# Smallest Rotation With Highest Score Problem

## Description

LeetCode Problem 798.

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, nums, …, 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:

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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:

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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 <= 10^5
• 0 <= nums[i] < nums.length

## Sample C++ Code

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class Solution {
public:
int bestRotation(vector<int>& nums) {
int n = size(nums);

vector<int> diff(n, 1);
for (int i = 0; i < n; ++i)
diff[(i + n - nums[i] + 1) % n]--;

int ans = 0, prefix = 0, mx = INT_MIN;
for (int i = 0; i < n; ++i) {
prefix += diff[i];
if (prefix > mx) {
mx = prefix;
ans = i;
}
}

return ans;
}
};
``````