Package g0701_0800.s0798_smallest_rotation_with_highest_score

Class Solution

java.lang.Object
g0701_0800.s0798_smallest_rotation_with_highest_score.Solution
public class Solution extends Object
798 - Smallest Rotation with Highest Score\. Hard 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`

  • Constructor Details

    • Solution

      public Solution()

  • Method Details

    • bestRotation

      public int bestRotation(int[] nums)