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# 0798. Smallest Rotation with Highest Score

<https://leetcode.com/problems/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:** 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`

## ac

```java
```