396. Rotate Function

problem

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.

solution

先遍历下标，然后遍历元素，性能没过关；嵌套次反过来呢？,

怎么嵌套大致都是O(n²)，差别不大

class Solution(object):
    def maxRotateFunction(self, A):
        """
        :type A: List[int]
        :rtype: int
        """
        sums = []
        length = len(A)
        for i in range(length):
            sum = 0
            for index,item in enumerate(A):
                sum += (index+i)%length * item
            sums.append(sum)
        if  not len(sums):
            sums.append(0)
        return max(sums)

class Solution(object):
    def maxRotateFunction(self, A):
        """
        :type A: List[int]
        :rtype: int
        """

        length = len(A)
        sums = [0]*length
        
        for index,item in enumerate(A):
            for offset in range(length):
                sums[offset] += (index+offset)%length * item
        if  not len(sums):
            sums.append(0)
        return max(sums)

discuss

数学寻找关联，然后求解

F(k) = 0 * Bk[0] + 1 * Bk[1] + ... + (n-1) * Bk[n-1]
F(k-1) = 0 * Bk-1[0] + 1 * Bk-1[1] + ... + (n-1) * Bk-1[n-1]
       = 0 * Bk[1] + 1 * Bk[2] + ... + (n-2) * Bk[n-1] + (n-1) * Bk[0]
Then,

F(k) - F(k-1) = Bk[1] + Bk[2] + ... + Bk[n-1] + (1-n)Bk[0]
              = (Bk[0] + ... + Bk[n-1]) - nBk[0]
              = sum - nBk[0]
Thus,

F(k) = F(k-1) + sum - nBk[0]
What is Bk[0]?

k = 0; B[0] = A[0];
k = 1; B[0] = A[len-1];
k = 2; B[0] = A[len-2];
...

int maxRotateFunction(int* A, int ASize) {
    int i, last, sum,m;
    
    if(ASize<=1) return 0;
    
    last=0;
    sum=0;
    for(i=0;i<ASize;i++){
        last += A[i]*i;
        sum+=A[i];
    }
    m=last ;
    for(i=ASize-1;i>=1;i--){
        last += sum - (ASize*A[i]);
        if (last > m) m= last;
    }
    
    return m;
}