Smallest Rotation With Highest Score - Leetcode Solution

The Smallest Rotation With Highest Score problem on Leetcode is a challenging problem that requires careful analysis and optimization to find the smallest rotation of an array that results in the highest score. The problem statement is as follows:

You are given an array of integers A. A rotation of the array means that each element is shifted right by one index, and the last element of the array is moved to the beginning. For example, the rotation of [0, 1, 2, 3] would be [3, 0, 1, 2].

A score is determined for each rotation by counting the number of elements in the rotated array that are equal to or greater than their index. For example, the score of the rotation [3, 0, 1, 2] would be 3, because 3 is greater than or equal to its index, and so are 0 and 1.

Your task is to find the smallest rotation of the array A that results in the highest score.

To solve this problem, we first need to understand the constraints and the solution approach. Here are the steps to solve this problem:

Step 1: Understand the problem constraints

The problem input contains an array of integers, A, such that 1 <= A.length <= 2*10^5. Therefore, we need to come up with an algorithm that can efficiently handle arrays of big sizes. Additionally, since we are looking for the smallest rotation of the array that gives the highest score, we need to define the relationship between the score of a rotation and its index explicitly.

Step 2: Define the score relationship with the rotation index

In general, the score of a rotation depends on the elements in the rotated array that are greater than or equal to their index. Therefore, we can use the following formula to calculate the score of a rotation:

score(i) = i - count(j, A[j] > i)

where i is the index of the rotation, j is the index of the array, A is the input integer array, and count(j, A[j] > i) represents the number of elements in A that are greater than i.

Step 3: Create a helper function to calculate the score of all rotations

Using the score formula we defined earlier, we can create a helper function to calculate the score of all rotations of the input array. The helper function takes the array A as input and returns the scores of all the rotations as an array.

Here is how the helper function looks like:

def calculate_score(A): n = len(A) count = [0] * n for i in range(n): count[(i - A[i] + n) % n] += 1 res = [n - count[i] + 2 * i for i in range(n)] return res

The helper function uses an efficient way of calculating the count of elements in A that are greater than i. First, it initializes a count array of size n with all zeros. Then, it iterates over the array A and calculates the index of the element in the rotated array using the following formula: (i - A[i] + n) % n. Finally, it updates the count array by incrementing the count at the calculated index.

After calculating the count of elements that are greater than i, the helper function uses the score formula we defined earlier to calculate the score of each rotation. Specifically, it iterates over all rotations, i, and calculates the score using the following formula: n - count[i] + 2 * i.

Step 4: Find the smallest rotation with the highest score

Now that we have a helper function to calculate the score of all rotations, we can use it to find the smallest rotation with the highest score. To do that, we can iterate over all rotations and find the one that has the highest score. However, this approach would take O(n^2) time, which is not efficient.

Instead, we can use a binary search approach to find the smallest rotation with the highest score. We can observe that if we rotate the array by x positions, then the scores of all subsequent rotations will be shifted by x positions. Therefore, we can use binary search to find the value of x that results in the highest score.

Here is how the main function for solving this problem looks like:

def bestRotation(A): n = len(A) left, right = 0, n - 1 scores = calculate_score(A) while left < right: mid = (left + right) // 2 if scores[mid] >= scores[mid + 1]: right = mid else: left = mid + 1 return left

The main function first initializes the left and right indices for binary search. It also calculates the score of all the rotations using the helper function we defined earlier.

Then, it starts the binary search loop by checking if the left index is less than the right index. If it is, then it calculates the mid index and compares the score of mid and mid+1 rotations. If the score of the mid rotation is greater than or equal to the score of mid+1 rotation, it means that the highest score can be achieved by rotating the array by mid positions or less. Therefore, it updates the right index to mid. Otherwise, it means that the highest score can be achieved by rotating the array by more than mid positions, so it updates the left index to mid+1.

Finally, the main function returns the left index, which corresponds to the smallest rotation with the highest score.

Overall, the time complexity of the solution is O(n log n), due to the binary search approach. The space complexity is also O(n), because we need to store the score of all rotations in an array.

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