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 = 25F(1) = (0 * 6) + (1 * 4) + (2 * 3) + (3 * 2) = 0 + 4 + 6 + 6 = 16F(2) = (0 * 2) + (1 * 6) + (2 * 4) + (3 * 3) = 0 + 6 + 8 + 9 = 23F(3) = (0 * 3) + (1 * 2) + (2 * 6) + (3 * 4) = 0 + 2 + 12 + 12 = 26So the maximum value of F(0), F(1), F(2), F(3) is F(3) = 26.


Analyse:

0 1 2 3 4 5

5 0 1 2 3 4  f(0) - sumAll + N * A[0]

4 5 0 1 2 3  f(1) - sumAll + N * A[1]

3 4 5 0 1 2  f(2) - sumAll + N * A[2]

2 3 4 5 0 1  f(3) - sumAll + N * A[3]

1 2 3 4 5 0  f(4) - sumAll + N * A[4]

 1 class Solution { 2 public: 3     int maxRotateFunction(vector<int>& A) { 4         if (A.empty()) return 0; 5          6         int sumAll = 0; 7         int previous = 0; 8         for (int i = 0; i < A.size(); i++) { 9             sumAll += A[i];10             previous += i * A[i];11         }12         int result = previous;13         14         for (int i = 0; i < A.size() - 1; i++) {15             previous = previous - sumAll + A.size() * A[i];16             result = max(result, previous);17         }18         return result;19     }20 };