Minimum Average of Smallest and Largest Elements

This approach directly simulates the process described in the problem. We use a data structure that allows for dynamic resizing and element removal, such as an ArrayList. In each step, we iterate through the current list to find the minimum and maximum elements, calculate their average, and then remove them. We repeat this process until the list is empty, keeping track of the minimum average found.

• Convert the input array nums into a List<Integer> to facilitate easy element removal.
• Initialize a variable minAverage to Double.MAX_VALUE.
• Loop n / 2 times, where n is the initial size of nums.
• In each iteration:
  • Find the minimum (minElement) and maximum (maxElement) values in the current list. This can be done by iterating through the list or using Collections.min() and Collections.max().
  • Calculate the average: currentAverage = (minElement + maxElement) / 2.0.
  • Update minAverage = Math.min(minAverage, currentAverage).
  • Remove the minElement and maxElement from the list. Note that removing by value requires careful handling, especially if there are duplicates. It's safer to remove by index or use list.remove(Integer.valueOf(value)).
• After the loop completes, return minAverage.

import java.util.ArrayList;
import java.util.Collections;
import java.util.List;

class Solution {
    public double minimumAverage(int[] nums) {
        List<Integer> list = new ArrayList<>();
        for (int num : nums) {
            list.add(num);
        }

        double minAverage = Double.MAX_VALUE;
        int n = nums.length;

        for (int i = 0; i < n / 2; i++) {
            int minElement = Collections.min(list);
            int maxElement = Collections.max(list);

            double currentAverage = (minElement + maxElement) / 2.0;
            minAverage = Math.min(minAverage, currentAverage);

            // Remove one occurrence of minElement and maxElement
            list.remove(Integer.valueOf(minElement));
            list.remove(Integer.valueOf(maxElement));
        }

        return minAverage;
    }
}