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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