Problem

Given an integer array arr, return the mean of the remaining integers after removing the smallest5% and the largest 5% of the elements.

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

Examples

Example 1

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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

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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

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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] <= 10^5

Solution

Method 1 – Sorting and Slicing

Intuition

To remove the smallest and largest 5% of elements, sort the array and slice off the required number of elements from both ends. The mean of the remaining elements is the answer.

Approach

  1. Sort the array.
  2. Compute k = n // 20 (5% of the array size).
  3. Remove the first k and last k elements.
  4. Calculate the mean of the remaining elements.
  5. Return the mean as a float.

Code

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class Solution {
public:
    double trimMean(vector<int>& arr) {
        int n = arr.size(), k = n / 20;
        sort(arr.begin(), arr.end());
        double sum = 0;
        for (int i = k; i < n - k; ++i) sum += arr[i];
        return sum / (n - 2 * k);
    }
};
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func trimMean(arr []int) float64 {
    n := len(arr)
    k := n / 20
    sort.Ints(arr)
    sum := 0
    for i := k; i < n-k; i++ {
        sum += arr[i]
    }
    return float64(sum) / float64(n-2*k)
}
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class Solution {
    public double trimMean(int[] arr) {
        int n = arr.length, k = n / 20;
        Arrays.sort(arr);
        double sum = 0;
        for (int i = k; i < n - k; i++) sum += arr[i];
        return sum / (n - 2 * k);
    }
}
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class Solution {
    fun trimMean(arr: IntArray): Double {
        val n = arr.size
        val k = n / 20
        arr.sort()
        var sum = 0.0
        for (i in k until n - k) sum += arr[i]
        return sum / (n - 2 * k)
    }
}
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def trim_mean(arr: list[int]) -> float:
    n = len(arr)
    k = n // 20
    arr.sort()
    return sum(arr[k:n-k]) / (n - 2 * k)
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impl Solution {
    pub fn trim_mean(arr: Vec<i32>) -> f64 {
        let n = arr.len();
        let k = n / 20;
        let mut arr = arr;
        arr.sort();
        let sum: i32 = arr[k..n-k].iter().sum();
        sum as f64 / (n - 2 * k) as f64
    }
}
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class Solution {
    trimMean(arr: number[]): number {
        const n = arr.length, k = Math.floor(n / 20);
        arr.sort((a, b) => a - b);
        let sum = 0;
        for (let i = k; i < n - k; ++i) sum += arr[i];
        return sum / (n - 2 * k);
    }
}

Complexity

  • ⏰ Time complexity: O(n log n), due to sorting the array.
  • 🧺 Space complexity: O(1) (in-place sort), or O(n) if not in-place.