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# 798. Smallest Rotation with Highest Score

## Description

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: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:nums = [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.

**Constraints:**

`1 <= nums.length <= 10`

^{5}`0 <= nums[i] < nums.length`