Smallest rotation with highest score

/*
 * You are given an array nums. You can rotate it by a non-negative integer k so that the array becomes [nums[k], nums[k + 1], ... nums[nums.length - 1], nums[0], nums[1], ..., nums[k-1]]. Afterward, any entries that are less than or equal to their index are worth one point.
 * For example, if we have nums = [2,4,1,3,0], and we rotate by k = 2, it becomes [1,3,0,2,4]. This is worth 3 points because 1 > 0 [no points], 3 > 1 [no points], 0 <= 2 [one point], 2 <= 3 [one point], 4 <= 4 [one point].
 * Return the rotation index k that corresponds to the highest score we can achieve if we rotated nums by it. If there are multiple answers, return the smallest such index k.
 * Example 1:
 * Input: nums = [2,3,1,4,0]
 * Output: 3
 * Explanation: Scores for each k are listed below:
 * k = 0,  nums = [2,3,1,4,0],    score 2
 * k = 1,  nums = [3,1,4,0,2],    score 3
 * k = 2,  nums = [1,4,0,2,3],    score 3
 * k = 3,  nums = [4,0,2,3,1],    score 4
 * k = 4,  nums = [0,2,3,1,4],    score 3
 * So we should choose k = 3, which has the highest score.
 * Example 2:
 * Input: nums = [1,3,0,2,4]
 * Output: 0
 * Explanation: nums will always have 3 points no matter how it shifts.
 * So we will choose the smallest k, which is 0.
 * Constraints:
 * 1 <= nums.length <= 105
 * 0 <= nums[i] < nums.length
 */
impl Solution {
  pub fn best_rotation(nums: Vec<i32>) -> i32 {
    let n = nums.len();
    let mut delta = vec![0i32; n];
    for (i, &v) in nums.iter().enumerate() {
      let v = v as usize;
      delta[(i + 1) % n] += 1;
      delta[(i + 1 + n - v) % n] -= 1;
    }
    let score0 = nums.iter().enumerate().filter(|&(i, &v)| i >= v as usize).count() as i32;
    let mut best_score = score0;
    let mut best_k = 0i32;
    let mut score = score0;
    for k in 1..n {
      score += delta[k];
      if score > best_score {
        best_score = score;
        best_k = k as i32;
      }
    }
    best_k
  }
}