Leetcode 798: Smallest Rotation with Highest Score

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

Problem Analysis:

When k = 1, we observe when the item in the nums array gets a point or loses a point after a rotation.

Gets a point: the last item rotate to index 0, no matter what item value it is

Lost a point: we can find it happens when new index is item value + 1

So we iterate the k, every k state can compute points from k-1 state, so we get every k states change. Then we can get the biggest change, when the k value is the answer

Solution

class Solution {
public:
    int bestRotation(vector<int>& nums) {
        vector<int> change(nums.size());
        for (int i = 0; i < nums.size(); ++i)
            change[(i - nums[i] + 1 + (int)nums.size()) % (int)nums.size()] -= 1;
        for (int i = 1; i < nums.size(); ++i)
            change[i] += change[i - 1] + 1;
        return distance(change.begin(), max_element(change.begin(), change.end()));
    }
};

Time complexity is O(nums)

Space complexity is O(nums)