Question

1619. Mean of Array After Removing Some Elements

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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

Solution 1

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)

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);
  }
};

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)
  }
}