Question

Formatted question description: https://leetcode.ca/all/396.html

 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.

 @tag-array

Algorithm

Abstract the concrete numbers as A, B, C, D, then we can get:

F(0) = 0A + 1B + 2C +3D

F(1) = 0D + 1A + 2B +3C

F(2) = 0C + 1D + 2A +3B

F(3) = 0B + 1C + 2D +3A

Then, through careful observation, we can draw the following law:

sum = 1A + 1B + 1C + 1D

F(1) = F(0) + sum-4D

F(2) = F(1) + sum-4C

F(3) = F(2) + sum-4B

Then we have found the rule, F(i) = F(i-1) + sum-n*A[n-i]

Code

Java

public class Rotate_Function {

    class Solution {
        public int maxRotateFunction(int[] A) {
            int prevValue = 0;
            int sum = 0;
            int n = A.length;

            for (int i = 0; i < n; ++i) {
                sum += A[i]; // get: 1A+1B+1C+1D+...
                prevValue += i * A[i]; // get: F(0) first
            }

            int result = prevValue;
            for (int i = 1; i < n; i++) { // start from index=1
                prevValue = prevValue + sum - n * A[n - i];
                result = Math.max(result, prevValue);
            }

            return result;
        }
    }
}

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