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" 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.

Solution:

Example: A = [A, B, C, D]

F(0) = 0A + 1B + 2C + 3D

F(1) = 3A + 0B + 1C + 2D

F(2) = 2A + 3B + 0C + 1D

F(3) = 1A + 2B + 3C + 0D

We can see

F(1) - F(0) = 4A - A - B - C - D = 4A - sum

F(2) - F(1) = 4B - A - B - C - D = 4B - sum

F(3) - F(2) = 4C - sum

Therefore,

F(k) = F(k - 1) - sum + n x A[i - 1]

The time complexity is O(n) and the space complexity is O(1).



Code:

public class Solution {
    public int maxRotateFunction(int[] A) {
        int n = A.length;
        int sum = 0;
        int F = 0;
        for (int i = 0; i < n; i++) {
            sum += A[i];
            F += i * A[i];
        }
        int max = F;
        for (int i = 1; i < n; i++) {
            F = F - sum + n * A[i - 1];
            max = Math.max(max, F);
        }
        return max;
    }
}