3194. Minimum Average of Smallest and Largest Elements

Problem Link

Description

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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

Solution

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Python3

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