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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 <= 105
  • 0 <= nums[i] < nums.length

Solution Explanation

This problem asks to find the rotation index k that maximizes the score after rotating the array nums. The score is calculated by counting the number of elements nums[i] such that nums[i] <= i after rotation. A brute-force approach would try all possible rotations, which has a time complexity of O(n²), where n is the length of nums. However, a more efficient solution exists using the concept of prefix sums and difference arrays.

Approach:

Instead of directly calculating the score for each rotation, we use a clever observation: the change in score when rotating from k to k+1 can be efficiently calculated.

  1. Difference Array: Create a difference array d of size n. This array will store the change in score when moving from one rotation to the next. Initially, d is filled with zeros.

  2. Calculate Difference: For each element nums[i] in the original array:

    • Identify the indices l and r where nums[i] contributes to the score.
    • l = (i + 1) % n: The index where nums[i] starts contributing to the score after the rotation.
    • r = (n + i + 1 - nums[i]) % n: The index where nums[i] stops contributing to the score after the rotation.
    • Increment d[l] (to account for the added contribution) and decrement d[r] (to account for the removed contribution). This is because nums[i] contributes to the score from index l to r-1 (inclusive).

  3. Prefix Sum: Calculate the prefix sum of d. The prefix sum s[k] represents the total score after rotating by k.

  4. Find Maximum: Iterate through the prefix sum array s to find the maximum score mx and the corresponding index ans (smallest index with the maximum score).

Time Complexity Analysis:

  • Creating the difference array d: O(n)
  • Calculating the prefix sum of d: O(n)
  • Finding the maximum score: O(n)

Therefore, the overall time complexity of this solution is O(n).

Space Complexity Analysis:

The space complexity is dominated by the difference array d, which has a size of n. Thus, the space complexity is O(n).

Code Explanation (Python3)

class Solution:
    def bestRotation(self, nums: List[int]) -> int:
        n = len(nums)
        d = [0] * n  # Difference array
        for i, v in enumerate(nums):
            l, r = (i + 1) % n, (n + i + 1 - v) % n
            d[l] += 1  # Increment at l
            d[r] -= 1  # Decrement at r
 
        s = 0  # Prefix sum
        mx = -1  # Maximum score
        ans = n  # Index of maximum score (initialized to n)
 
        for k, t in enumerate(d):
            s += t  # Calculate prefix sum
            if s > mx:
                mx = s
                ans = k  # Update index if a better score is found
 
        return ans

The code directly implements the steps described in the approach section. The use of the modulo operator % handles the circular nature of array rotation effectively. The prefix sum efficiently accumulates the score changes, leading to the O(n) time complexity. The other languages (Java, C++, Go) follow a very similar structure and logic.