Hex.*

LC396 - Rorate Function

Posted on In
Leetcode: 114/2658 (51+50+13)

Description

You are given an integer array nums of length n.

Assume arrk to be an array obtained by rotating nums by k positions clock-wise. We define the rotation function F on nums as follow:

F(k) = 0 * arrk[0] + 1 * arrk[1] + … + (n - 1) * arrk[n - 1].
Return the maximum value of F(0), F(1), …, F(n-1).

The test cases are generated so that the answer fits in a 32-bit integer.

Example 1:

Input: nums = [4,3,2,6]
Output: 26
Explanation:
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.
Example 2:

Input: nums = [100]
Output: 0

Constraints:

n == nums.length
1 <= n <= 105
-100 <= nums[i] <= 100

Solution

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
nums: [A0,A1,A2,A3]

 F0 = 0*A0 + 1*A1 + 2*A2 + 3*A3

 F1 = 0*A3 + 1*A0 + 2*A1 + 3*A2
    = F0 + A0 + A1 + A2 - 3*A3
    = F0 + sum-A3 - 3*A3
    = F0 + sum - 4*A3

 F2 = 0*A2 + 1*A3 + 2*A0 + 3*A1
    = F1 + A3 + A0 + A1 - 3*A2
    = F1 + sum - 4*A2

 F3 = 0*A1 + 1*A2 + 2*A3 + 3*A0
    = F2 + A2 + A3 + A0 - 3*A1
    = F2 + sum - 4*A1

Formula：
F(i) = F(i-1) + sum - n * A(n-i)
1
2
3
4
5
6
7
8
9
10
11
12
13
14
# O(n) time | O(1) space
class Solution:
    def maxRotateFunction(self, nums: List[int]) -> int:
        summ = sum(nums)
        length = len(nums)
        f_pre=0
        for i, v in enumerate(nums):
            f_pre += i*v

        res = f_pre
        for i in range(length):
            f_pre = f_pre + summ - length * nums[length-1-i]
            res = max(res, f_pre)
        return res