Rotate function

Time: O(N); Space: O(1); easy

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 1:

Input: A = [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.

[1]:

class Solution1(object):
    def maxRotateFunction(self, A):
        """
        :type A: List[int]
        :rtype: int
        """
        s = sum(A)
        fi = 0
        for i in range(len(A)):
            fi += i * A[i]

        result = fi
        for i in range(1, len(A)+1):
            fi += s - len(A) * A[-i]
            result = max(result, fi)
        return result

[3]:

s = Solution1()
A = [4, 3, 2, 6]
assert s.maxRotateFunction(A) == 26