**source:**https://leetcode.com/problems/smallest-rotation-with-highest-score/**C/C++**

**Solution to LeetCode**problem

**798**.

**Smallest Rotation with Highest Score**.

## Problem

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`

.

## Examples

**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 <= 10`

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

## Solution

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class Solution {
public:
int bestRotation(vector<int>& nums) {
vector<int> points(2 * nums.size() + 1, 0);
int k = 0;
int pMax = 0;
for (int i = 0; i < nums.size(); i += 1) {
if (nums[i] <= i) {
points[0] += 1;
points[(i - nums[i]) + 1] -= 1;
}
points[i + 1] += 1;
points[i + (nums.size() - nums[i]) + 1] -= 1;
}
pMax = points[0];
for (int i = 1; i < points.size(); i += 1) {
points[i] += points[i - 1];
if (points[i] > pMax) {
pMax = points[i];
k = i;
}
}
return k;
}
};