Welcome to Subscribe On Youtube

1619. Mean of Array After Removing Some Elements

Description

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

 

Constraints:

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

Solutions

  • class Solution {
        public double trimMean(int[] arr) {
            Arrays.sort(arr);
            int n = arr.length;
            double s = 0;
            for (int start = (int) (n * 0.05), i = start; i < n - start; ++i) {
                s += arr[i];
            }
            return s / (n * 0.9);
        }
    }
    
  • class Solution {
    public:
        double trimMean(vector<int>& arr) {
            sort(arr.begin(), arr.end());
            int n = arr.size();
            double s = 0;
            for (int start = (int) (n * 0.05), i = start; i < n - start; ++i)
                s += arr[i];
            return s / (n * 0.9);
        }
    };
    
  • class Solution:
        def trimMean(self, arr: List[int]) -> float:
            n = len(arr)
            start, end = int(n * 0.05), int(n * 0.95)
            arr.sort()
            t = arr[start:end]
            return round(sum(t) / len(t), 5)
    
    
  • func trimMean(arr []int) float64 {
    	sort.Ints(arr)
    	n := len(arr)
    	sum := 0.0
    	for i := n / 20; i < n-n/20; i++ {
    		sum += float64(arr[i])
    	}
    	return sum / (float64(n) * 0.9)
    }
    
  • function trimMean(arr: number[]): number {
        arr.sort((a, b) => a - b);
        let n = arr.length,
            rmLen = n * 0.05;
        let sum = 0;
        for (let i = rmLen; i < n - rmLen; i++) {
            sum += arr[i];
        }
        return sum / (n * 0.9);
    }
    
    
  • impl Solution {
        pub fn trim_mean(mut arr: Vec<i32>) -> f64 {
            arr.sort();
            let n = arr.len();
            let count = ((n as f64) * 0.05).floor() as usize;
            let mut sum = 0;
            for i in count..n - count {
                sum += arr[i];
            }
            (sum as f64) / ((n as f64) * 0.9)
        }
    }
    
    

All Problems

All Solutions