# 396-rotate-function

## Question

https://leetcode.com/problems/rotate-function/description/

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 + 1 * Bk + ... + (n-1) * Bk[n-1].

Calculate the maximum value of F(0), F(1), ..., F(n-1).

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.
``````

## Thought Process

1. Math
1. As we rotate the number clockwise, the value is increased by sum - the length * preLastNum
2. We start from the back, since every rotation will move the last element to the first
3. Time complexity O(n)
4. Space complexity O(1)

## Solution

``````class Solution {
public int maxRotateFunction(int[] A) {
// F(k) = 0 * Bk + 1 * Bk + ... + (n-1) * Bk[n-1]
// F(k+1) = 0 * Bk+1 + 1 * Bk+1 + ... + (n-1) * Bk+1[n-1]
//        = 0 * Bk[n - 1] + 1 * Bk + ... + (n -1) * Bk[n-2]
// F(k+1) - F(k) = Bk + Bk + .. Bk[n -2] - (n-1) * Bk[n-1]
//               = sum(0 to n - 1) - n * Bk[n - 1]
int sum = 0, F = 0, n = A.length;
for (int i = 0; i < n; i++) {
sum += A[i];
F += i * A[i];
}
int max = F;
for (int i = n - 1; i > 0; i--) {
F = F + sum - n * A[i];
max = Math.max(max, F);
}
return max;
}
}
``````