Leetcode 396. Rotate Function (Easy) (java)

Leetcode 396. Rotate Function (Easy) (java)

Tag: Array

Difficulty: Easy

/*

396. Rotate Function (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 10^5.

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.

*/
public class Solution {
public int maxRotateFunction(int[] A) {
if (A == null || A.length == 0) return 0;
int sum = 0, res = Integer.MIN_VALUE, temp = 0;
for (int i = 0; i < A.length; i++) {
sum += A[i];
temp += i * A[i];
}
res = temp;
for (int i = 1; i < A.length; i++) {
temp = sum - A.length * A[A.length - i] + temp;
res = Math.max(res, temp);
}
return res;
}
};