RotateFunction.java

package math;

/**
 * Created by gouthamvidyapradhan on 18/03/2017.
 * Given an array of integers A and let n to be its length.
 * <p>
 * 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:
 * <p>
 * F(k) = 0 * Bk[0] + 1 * Bk[1] + ... + (n-1) * Bk[n-1].
 * <p>
 * Calculate the maximum value of F(0), F(1), ..., F(n-1).
 * <p>
 * Note:
 * n is guaranteed to be less than 105.
 * <p>
 * Example:
 * <p>
 * A = [4, 3, 2, 6]
 * <p>
 * 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
 * <p>
 * So the maximum value of F(0), F(1), F(2), F(3) is F(3) = 26.
 */
public class RotateFunction {
    /**
     * Main method
     *
     * @param args
     * @throws Exception
     */
    public static void main(String[] args) throws Exception {
        int[] a = {4, 3, 2, 6};
        System.out.println(new RotateFunction().maxRotateFunction(a));
    }

    public int maxRotateFunction(int[] A) {
        if (A.length == 0 || A.length == 1) return 0;
        int max = Integer.MIN_VALUE;
        int l = A.length;
        int sum = 0, prodSum = 0;
        for (int i = 0; i < l; i++) {
            prodSum += (A[i] * i);
            sum += A[i];
        }
        max = Math.max(max, prodSum);
        for (int i = 0; i < l - 1; i++) {
            prodSum = (prodSum - sum + A[i] + ((l - 1) * A[i]));
            max = Math.max(max, prodSum);
        }
        return max;
    }
}