Introduction

In this blog post, we will explore how to solve the LeetCode problem "2090" 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 a 0-indexed array nums of n integers, and an integer k. The k-radius average for a subarray of nums centered at some index i with the radius k is the average of all elements in nums between the indices i - k and i + k (inclusive). If there are less than k elements before or after the index i, then the k-radius average is -1. Build and return an array avgs of length n where avgs[i] is the k-radius average for the subarray centered at index i. The average of x elements is the sum of the x elements divided by x, using integer division. The integer division truncates toward zero, which means losing its fractional part. For example, the average of four elements 2, 3, 1, and 5 is (2 + 3 + 1 + 5) / 4 = 11 / 4 = 2.75, which truncates to 2. Example 1: Input: nums = [7,4,3,9,1,8,5,2,6], k = 3 Output: [-1,-1,-1,5,4,4,-1,-1,-1] Explanation: - avg[0], avg[1], and avg[2] are -1 because there are less than k elements before each index. - The sum of the subarray centered at index 3 with radius 3 is: 7 + 4 + 3 + 9 + 1 + 8 + 5 = 37. Using integer division, avg[3] = 37 / 7 = 5. - For the subarray centered at index 4, avg[4] = (4 + 3 + 9 + 1 + 8 + 5 + 2) / 7 = 4. - For the subarray centered at index 5, avg[5] = (3 + 9 + 1 + 8 + 5 + 2 + 6) / 7 = 4. - avg[6], avg[7], and avg[8] are -1 because there are less than k elements after each index. Example 2: Input: nums = [100000], k = 0 Output: [100000] Explanation: - The sum of the subarray centered at index 0 with radius 0 is: 100000. avg[0] = 100000 / 1 = 100000. Example 3: Input: nums = [8], k = 100000 Output: [-1] Explanation: - avg[0] is -1 because there are less than k elements before and after index 0. Constraints: n == nums.length 1 <= n <= 105 0 <= nums[i], k <= 105

Explanation

Here's an efficient solution to calculate the k-radius averages for a given array.

• Sliding Window: The core idea is to use a sliding window of size 2*k + 1 to efficiently calculate the sum of elements within the radius k of each index. This avoids redundant calculations by reusing the sum from the previous window.
• Prefix Sum Optimization: We can also use a modified approach employing prefix sums to calculate the sum of elements between i - k and i + k.

• Handle Edge Cases: If i - k or i + k goes out of bounds, the average for index i is -1.

• Time Complexity: O(n), Space Complexity: O(1) (excluding the output array)

Code

    def getAverages(nums, k):
    n = len(nums)
    avgs = [-1] * n

    if 2 * k + 1 > n:
        return avgs

    window_sum = 0
    window_size = 2 * k + 1

    # Calculate the initial window sum
    for i in range(window_size):
        window_sum += nums[i]

    # Calculate the first average
    avgs[k] = window_sum // window_size

    # Slide the window
    for i in range(k + 1, n - k):
        window_sum -= nums[i - k - 1]  # Remove the leftmost element
        window_sum += nums[i + k]      # Add the rightmost element
        avgs[i] = window_sum // window_size

    return avgs