- Difficulty: Hard
- Tags: LeetCode, Hard, Array, Prefix Sum, leetcode-798, Greedy, O(n), O(1)
Problem
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 byk = 2
, it becomes[1,3,0,2,4]
. This is worth3
points because1 > 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