## Algorithm

Problem Name: 798. 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`

## Code Examples

### #1 Code Example with C++ Programming

```Code - C++ Programming```

``````
class Solution {
public:
int bestRotation(vector<int>& A) {
int n = A.size();
vector<int>k(n);
for(int i = 0; i  <  n; i++){
for(int j = 0; j  < = i - A[i]; j++) k[j]++;
for(int j = i + 1; j  < = i + n - A[i] && j < n; j++) k[j]++;
}
return max_element(k.begin(), k.end()) - k.begin();
}
};
``````
Copy The Code &

Input

cmd
nums = [2,3,1,4,0]

Output

cmd
3

### #2 Code Example with Javascript Programming

```Code - Javascript Programming```

``````
const bestRotation = function(A) {
const N = A.length
for (let i = 0; i  <  N; ++i) {
let left = (i - A[i] + 1 + N) % N
let right = (i + 1) % N
}

let best = -N
let ans = 0,
cur = 0
for (let i = 0; i  <  N; ++i) {
if (cur > best) {
best = cur
ans = i
}
}
return ans
}
``````
Copy The Code &

Input

cmd
nums = [2,3,1,4,0]

Output

cmd
3

### #3 Code Example with Python Programming

```Code - Python Programming```

``````
class Solution:
def bestRotation(self, A):
N = len(A)
change = [1] * N
for i in range(N): change[(i - A[i] + 1) % N] -= 1
for i in range(1, N): change[i] += change[i - 1]
return change.index(max(change))
``````
Copy The Code &

Input

cmd
nums = [1,3,0,2,4]