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396. 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 = 25F(1) = (0 * 6) + (1 * 4) + (2 * 3) + (3 * 2) = 0 + 4 + 6 + 6 = 16F(2) = (0 * 2) + (1 * 6) + (2 * 4) + (3 * 3) = 0 + 6 + 8 + 9 = 23F(3) = (0 * 3) + (1 * 2) + (2 * 6) + (3 * 4) = 0 + 2 + 12 + 12 = 26So the maximum value of F(0), F(1), F(2), F(3) is F(3) = 26.
public int MaxRotateFunction(int[] A) { int size = A.Count(); int res = 0; for(int i = 0; i< size; i++) { int sum = 0; int k = i; for(int j = 0;j< size;j++) { sum += A[j]*((j+k)%size); } res = i==0? sum:Math.Max(res,sum); } return res; }
DP
public int MaxRotateFunction(int[] A) { int size = A.Count(); if(size == 0) return 0; var f = new int[size]; int sum = 0; int add = 0; for(int i = 0; i< size; i++) { sum += A[i]*i; add += A[i]; } f[0] = sum; int res = sum; for(int i = 1; i< size; i++) { f[i] = f[i-1] + add - size*A[size-i]; res = Math.Max(res,f[i]); } return res; }
396. Rotate Function
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