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Minimum Average of Smallest and Largest Elements

Difficulty: Easy

Problem Description

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.

Key Insights

  • The problem requires repeated removal of the smallest and largest elements in the array.
  • The average of these two elements needs to be stored and the minimum of these averages must be returned.
  • Since the input size is small (up to 50), sorting or using a priority queue can be efficient.
  • The solution requires careful handling of floating-point arithmetic.

Space and Time Complexity

Time Complexity: O(n log n) - due to sorting the array. Space Complexity: O(n) - for storing the averages.

Solution

To solve this problem, we can follow these steps:

  1. Sort the input array nums. This allows us to easily access the smallest and largest elements.
  2. Use a loop to perform the averaging operation n / 2 times.
  3. In each iteration, compute the average of the smallest and largest elements.
  4. Keep track of the minimum average encountered during the iterations.
  5. Return the minimum average.

Data Structures Used:

  • An array for the input numbers.
  • A list to store the averages.

Code Solutions

def minimum_average(nums):
    nums.sort()  # Sort the array
    averages = []  # List to store averages
    n = len(nums)

    for i in range(n // 2):
        minElement = nums[i]  # Smallest element
        maxElement = nums[n - 1 - i]  # Largest element
        average = (minElement + maxElement) / 2  # Calculate average
        averages.append(average)  # Add to averages list

    return min(averages)  # Return the minimum of averages
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