396. Rotate Function #

Problem #

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

Problem Restatement #

Given an integer array nums of length n, let arrk be the array obtained by rotating nums clockwise by k positions.

Define the rotation function F of nums as:

• F(k) = 0 * arrk[0] + 1 * arrk[1] + ... + (n - 1) * arrk[n - 1]

Return the maximum value among F(0), F(1), ..., F(n-1).

Solution Approach #

Abstract observation:

nums = [A0, A1, A2, A3]

sum  = A0 + A1 + A2+ A3
F(0) = 0*A0 +0*A0 + 1*A1 + 2*A2 + 3*A3
    
F(1) = 0*A3 + 1*A0 + 2*A1 + 3*A2 
     = F(0) + (A0 + A1 + A2) - 3*A3 
     = F(0) + (sum-A3) - 3*A3 
     = F(0) + sum - 4*A3

F(2) = 0*A2 + 1*A3 + 2*A0 + 3*A1 
     = F(1) + A3 + A0 + A1 - 3*A2 
     = F(1) + sum - 4*A2

F(3) = 0*A1 + 1*A2 + 2*A3 + 3*A0 
     = F(2) + A2 + A3 + A0 - 3*A1 
     = F(2) + sum - 4*A1
    
// Let sum be the sum of all elements in the nums array
// It can be conjectured that when 0 ≤ i < n, the following formula holds:
F(i) = F(i-1) + sum - n * A(n-i)

Proof of the recurrence formula by mathematical induction:

According to the rotation function formula given in the problem, we can obtain the known conditions:

• F(0) = 0×nums[0] + 1×nums[1] + ... + (n−1)×nums[n−1];

• F(1) = 1×nums[0] + 2×nums[1] + ... + 0×nums[n-1].

Let the sum of all elements in the array nums be sum. Use mathematical induction to verify that when 1 ≤ k < n, F(k) = F(k-1) + sum - n×nums[n-k] holds.

Base case: Prove that the proposition holds when k=1.

F(1) = 1×nums[0] + 2×nums[1] + ... + 0×nums[n-1]
     = F(0) + sum - n×nums[n-1]

Induction hypothesis: Assume that F(k) = F(k-1) + sum - n×nums[n-k] holds.

Inductive step: From the induction hypothesis, derive that F(k+1) = F(k) + sum - n×nums[n-(k+1)] holds, so the assumed recurrence formula holds.

F(k+1) = (k+1)×nums[0] + k×nums[1] + ... + 0×nums[n-1]
       = F(k) + sum - n×nums[n-(k+1)] 

Therefore, the recurrence formula can be obtained:

• When n = 0, F(0) = 0×nums[0] + 1×nums[1] + ... + (n−1)×nums[n−1]
• When 1 ≤ k < n, F(k) = F(k-1) + sum - n×nums[n-k] holds.

Loop through 0 ≤ k < n, compute the different F(k) values, and continuously update the maximum value to obtain the maximum value among F(0), F(1), ..., F(n-1).

Code #

package leetcode

func maxRotateFunction(nums []int) int {
	n := len(nums)
	var sum, f int
	for i, num := range nums {
		sum += num
		f += i * num // F(0)
	}
	ans := f
	for i := 1; i < n; i++ {
		f += sum - n*nums[n-i] // F(i) = F(i-1) + sum - n*nums[n-i]
		if f > ans {
			ans = f
		}
	}
	return ans
}