1619. Mean of Array After Removing Some Elements Easy
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Share 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
public class Solution {
public double TrimMean(int[] arr) {
Array.Sort(arr);
int len = arr.Length;
double sum = 0;
for(int i= len/20 ; i<arr.Length-(len/20);i++){
sum+=arr[i];
}
return sum/(len-(len/10));
}
}Time Complexity: O(nlogn)
Space Complexity: O(1)
