LeetCode in Kotlin

3194. Minimum Average of Smallest and Largest Elements

Easy

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

import kotlin.math.min

class Solution {
    fun minimumAverage(nums: IntArray): Double {
        nums.sort()
        var m = 102.0
        var i = 0
        val l = nums.size
        while (i < l / 2) {
            m = min(m, nums[i] + nums[l - i - 1].toDouble())
            i++
        }
        return m / 2.0
    }
}

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