# Question

Formatted question description: https://leetcode.ca/all/1619.html

 1619. Mean of Array After Removing Some Elements

Given an integer array arr,
return the mean of the remaining integers after removing the smallest 5% and the largest 5% of the elements.

Example 1:

Input: arr = [1,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,3]
Output: 2.00000
Explanation: After erasing the minimum and the maximum values of this array, all elements are equal to 2, so the mean is 2.

Example 2:

Input: arr = [6,2,7,5,1,2,0,3,10,2,5,0,5,5,0,8,7,6,8,0]
Output: 4.00000

Example 3:

Input: arr = [6,0,7,0,7,5,7,8,3,4,0,7,8,1,6,8,1,1,2,4,8,1,9,5,4,3,8,5,10,8,6,6,1,0,6,10,8,2,3,4]
Output: 4.77778

Example 4:

Input: arr = [9,7,8,7,7,8,4,4,6,8,8,7,6,8,8,9,2,6,0,0,1,10,8,6,3,3,5,1,10,9,0,7,10,0,10,4,1,10,6,9,3,6,0,0,2,7,0,6,7,2,9,7,7,3,0,1,6,1,10,3]
Output: 5.27778

Example 5:

Input: arr = [4,8,4,10,0,7,1,3,7,8,8,3,4,1,6,2,1,1,8,0,9,8,0,3,9,10,3,10,1,10,7,3,2,1,4,9,10,7,6,4,0,8,5,1,2,1,6,2,5,0,7,10,9,10,3,7,10,5,8,5,7,6,7,6,10,9,5,10,5,5,7,2,10,7,7,8,2,0,1,1]
Output: 5.29167

Constraints:
20 <= arr.length <= 1000
arr.length is a multiple of 20.
0 <= arr[i] <= 105


# Algorithm

Sort first, then calculate the 5% position according to the length, traverse the part of the array that removes the first and last trim length, and calculate the average.

Time complexity: O(nlogn)

Space complexity: O(1)

# Code

Java

Java

class Solution {
public double trimMean(int[] arr) {
Arrays.sort(arr);
int length = arr.length;
int removeCount = length / 20;
int start = removeCount, end = length - removeCount;
double sum = 0;
for (int i = start; i < end; i++)
sum += arr[i];
return sum / (length - removeCount * 2);
}
}