3194. Minimum Average of Smallest and Largest Elements LeetCode Solution

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Table of Contents

1. Problem Statement
2. Complexity Analysis
3. Minimum Average of Smallest and Largest Elements solution in C++
4. Minimum Average of Smallest and Largest Elements solution in Java
5. Minimum Average of Smallest and Largest Elements solution in Python
6. Additional Resources

Problem Statement of Minimum Average of Smallest and Largest Elements

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

Complexity Analysis

• Time Complexity: O(\texttt{sort})
• Space Complexity: O(\texttt{sort})

3194. Minimum Average of Smallest and Largest Elements LeetCode Solution in C++

class Solution {
 public:
  double minimumAverage(vector<int>& nums) {
    constexpr int kMax = 50;
    double ans = kMax;
    int i = 0;
    int j = nums.size() - 1;

    ranges::sort(nums);

    while (i < j)
      ans = min(ans, (nums[i++] + nums[j--]) / 2.0);

    return ans;
  }
};
/* code provided by PROGIEZ */

3194. Minimum Average of Smallest and Largest Elements LeetCode Solution in Java

class Solution {
  public double minimumAverage(int[] nums) {
    final int kMax = 50;
    double ans = kMax;
    int i = 0;
    int j = nums.length - 1;

    Arrays.sort(nums);

    while (i < j)
      ans = Math.min(ans, (nums[i++] + nums[j--]) / 2.0);

    return ans;
  }
}
// code provided by PROGIEZ

3194. Minimum Average of Smallest and Largest Elements LeetCode Solution in Python

class Solution:
  def minimumAverage(self, nums: list[int]) -> float:
    nums.sort()
    return min((nums[i] + nums[~i]) / 2
               for i in range(len(nums) // 2 + 1))
 # code by PROGIEZ

Additional Resources

• Explore all LeetCode problem solutions at Progiez here
• Explore all problems on LeetCode website here

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