#798

Hard

LeetCode Problem Workspace

Smallest Rotation with Highest Score

Find the smallest rotation index with the highest score using array and prefix sum techniques.

Problem Statement

You are given an array nums and asked to find the smallest rotation index k such that rotating the array by k results in the highest score. A score is calculated by counting how many elements in the array are less than or equal to their index after the rotation. For each k, rotate the array and compute the score, then return the smallest k that maximizes the score.

For example, given nums = [2,3,1,4,0], rotating by k=0 gives a score of 2, rotating by k=1 gives a score of 3, and so on. The goal is to efficiently compute the highest score and return the smallest rotation index that achieves it.

Examples

Example 1

Input: nums = [2,3,1,4,0]

Output: 3

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

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 Approach

Prefix Sum Array

The key to solving this problem efficiently is using a prefix sum array to track scores dynamically. First, calculate the score for the initial rotation and then compute subsequent rotations in O(1) time by adjusting the previous score with the prefix sum.

Efficient Calculation with Sliding Window

Instead of recalculating the score for every possible rotation, use a sliding window technique to update the score as you move from one rotation to the next. This reduces redundant calculations, ensuring the solution is optimized for larger arrays.

Optimizing the Search for the Smallest k

Once the highest score is identified, it's important to track the smallest index k that achieves this score. Keep track of the smallest k during the iterations to ensure an optimal solution.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity of the solution depends on the method used to calculate the score for each rotation. With the sliding window approach, the time complexity is O(n), where n is the length of the array. The space complexity is also O(n) due to the prefix sum array used for efficient score computation.

What Interviewers Usually Probe

Tests for efficiency with large arrays (e.g., length 10^5).
Ability to implement sliding window and prefix sum techniques.
Understanding the trade-off between recalculating and updating scores.

Common Pitfalls or Variants

Common pitfalls

Inefficient recalculation of scores for each rotation instead of using prefix sums.
Forgetting to track the smallest k when multiple rotations yield the same highest score.
Not considering edge cases with small arrays or array with the same values.

Follow-up variants

Rotating arrays in other ways (e.g., left or right rotations).
Generalizing the score calculation to other comparison rules.
Applying this approach to arrays of different sizes and constraints.

FAQ

What is the smallest rotation index with the highest score?

It is the index k where rotating the array nums by k results in the maximum score, with ties broken by choosing the smallest k.

How do I optimize calculating the score for each rotation?

Use a prefix sum array to calculate the score for each rotation in O(1) time, instead of recalculating it from scratch each time.

What is the key pattern for solving Smallest Rotation with Highest Score?

The problem can be solved efficiently using an array plus prefix sum approach, along with a sliding window technique for score computation.

What are common pitfalls in solving this problem?

Common pitfalls include recalculating scores inefficiently, missing edge cases, and not tracking the smallest k in case of score ties.

Can the solution be optimized further?

The sliding window approach, combined with prefix sum optimization, provides a highly efficient solution with O(n) time complexity for this problem.

terminal

Solution

Solution 1

#### Python3

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class Solution:
    def bestRotation(self, nums: List[int]) -> int:
        n = len(nums)
        mx, ans = -1, n
        d = [0] * n
        for i, v in enumerate(nums):
            l, r = (i + 1) % n, (n + i + 1 - v) % n
            d[l] += 1
            d[r] -= 1
        s = 0
        for k, t in enumerate(d):
            s += t
            if s > mx:
                mx = s
                ans = k
        return ans

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