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.

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

class Solution {
    public int bestRotation(int[] A) {
        int[] count = new int[A.length + 1];
        int len = A.length;
        for(int index = 0; index < A.length; index ++){
            int num = A[index];
            if(index < num){
                int maxLenK = (index + 1) % len;
                int ori = (index + len - num) % len;
                count[maxLenK] ++;
                count[ori + 1] --;
            }else{      // index >= num
                int ori = (index + len - num) % len;
                count[0] ++;
                count[ori + 1] --;
                if(index != len - 1){
                    count[(index + 1) % len] ++;
                    count[len] --;
                }
            }
        }
        int max = count[0];
        int res = 0;
        for(int i = 1; i < len; i ++){
            count[i] = count[i] + count[i - 1];
            if(count[i] > max){
                max = count[i];
                res = i;
            }            
        }
        return res;
    }
}

Explain:

nope.

Complexity:

Time complexity : O(n).
Space complexity : O(n).