Introduction

In this blog post, we will explore how to solve the LeetCode problem "2465" 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 integer array nums of even length. As long as nums is not empty, you must repetitively: Find the minimum number in nums and remove it. Find the maximum number in nums and remove it. Calculate the average of the two removed numbers. The average of two numbers a and b is (a + b) / 2. For example, the average of 2 and 3 is (2 + 3) / 2 = 2.5. Return the number of distinct averages calculated using the above process. Note that when there is a tie for a minimum or maximum number, any can be removed. Example 1: Input: nums = [4,1,4,0,3,5] Output: 2 Explanation: 1. Remove 0 and 5, and the average is (0 + 5) / 2 = 2.5. Now, nums = [4,1,4,3]. 2. Remove 1 and 4. The average is (1 + 4) / 2 = 2.5, and nums = [4,3]. 3. Remove 3 and 4, and the average is (3 + 4) / 2 = 3.5. Since there are 2 distinct numbers among 2.5, 2.5, and 3.5, we return 2. Example 2: Input: nums = [1,100] Output: 1 Explanation: There is only one average to be calculated after removing 1 and 100, so we return 1. Constraints: 2 <= nums.length <= 100 nums.length is even. 0 <= nums[i] <= 100

Explanation

• Sorting: Sort the input array nums in ascending order. This allows us to easily identify the minimum and maximum elements at each step by accessing the beginning and end of the array.
  • Two Pointers: Use two pointers, one at the beginning (left) and one at the end (right) of the sorted array. Calculate the average of the elements pointed to by these pointers, add the average to a set to track distinct averages, and then move the pointers inward.
  • Set for Distinct Averages: Store the calculated averages in a set. Sets automatically handle uniqueness, ensuring that we only count distinct averages.

• Runtime Complexity: O(n log n) due to the sorting step.
• Storage Complexity: O(n) in the worst case for the set of distinct averages (although in practice, it's bounded by nums.length/2).

Code

    def distinctAverages(nums):
    """
    Calculates the number of distinct averages obtained by repeatedly removing
    the minimum and maximum elements from the input array.

    Args:
        nums: A list of integers.

    Returns:
        The number of distinct averages.
    """

    nums.sort()
    left = 0
    right = len(nums) - 1
    distinct_averages = set()

    while left < right:
        average = (nums[left] + nums[right]) / 2
        distinct_averages.add(average)
        left += 1
        right -= 1

    return len(distinct_averages)