LeetCode-in-Java

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 java.util.Arrays;

public class Solution {
    public double minimumAverage(int[] nums) {
        Arrays.sort(nums);
        double m = 102.0;
        for (int i = 0, l = nums.length; i < l / 2; i++) {
            m = Math.min(m, nums[i] + (double) nums[l - i - 1]);
        }
        return m / 2.0;
    }
}

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