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

Solution: Prefix Sum

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

 

JavaScript

/**
 * @param {number[]} nums
 * @return {number}
 */
const bestRotation = function (nums) {
  const n = nums.length;
  const change = new Array(n).fill(0);
  let result = 0;

  for (let index = 0; index < n; index++) {
    const num = nums[index];

    change[(index - num + n + 1) % n] -= 1;
  }
  for (let index = 1; index < n; index++) {
    change[index] += change[index - 1] + 1;

    if (change[index] <= change[result]) continue;
    result = index;
  }
  return result;
};