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.

 Answers within 10^-5 of the actual answer will be considered accepted.


 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

public class Mean_of_Array_After_Removing_Some_Elements {
    class Solution {
        public double trimMean(int[] arr) {
            int n = arr.length;
            double sum = 0d;
            Arrays.sort(arr);
            for (int i = n / 20; i < n - n / 20; ++i) {
                sum += arr[i];
            }
            return sum / (n * 9 / 10);
        }

    }
}

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);
    }
}

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