LeetCode in Kotlin

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

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.
    fun bestRotation(nums: IntArray): Int {
        val n = nums.size
        var res = 0
        val change = IntArray(n)
        for (i in 0 until n) {
            change[(i - nums[i] + 1 + n) % n]--
        }
        for (i in 1 until n) {
            change[i] += change[i - 1] + 1
            res = if (change[i] > change[res]) i else res
        }
        return res
    }
}

This site is open source. Improve this page.