# Solving Leetcode Interviews in Seconds with AI: Sum of Good Numbers


	# 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](https://www.chatmagic.app), 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
	```python
	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
	```