Introduction

In this blog post, we will explore how to solve the LeetCode problem "798" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

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

Explanation

Here's the solution:

• Difference Array: The core idea is to use a difference array to efficiently track how the score changes with each rotation. For each element nums[i], we determine the range of rotations where it contributes to the score. This range is then used to update the difference array.
• Cumulative Sum: After calculating the difference array, we compute its cumulative sum. This cumulative sum array represents the score for each possible rotation k.

• Find Maximum: Finally, we iterate through the cumulative sum array to find the maximum score and its corresponding rotation index k. We return the smallest such index in case of ties.

• Runtime Complexity: O(n), Storage Complexity: O(n)

Code

    def bestRotation(nums):
    n = len(nums)
    diffs = [0] * n
    for i in range(n):
        low = (i + 1) % n
        high = (i - nums[i] + n) % n
        if low <= high:
            diffs[low] += 1
            if high + 1 < n:
                diffs[high + 1] -= 1
        else:
            diffs[0] += 1
            diffs[high + 1] -= 1
            diffs[low] += 1

    score = 0
    max_score = 0
    best_rotation = 0
    for i in range(n):
        score += diffs[i]
        if score > max_score:
            max_score = score
            best_rotation = i

    return best_rotation