Given an array `nums`

, we may rotate it by a non-negative integer `k`

so that the array becomes `nums[k], nums[k+1], nums[k+2], ... 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 1 point.

For example, if we have `[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].

Over all possible rotations, return the rotation index k that corresponds to the highest score we could receive. If there are multiple answers, return the smallest such index k.

Example 1:Input:[2, 3, 1, 4, 0]Output:3Explanation: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:[1, 3, 0, 2, 4]Output:0Explanation:nums will always have 3 points no matter how it shifts. So we will choose the smallest k, which is 0.

**Note:**

`nums`

will have length at most`20000`

.`nums[i]`

will be in the range`[0, nums.length]`

.

class Solution {
public int bestRotation(int[] nums) {
}
}