# 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: 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

## Solutions

• class Solution {
public int bestRotation(int[] nums) {
int n = nums.length;
int[] d = new int[n];
for (int i = 0; i < n; ++i) {
int l = (i + 1) % n;
int r = (n + i + 1 - nums[i]) % n;
++d[l];
--d[r];
}
int mx = -1;
int s = 0;
int ans = n;
for (int k = 0; k < n; ++k) {
s += d[k];
if (s > mx) {
mx = s;
ans = k;
}
}
return ans;
}
}

• class Solution {
public:
int bestRotation(vector<int>& nums) {
int n = nums.size();
int mx = -1, ans = n;
vector<int> d(n);
for (int i = 0; i < n; ++i) {
int l = (i + 1) % n;
int r = (n + i + 1 - nums[i]) % n;
++d[l];
--d[r];
}
int s = 0;
for (int k = 0; k < n; ++k) {
s += d[k];
if (s > mx) {
mx = s;
ans = k;
}
}
return ans;
}
};

• class Solution:
def bestRotation(self, nums: List[int]) -> int:
n = len(nums)
mx, ans = -1, n
d = [0] * n
for i, v in enumerate(nums):
l, r = (i + 1) % n, (n + i + 1 - v) % n
d[l] += 1
d[r] -= 1
s = 0
for k, t in enumerate(d):
s += t
if s > mx:
mx = s
ans = k
return ans


• func bestRotation(nums []int) int {
n := len(nums)
d := make([]int, n)
for i, v := range nums {
l, r := (i+1)%n, (n+i+1-v)%n
d[l]++
d[r]--
}
mx, ans, s := -1, n, 0
for k, t := range d {
s += t
if s > mx {
mx = s
ans = k
}
}
return ans
}