Leetcode 396.Rotate Function

Question:

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 = 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.

Explanation: OK,数学题Again。如果按照原题意解，会出现TLE的问题。

借用Top solution的思路分析下：

F(k) = 0 * Bk[0] + 1 * Bk[1] + ... + (n-1) * Bk[n-1]
F(k-1) = 0 * Bk-1[0] + 1 * Bk-1[1] + ... + (n-1) * Bk-1[n-1]
       = 0 * Bk[1] + 1 * Bk[2] + ... + (n-2) * Bk[n-1] + (n-1) * Bk[0]


 F(k) - F(k-1) = Bk[1] + Bk[2] + ... + Bk[n-1] + (1-n)Bk[0]
               = (Bk[0] + ... + Bk[n-1]) - nBk[0]
               = sum - nBk[0]
So:
F(k) = F(k-1) + sum - nBk[0]

What is Bk[0]?

k = 0; B[0] = A[0];
k = 1; B[0] = A[len-1];
k = 2; B[0] = A[len-2];
...

Code

class Solution(object):
    def maxRotateFunction(self, A):
        """
        :type A: List[int]
        :rtype: int
        """
        sumA = sum(A)
        countA = 0
        for index, value in enumerate(A):
            countA += index * value
        result = countA
        for i in range(len(A)-1,0,-1):
            countA = countA + sumA - len(A) * A[i]
            result = max(countA, result)
        return result