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.

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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


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Constraints:


	n == nums.length
	1 &lt;= n &lt;= 105
	-100 &lt;= nums[i] &lt;= 100


class Solution:
    def maxRotateFunction(self, nums: List[int]) -> int:
        preSum, cur = 0, 0
        for i in range(len(nums)):
            cur += i * nums[i]
            preSum += nums[i]
        ans = cur
        for i in range(1, len(nums)):
            cur -= len(nums) * nums[len(nums) - i]
            cur += preSum
            ans = max(ans, cur)
        return ans
class Solution {
    public int maxRotateFunction(int[] nums) {
        int sum1 =0,sum2 = 0;
        for(int i=0;i<nums.length;i++){
            sum1 += nums[i];
            sum2 += i*nums[i];
        }
        int result = sum2;
        for(int i=0;i<nums.length;i++){
            sum2 = sum2-sum1+(nums.length)*nums[i];
            result = Math.max(result,sum2);
        }
        return result;
    }
}
class Solution {
public:
    int maxRotateFunction(vector<int>& A) {
        long sum = 0, fn = 0;
        int len = A.size();
        for(int i=0;i<len;i++) {
            sum += A[i];
            fn += (i * A[i]);
        }

        long l = 1, r;
        long newfn = fn;
        
        while(l < len) {
            r = l + len - 1;
            
            long removed = (l-1) * A[l-1];
            long added = r * A[r%len];
            
            newfn = newfn - removed + added - sum;
            
            fn = max(fn, newfn);
            
            l++;
        }
        
        return (int)fn;
    }
};
var maxRotateFunction = function(nums) { 
    let n = nums.length;
    let dp = 0;
    
    let sum = 0; 
    for (let i=0; i<n;i++) {
        sum += nums[i];
        dp += i*nums[i];
    }
    let max = dp;
    for (let i=1; i<n;i++) {
        dp += sum - nums[n-i]*n;
        max = Math.max(max, dp);
    } 
    return max;
};
Rotate Function