396. Rotate Function

Q:

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 = 25F(1) = (0 * 6) + (1 * 4) + (2 * 3) + (3 * 2) = 0 + 4 + 6 + 6 = 16F(2) = (0 * 2) + (1 * 6) + (2 * 4) + (3 * 3) = 0 + 6 + 8 + 9 = 23F(3) = (0 * 3) + (1 * 2) + (2 * 6) + (3 * 4) = 0 + 2 + 12 + 12 = 26So the maximum value of F(0), F(1), F(2), F(3) is F(3) = 26.

A:

Consider we have an Array containing 5 items A,B,C,D,E.

According to the problem statement

F(0) = (0A) + (1B) + (2C) + (3D) + (4E)

F(1) = (4A) + (0B) + (1C) + (2D) + (3E) = F(0) - (A + B + C + D + E) + (5A) 

F(2) = (3A) + (4B) + (0C) + (1D) + (2E) = F(1) - (A + B + C + D + E) + (5B)

......

We can easily get the following conclusion:

F(0) = (0A) + (1B) + (2C) + (3D) + (4E)

F(n) = F(n-1) - (A + B + C + D + E) + (5*Array[n-1])

C:

var maxRotateFunction = function(A) {
    var sumA = A.reduce(function (prev, cur) {
        return prev + cur;
    }, 0);
    var prevRotationSum = A.reduce(function (prev, cur, index) {
        return prev + cur * index;
    }, 0);
    var maxSum = prevRotationSum;

    for(var i = 1; i < A.length ; ++i) {
        prevRotationSum += (A.length * A[i-1] - sumA);
        maxSum = Math.max(prevRotationSum, maxSum);
    }
    return maxSum;
};