# Question

Formatted question description: https://leetcode.ca/all/396.html

You are given an integer array nums of length n.

Assume arrk to be an array obtained by rotating nums by k positions clock-wise. We define the rotation function F on nums as follow:

• F(k) = 0 * arrk[0] + 1 * arrk[1] + ... + (n - 1) * arrk[n - 1].

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

The test cases are generated so that the answer fits in a 32-bit integer.

Example 1:

Input: nums = [4,3,2,6]
Output: 26
Explanation:
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.


Example 2:

Input: nums = [100]
Output: 0


Constraints:

• n == nums.length
• 1 <= n <= 105
• -100 <= nums[i] <= 100

# Algorithm

Abstract the concrete numbers as A, B, C, D, then we can get:

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

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

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

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


Then, through careful observation, we can draw the following law:

sum = 1A + 1B + 1C + 1D

F(1) = F(0) + sum-4D

F(2) = F(1) + sum-4C

F(3) = F(2) + sum-4B


Then we have found the rule, F(i) = F(i-1) + sum - n*A[n-i]

# Code

• 
public class Rotate_Function {

class Solution {
public int maxRotateFunction(int[] A) {
int prevValue = 0;
int sum = 0;
int n = A.length;

for (int i = 0; i < n; ++i) {
sum += A[i]; // get: 1A+1B+1C+1D+...
prevValue += i * A[i]; // get: F(0) first
}

int result = prevValue;
for (int i = 1; i < n; i++) { // start from index=1
prevValue = prevValue + sum - n * A[n - i];
result = Math.max(result, prevValue);
}

return result;
}
}
}

############

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

• // OJ: https://leetcode.com/problems/rotate-function/
// Time: O(N)
// Space: O(1)
class Solution {
public:
int maxRotateFunction(vector<int>& A) {
if (A.empty()) return 0;
long long f = 0, ans = INT_MIN, N = A.size(), sum = accumulate(A.begin(), A.end(), (long long)0);
for (int i = 0; i < N; ++i) f += i * A[i];
for (int i = N - 1; i >= 0; --i) ans = max(ans, f += (sum - N * A[i]));
return ans;
}
};

• class Solution:
def maxRotateFunction(self, nums: List[int]) -> int:
f = sum(i * v for i, v in enumerate(nums))
n, s = len(nums), sum(nums)
ans = f
for i in range(1, n): # starting at 1, not 0 which is f
f = f + s - n * nums[n - i]
ans = max(ans, f)
return ans

############

class Solution(object):
def maxRotateFunction(self, A):
"""
:type A: List[int]
:rtype: int
"""
if not A:
return 0

sumA = sum(A)
fk = 0
n = len(A)
for i, num in enumerate(A):
fk += i * num
idx = n - 1
ans = float("-inf")
for _ in range(n):
fk += sumA - n * A[idx]
ans = max(ans, fk)
idx -= 1
return ans


• func maxRotateFunction(nums []int) int {
f, s, n := 0, 0, len(nums)
for i, v := range nums {
f += i * v
s += v
}
ans := f
for i := 1; i < n; i++ {
f = f + s - n*nums[n-i]
if ans < f {
ans = f
}
}
return ans
}

• function maxRotateFunction(nums: number[]): number {
const n = nums.length;
const sum = nums.reduce((r, v) => r + v);
let res = nums.reduce((r, v, i) => r + v * i, 0);
let pre = res;
for (let i = 1; i < n; i++) {
pre = pre - (sum - nums[i - 1]) + nums[i - 1] * (n - 1);
res = Math.max(res, pre);
}
return res;
}


• impl Solution {
pub fn max_rotate_function(nums: Vec<i32>) -> i32 {
let n = nums.len();
let sum: i32 = nums.iter().sum();
let mut pre: i32 = nums.iter().enumerate().map(|(i, &v)| i as i32 * v).sum();
(0..n)
.map(|i| {
let res = pre;
pre = pre - (sum - nums[i]) + nums[i] * (n - 1) as i32;
res
})
.max()
.unwrap_or(0)
}
}