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.
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.
This is a math problem. :(
Do this:
f = 0 *A[0] + 1 * A[1] + ... + (n - 1) * A[n - 1]
sum = A[0] + A[1] + ... + A[n - 1]
f + sum = 1 * A[0] + 2 * A[1] + ... + n * A[n - 1]
k = 1 f = f + sum - n * A[n - k] = 1 * A[0] + 2 * A[1] + ... + n * A[n - 1] - n * A[n - 1]
= 1 * A[0] + 2 * A[1] + ... + 0 * A[n - 1] = f(n - 1)
k = 2 f = 2 * A[0] + 3 * A[1] + ... + n * A[n - 2] + 1 * A[n - 1] - n * A[n - 2] = f(n - 2)
.
.
.
public int maxRotateFunction(int[] A) { int len = A.length; int f = 0; int sum = 0; for (int i = 0; i < len; i++) { f += i * A[i]; sum += A[i]; } int max = f; for (int k = 1; k < len; k++) { f = f + sum - len * A[len - k]; max = Math.max(max, f); } return max; }
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