Rotate Function (Easy)

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保证不会超过10的5次方

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.

Java Solution O(n) with non mathametical explaination

Consider we have 5 coins A,B,C,D,E

According to the problem statement
F(0) = (0_A) + (1_B) + (2_C) + (3_D) + (4_E)
F(1) = (4_A) + (0_B) + (1_C) + (2_D) + (3_E)
F(2) = (3_A) + (4_B) + (0_C) + (1_D) + (2*E)

This problem at a glance seem like a difficult problem. I am not very strong in mathematics, so this is how I visualize this problem

We can construct F(1) from F(0) by two step:
Step 1. taking away one count of each coin from F(0), this is done by subtracting "sum" from "iteration" in the code below
after step 1 F(0) = (-1_A) + (0_B) + (1_C) + (2_D) + (3*E)

Step 2. Add n times the element which didn't contributed in F(0), which is A. This is done by adding "A[j-1]_len" in the code below.
after step 2 F(0) = (4_A) + (0_B) + (1_C) + (2_D) + (3_E)

At this point F(0) can be considered as F(1) and F(2) to F(4) can be constructed by repeating the above steps.

Hope this explanation helps, cheers!

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

        int sum =0, iteration = 0, len = A.length;

        for(int i=0; i<len; i++){
            sum += A[i];
            iteration += (A[i] * i);
        }

        int max = iteration;
        for(int j=1; j<len; j++){
            // for next iteration lets remove one entry value of each entry and the prev 0 * k
            iteration = iteration - sum + A[j-1]*len;
            max = Math.max(max, iteration);
        }

        return max;
}