得分最高的最小轮调

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

/* 
 * This function takes an array of integers and returns the index k that maximizes the score after rotating the array. 
 * The score is calculated based on the number of elements less than or equal to their indices. 
 * This solution uses Java Stream for the computation. 
 */
import java.util.stream.IntStream;
public int bestRotation(int[] nums) {
    int n = nums.length;
    int[] change = new int[n];
    // Calculate the changes needed to the score for each k
    IntStream.range(0, n).forEach(i -> change[(i - nums[i] + 1 + n) % n]--);
    // Calculate the prefix sum array
    IntStream.range(1, n).forEach(i -> change[i] += change[i - 1] + 1);
    // Find the index with the maximum score
    return IntStream.range(0, n).reduce((a, b) -> change[a] > change[b] ? a : b).getAsInt();
}

该题解采用了差分数组和前缀和的思想来解决问题。首先，创建一个长度为n的数组d，并将所有元素初始化为1。这是因为在不考虑特殊情况下，每个元素至少可以贡献1分。然后，对于数组nums中的每个元素x，计算它在轮调后可能导致分数减少的位置，即当x被移到其索引之后时，它将不再贡献分数。因此，我们在数组d的相应位置上减去1。接下来，使用累积和函数accumulate来计算数组d的前缀和，这样每个元素d[i]就表示在轮调k=i时的得分。最后，返回得分最高的轮调下标k，即d中最大元素的索引。

O(n)

O(n)