Problem Description

You have an integer array nums of length n. The problem asks you to find the maximum value among all possible rotation functions.

A rotation function F(k) is calculated for an array that has been rotated k positions clockwise. When you rotate an array by k positions clockwise, the last k elements move to the front of the array.

For each rotation k, the function F(k) is computed as:

• F(k) = 0 * arr_k[0] + 1 * arr_k[1] + 2 * arr_k[2] + ... + (n-1) * arr_k[n-1]

Where arr_k is the array after rotating the original nums by k positions.

Example to understand rotations:

• If nums = [4, 3, 2, 6]
• arr_0 = [4, 3, 2, 6] (no rotation)
• arr_1 = [6, 4, 3, 2] (rotated 1 position clockwise)
• arr_2 = [2, 6, 4, 3] (rotated 2 positions clockwise)
• arr_3 = [3, 2, 6, 4] (rotated 3 positions clockwise)

Calculating F(k) for each rotation:

• 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

The task is to return the maximum value among all F(k) where k ranges from 0 to n-1. In the example above, the answer would be 26.

Intuition

The key insight is to find a mathematical relationship between consecutive rotation functions F(k) and F(k+1) rather than computing each one from scratch.

Let's observe what happens when we go from F(k) to F(k+1):

• When rotating by one more position, the last element of the current array moves to the front
• Every other element shifts one position to the right, which means their multipliers increase by 1

For example, with nums = [A, B, C, D]:

• F(0) = 0*A + 1*B + 2*C + 3*D
• F(1) = 0*D + 1*A + 2*B + 3*C

Notice that when going from F(0) to F(1):

• Elements A, B, C each get multiplied by a coefficient that's 1 higher
• Element D goes from being multiplied by 3 to being multiplied by 0

This means: F(1) = F(0) + A + B + C - 3*D

More generally: F(1) = F(0) + (A + B + C + D) - 4*D

This can be written as: F(1) = F(0) + sum(nums) - n * nums[n-1]

For any transition from F(k) to F(k+1): F(k+1) = F(k) + sum(nums) - n * nums[n-k-1]

This pattern holds because:

1. Adding sum(nums) accounts for increasing each element's coefficient by 1
2. Subtracting n * nums[n-k-1] corrects for the element that wraps around from position (n-1) to position 0, losing its multiplier of (n-1)

With this recurrence relation, we can compute all F(k) values in O(n) time by:

1. Computing F(0) directly
2. Computing the sum of all elements once
3. Using the formula to derive each subsequent F(k) from the previous one
4. Tracking the maximum value encountered

Solution Approach

The implementation follows the mathematical relationship we discovered between consecutive rotation functions:

1. Initialize F(0): First, we calculate F(0) directly using the formula:

   f = sum(i * v for i, v in enumerate(nums))

   This computes 0 * nums[0] + 1 * nums[1] + ... + (n-1) * nums[n-1]

2. Precompute constants: We store two values that remain constant throughout:

   n, s = len(nums), sum(nums)

   • n: the length of the array
   • s: the sum of all elements in the array

3. Initialize the answer: Set the initial maximum to F(0):

   ans = f

4. Iterate through remaining rotations: For each rotation from 1 to n-1, we use the recurrence relation:

   for i in range(1, n):
       f = f + s - n * nums[n - i]
       ans = max(ans, f)

   In each iteration:

   • f = f + s - n * nums[n - i] updates F(i) based on F(i-1)
   • The term nums[n - i] identifies the element that was at the last position in the previous rotation and moves to the front
   • s - n * nums[n - i] represents adding 1 to each element's coefficient (via s) and then correcting for the element that wraps around (via - n * nums[n - i])
   • We update ans with the maximum value seen so far

5. Return the result: After checking all rotations, return the maximum value found:

   return ans

Time Complexity: O(n) - We compute F(0) in O(n) time and then compute each subsequent F(k) in O(1) time.

Space Complexity: O(1) - We only use a constant amount of extra space for variables f, n, s, and ans.

Example Walkthrough

Let's walk through the solution with nums = [4, 3, 2, 6].

Step 1: Calculate F(0)

F(0) = 0*4 + 1*3 + 2*2 + 3*6 = 0 + 3 + 4 + 18 = 25

So f = 25

Step 2: Precompute constants

n = 4 (length of array)
s = 4 + 3 + 2 + 6 = 15 (sum of all elements)

Step 3: Initialize answer

ans = 25 (start with F(0))

Step 4: Calculate F(1) using the recurrence relation

When we rotate from position 0 to position 1, the element at index n-1 = 3 (which is 6) moves from the last position to the first position.

Using our formula: F(1) = F(0) + s - n * nums[n-1]

F(1) = 25 + 15 - 4*6
F(1) = 25 + 15 - 24
F(1) = 16

Update: f = 16, ans = max(25, 16) = 25

Step 5: Calculate F(2)

The element at index n-2 = 2 (which is 2) was at the last position in rotation 1, now moves to the front.

Using our formula: F(2) = F(1) + s - n * nums[n-2]

F(2) = 16 + 15 - 4*2
F(2) = 16 + 15 - 8
F(2) = 23

Update: f = 23, ans = max(25, 23) = 25

Step 6: Calculate F(3)

The element at index n-3 = 1 (which is 3) was at the last position in rotation 2, now moves to the front.

Using our formula: F(3) = F(2) + s - n * nums[n-3]

F(3) = 23 + 15 - 4*3
F(3) = 23 + 15 - 12
F(3) = 26

Update: f = 26, ans = max(25, 26) = 26

Step 7: Return the result

return ans = 26

The maximum value among all rotation functions is 26, achieved at rotation k=3.

Solution Implementation

Time and Space Complexity

Time Complexity: O(n)

The algorithm consists of:

• Initial calculation of f using list comprehension with enumerate(): O(n)
• Calculating the sum of all elements s: O(n)
• A for loop that runs n-1 times: O(n)
  • Inside the loop, each operation (arithmetic and max comparison) takes O(1)

Since these operations are sequential, the overall time complexity is O(n) + O(n) + O(n) = O(n).

Space Complexity: O(1)

The algorithm uses only a constant amount of extra space:

• Variable f to store the current rotation function value
• Variables n and s to store the length and sum
• Variable ans to track the maximum value
• Loop variable i

All these variables use constant space regardless of the input size, resulting in O(1) space complexity.

Common Pitfalls

1. Incorrect Index in Recurrence Relation

One of the most common mistakes is using the wrong index when identifying which element moves from the last position to the first during rotation.

Incorrect Implementation:

for i in range(1, n):
    f = f + s - n * nums[i]  # Wrong! Uses nums[i] instead of nums[n-i]
    ans = max(ans, f)

Why it's wrong: When rotating by i positions, the element that was at the last position in the (i-1)th rotation and moves to the front is nums[n-i], not nums[i].

Correct Implementation:

for i in range(1, n):
    f = f + s - n * nums[n - i]  # Correct index
    ans = max(ans, f)

2. Misunderstanding Rotation Direction

Another pitfall is confusing the rotation direction. The problem specifies clockwise rotation, meaning the last elements move to the front.

Example of confusion:

• Original: [4, 3, 2, 6]
• Rotate 1 clockwise: [6, 4, 3, 2] (last element 6 moves to front)
• NOT: [3, 2, 6, 4] (this would be counter-clockwise)

3. Integer Overflow in Large Arrays

For very large arrays with large values, the rotation function calculations might overflow in languages with fixed integer sizes.

Solution for overflow-prone languages:

• Use appropriate data types (e.g., long long in C++)
• In Python, this isn't an issue due to arbitrary precision integers

4. Off-by-One Errors in Loop Range

Forgetting that we need to check exactly n rotations (from 0 to n-1).

Incorrect:

for i in range(n):  # This would try to calculate F(n), which doesn't exist
    f = f + s - n * nums[n - i - 1]

Correct:

for i in range(1, n):  # Calculates F(1) through F(n-1)
    f = f + s - n * nums[n - i]