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.

Approach #1: Math. [Java]

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

　　

Analysis:

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

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

F(k) = F(k-1) + sum - nBk[0]
What is Bk[0]?

k = 0; B[0] = A[0];
k = 1; B[0] = A[len-1];
k = 2; B[0] = A[len-2];

Reference:

https://leetcode.com/problems/rotate-function/discuss/87853/Java-O(n)-solution-with-explanation