Introduction

In this blog post, we will explore how to solve the LeetCode problem "3452" 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

Given an array of integers nums and an integer k, an element nums[i] is considered good if it is strictly greater than the elements at indices i - k and i + k (if those indices exist). If neither of these indices exists, nums[i] is still considered good. Return the sum of all the good elements in the array. Example 1: Input: nums = [1,3,2,1,5,4], k = 2 Output: 12 Explanation: The good numbers are nums[1] = 3, nums[4] = 5, and nums[5] = 4 because they are strictly greater than the numbers at indices i - k and i + k. Example 2: Input: nums = [2,1], k = 1 Output: 2 Explanation: The only good number is nums[0] = 2 because it is strictly greater than nums[1]. Constraints: 2 <= nums.length <= 100 1 <= nums[i] <= 1000 1 <= k <= floor(nums.length / 2)

Explanation

• Iterate through the nums array.
  • For each element nums[i], check if it's greater than nums[i-k] and nums[i+k] (if these indices are valid).
  • Accumulate the sum of good elements.

• Time Complexity: O(n), where n is the length of the nums array. Space Complexity: O(1).

Code

    def sum_of_good_numbers(nums, k):
    """
    Calculates the sum of "good" elements in an array.

    Args:
        nums: A list of integers.
        k: An integer representing the distance to check for neighbors.

    Returns:
        The sum of all good elements in the array.
    """

    n = len(nums)
    good_sum = 0

    for i in range(n):
        is_good = True

        if i - k >= 0 and nums[i] <= nums[i - k]:
            is_good = False
        if i + k < n and nums[i] <= nums[i + k]:
            is_good = False

        if is_good:
            good_sum += nums[i]

    return good_sum