Leetcode daily question (396. rotate function)

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

First of all, calculate the sum, Then calculate the value when the rotation operation is not performed curr, Each subsequent rotation operation takes the last one to the first , Because the coefficient in the first place is 0, So we subtract the value of the last digit by the coefficient given to it before the rotation length-1, therefore curr = curr - nums[last] _ (length-1), Then except for the last one , The coefficients of other numbers have increased 1, We calculated earlier sum Yes, all numerical coefficients are 1 And , So we just need to subtract the last number to be the increase , curr = curr + sum - nums[last], So the final , curr = curr - nums[last] _ (length-1) + sum - nums[last]

impl Solution {
    
    pub fn max_rotate_function(nums: Vec<i32>) -> i32 {
    
        let sum = nums.iter().map(|v| *v).sum::<i32>();
        let mut curr = nums.iter().enumerate().map(|(i, v)| i as i32 * v).sum::<i32>();
        let mut ans = curr;
        let length = nums.len() as i32;
        for v in nums.into_iter().rev() {
    
            curr = curr - v * (length - 1) + sum - v;
            ans = ans.max(curr);
        }
        ans
    }
}