396. Rotate Function

Problem statement

You are given an integer array nums of length n.

Assume arrk to be an array obtained by rotating nums by k positions clockwise. We define the rotation function F on nums as follows:

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.

Examples

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 <= 10^5
-100 <= nums[i] <= 100

Explanation

Simulating each rotation step by step is the easiest way to understand this problem.

First, calculate F(0):

F(0) = 0*a[0] + 1*a[1] + 2*a[2] + ... + (n-1)*a[n-1]

For the next rotation, the last element moves to the front and everything else shifts right by one:

F(1) = 0*a[n-1] + 1*a[0] + 2*a[1] + ... + (n-1)*a[n-2]
F(2) = 0*a[n-2] + 1*a[n-1] + 2*a[0] + ... + (n-1)*a[n-3]

Here’s the key insight: when we rotate, the element at position i (from the end) becomes the first element, so its contribution drops to 0. All other elements now contribute +1 more than before because their positions increased by 1.

So the change from one rotation to the next is:

We lose n * a[i] (the element that moved to position 0)
We gain sum (every other element adds 1 more)

This gives us the recurrence:

F = F + sum - n * a[i]

Just store the total sum of the array, then iterate backwards updating F using this formula. Keep track of the maximum F you see.

Time: O(n) — one pass to compute sum and F(0), one pass for rotations
Space: O(1) — just a few variables

Simulation

Code

#define ll long long

class Solution {
public:
    int maxRotateFunction(vector<int>& nums) {
        ll ans = INT_MIN, sum = 0, F = 0, n = nums.size();

        for (int i = 0; i < n; i++) {
            sum += nums[i];
            F += i * nums[i];
        }

        for (int i = n - 1; i >= 0; i--) {
            int rem = n * nums[i];
            F = F + sum - rem;
            ans = max<ll>(ans, F);
        }

        return ans;
    }
};