LeetCode 798. Smallest Rotation with Highest Score

原题链接在这里：https://leetcode.com/problems/smallest-rotation-with-highest-score/

题目：

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

题解：

For the example 1 in [2,3,1,4,0]. The index of 1 is 2. 1 <= index, it is counted as 1 point.

It is rotating 2 times to index 0, it is going to lose 1 point.

It is rotated from i to A[i] - 1, which means rotates i - (A[i] - 1) times.

And when it rotate from index 0 to n - 1, it is going to gain one point.

Thus we have diff to maintain the decreasing point for each step. 

The second for loop is to accumlate the change point. totoal[3] = total[2] + diff[3] + 1.

For 0 and n, it would never add or minus point, we could think it from 0 to n - 1, first minus 1 point and then add 1 point.

Time Complexity: O(n). n = nums.length.

Space: O(n).

AC Java:

class Solution {
    public int bestRotation(int[] nums) {
        int n = nums.length;
        int res = 0;
        int [] diff = new int[n];
        for(int i = 0; i < n; i++){
            diff[(i - nums[i] + 1 + n) % n]--;
        }
        
        for(int k = 1; k < n; k++){
            diff[k] += diff[k - 1] + 1;
            if(diff[res] < diff[k]){
                res = k;
            }
        }
        
        return res;
    }
}