# Solving Leetcode Interviews in Seconds with AI: Minimum Average of Smallest and Largest Elements


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "3194" 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
	> You have an array of floating point numbers averages which is initially empty. You are given an array nums of n integers where n is even. You repeat the following procedure n / 2 times:  Remove the smallest element, minElement, and the largest element maxElement, from nums. Add (minElement + maxElement) / 2 to averages.  Return the minimum element in averages.   Example 1:  Input: nums = [7,8,3,4,15,13,4,1] Output: 5.5 Explanation:    step nums averages   0 [7,8,3,4,15,13,4,1] []   1 [7,8,3,4,13,4] [8]   2 [7,8,4,4] [8,8]   3 [7,4] [8,8,6]   4 [] [8,8,6,5.5]    The smallest element of averages, 5.5, is returned. Example 2:  Input: nums = [1,9,8,3,10,5] Output: 5.5 Explanation:    step nums averages   0 [1,9,8,3,10,5] []   1 [9,8,3,5] [5.5]   2 [8,5] [5.5,6]   3 [] [5.5,6,6.5]     Example 3:  Input: nums = [1,2,3,7,8,9] Output: 5.0 Explanation:    step nums averages   0 [1,2,3,7,8,9] []   1 [2,3,7,8] [5]   2 [3,7] [5,5]   3 [] [5,5,5]       Constraints:  2 <= n == nums.length <= 50 n is even. 1 <= nums[i] <= 50  

	# Explanation
	Here's the solution to the problem, explained and implemented:

*   **High-Level Approach:**
    *   Sort the input array `nums`. This allows easy access to the minimum and maximum elements in each iteration.
    *   Iterate `n / 2` times, calculating the average of the current minimum and maximum elements, and storing them.
    *   Find and return the minimum element in the `averages` list.

*   **Complexity:**
    *   Runtime: O(n log n) due to the initial sorting step. The rest of the operations take O(n).
    *   Storage: O(n) for the `averages` list.

	
	# Code
	```python
	def min_average_pair(nums):
    """
    Calculates the minimum average of pairs formed by repeatedly removing the smallest and largest elements from nums.

    Args:
        nums: A list of integers.

    Returns:
        The minimum average value.
    """
    nums.sort()
    averages = []
    n = len(nums)
    for i in range(n // 2):
        averages.append((nums[i] + nums[n - 1 - i]) / 2)
    return min(averages)
	```