396. Rotate Function

Posted 2020-08-09 咖啡中不塌缩的方糖

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.

 

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;
    }