# 3194. Minimum Average of Smallest and Largest Elements

## 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.

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

## Solutions

### Solution 1: Sorting

First, we sort the array $\text{nums}$. Then, we start taking elements from both ends of the array, calculating the sum of the two elements, and taking the minimum value. Finally, we return the minimum value divided by 2 as the answer.

The time complexity is $O(n \log n)$, and the space complexity is $O(\log n)$, where $n$ is the length of the array $\text{nums}$.

• class Solution {
public double minimumAverage(int[] nums) {
Arrays.sort(nums);
int n = nums.length;
int ans = 1 << 30;
for (int i = 0; i < n / 2; ++i) {
ans = Math.min(ans, nums[i] + nums[n - i - 1]);
}
return ans / 2.0;
}
}

• class Solution {
public:
double minimumAverage(vector<int>& nums) {
sort(nums.begin(), nums.end());
int ans = 1 << 30, n = nums.size();
for (int i = 0; i < n; ++i) {
ans = min(ans, nums[i] + nums[n - i - 1]);
}
return ans / 2.0;
}
};

• class Solution:
def minimumAverage(self, nums: List[int]) -> float:
nums.sort()
n = len(nums)
return min(nums[i] + nums[n - i - 1] for i in range(n // 2)) / 2


• func minimumAverage(nums []int) float64 {
sort.Ints(nums)
n := len(nums)
ans := 1 << 30
for i, x := range nums[:n/2] {
ans = min(ans, x+nums[n-i-1])
}
return float64(ans) / 2
}

• function minimumAverage(nums: number[]): number {
nums.sort((a, b) => a - b);
const n = nums.length;
let ans = Infinity;
for (let i = 0; i * 2 < n; ++i) {
ans = Math.min(ans, nums[i] + nums[n - 1 - i]);
}
return ans / 2;
}