || A.length == 0) return 0;

  int sum = 0, F = 0, n = A.length;
  for (int i = 0; i < n; i++) {
    sum += A[i];
    F += i * A[i];
  }

  int res = F;
  for (int i = 1; i < n; i++) {
    F = F + sum - n * A[n - i];
    res = Math.max(res, F);
  }

  return res;
}