How do you solve Smallest Rotation with Highest Score on LeetCode?

Smallest Rotation with Highest Score (LeetCode #798) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: The score for each rotation can be efficiently calculated using a score change array.

What is the brute force solution for Smallest Rotation with Highest Score?

The brute force approach for Smallest Rotation with Highest Score is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves calculating the score for every possible rotation of the array. For each rotation, we check how many elements are less than or equal to their indices and keep track of the maximum score found. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Smallest Rotation with Highest Score?

Smallest Rotation with Highest Score uses the following concepts: Array, Prefix Sum. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Smallest Rotation with Highest Score?

Always consider edge cases, such as arrays of length 1. Explain your thought process clearly when transitioning from brute force to optimal solutions. Practice explaining your code as if you were teaching someone else.

What are common mistakes when solving Smallest Rotation with Highest Score?

Not recognizing that scores can be derived from previous calculations. Failing to account for index wrapping when rotating the array.

#798

Smallest Rotation with Highest Score

Hard

ArrayPrefix SumHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

Instead of recalculating the score from scratch for each rotation, we can use a clever observation about how scores change as we rotate the array. We can maintain a score array that tracks the score changes efficiently.

⚙️

Algorithm

4 steps

1Step 1: Initialize an array to keep track of score changes for each rotation.
2Step 2: For each element in nums, determine how it affects the score for each possible rotation.
3Step 3: Calculate the cumulative score for each rotation using the score change array.
4Step 4: Find the rotation with the maximum score and return its index.

solution.py20 lines

1# Full working Python code
2
3def bestRotation(nums):
4    n = len(nums)
5    score = [0] * n
6    for i in range(n):
7        left = (i - nums[i] + 1 + n) % n
8        right = (i + 1) % n
9        score[left] += 1
10        if right != left:
11            score[right] -= 1
12    max_score = 0
13    best_k = 0
14    current_score = 0
15    for k in range(n):
16        current_score += score[k]
17        if current_score > max_score:
18            max_score = current_score
19            best_k = k
20    return best_k

ℹ

Complexity note: The time complexity is O(n) because we only loop through the array a constant number of times, and the space complexity is O(n) due to the score array used to track score changes.

1The score for each rotation can be efficiently calculated using a score change array.
2The optimal solution leverages cumulative sums to avoid recalculating scores from scratch.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.