# LeetCode Rotate Function

LeetCode Rotate Function

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.

class Solution {
public:
int maxRotateFunction(vector<int>& A) {
if (A.empty())return 0;
int ans = INT_MIN, n = A.size();
for (int k = 0; k < n; ++k) {
int cur = 0;
for (int i = 0; i < n; ++i) {
cur += ((k + i) % n)*A[i];
}
ans = max(ans, cur);
}
return ans;
}
};


class Solution {
public:
int maxRotateFunction(vector<int>& A) {
if (A.empty())return 0;
int n = A.size(), sum = 0, f0 = 0;
for (int i = 0; i < n; ++i) {
sum += A[i];
f0 += i*A[i];
}
int ans = f0, pre = f0;
for (int i = 1; i < n; ++i) {
int cur = pre + sum - n * A[n - i];
ans = max(ans, cur);
pre = cur;
}
return ans;
}
};