LeetCode-in-Java

798. Smallest Rotation with Highest Score

Hard

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

Solution

public class Solution {
    // nums[i] will be in the range [0, nums.length].
    // At which positions will we lose points? The answer is k = i - nums[i] + 1.
    // We need to accumulate points we have lost from previous rotations using prefix sum except one
    // we did not lose.
    public int bestRotation(int[] nums) {
        int n = nums.length;
        int res = 0;
        int[] change = new int[n];
        for (int i = 0; i < n; i++) {
            change[(i - nums[i] + 1 + n) % n]--;
        }
        for (int i = 1; i < n; i++) {
            change[i] += change[i - 1] + 1;
            res = change[i] > change[res] ? i : res;
        }
        return res;
    }
}

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