Rotation function of leetcode medium problem

subject

Given a length of n Array of integers for nums .
hypothesis arrk It's an array nums Clockwise rotation k Array after position , We define nums Of Rotation function F by ：
F(k) = 0 * arrk[0] + 1 * arrk[1] + … + (n - 1) * arrk[n - 1]
return F(0), F(1), …, F(n-1) Maximum of .
The generated test cases make the answers meet the requirements 32 position Integers .
Example 1:
Input : nums = [4,3,2,6]
Output : 26
explain :
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
therefore F(0), F(1), F(2), F(3) The maximum value in is F(3) = 26 .
Example 2:
Input : nums = [100]
Output : 0
Tips :
n == nums.length
1 <= n <= 10^5
-100 <= nums[i] <= 100
source ： Power button （LeetCode）

Their thinking

   The practice of the topic is easy to understand , As an example 1 It means that the subscript does not move, and each element of the array moves to the right or the element does not move , Move subscript left , We can complete the requirements of the subject by choosing one fixed and the other .

class Solution:
    def maxRotateFunction(self, nums: List[int]) -> int:
        index=deque([i for i in range(len(nums))]) # Rotation subscript 
        MAX=-float('inf')
        for i in range(len(nums)):
            s=0
            for j in range(len(nums)):
                s+=nums[j]*index[j]
            if s>MAX:
                MAX=s
            index.append(index.popleft()) # Move subscript left 
        return MAX

   Obviously, all arrays and their subscripts are evaluated each time , Complexity is O(n^2) So it's very easy to time out , Therefore, we need to use mathematical methods to reduce the amount of calculation , If you have no idea, you can write on the paper , When we will F(0),F(1) You'll find it when you write your expression , This is actually before high school n Items and common means , The essence of Gauss's calculation is the reduction of dislocation. .

class Solution:
    def maxRotateFunction(self, nums: List[int]) -> int:
        s,MAX,S,k=0,-float('inf'),sum(nums),1
        for i,j in enumerate(nums):
            s+=i*j
        for i in range(len(nums)):
            if s>MAX:
                MAX=s
            s+=S-len(nums)*nums[-k]
            k+=1
        return MAX