1619. 删除某些元素后的数组均值
1619. 删除某些元素后的数组均值
题目
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.
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 of20
.0 <= arr[i] <= 10^5
题目大意
给你一个整数数组 arr
,请你删除最小 5%
的数字和最大 5%
的数字后,求剩余数字的平均值。
与 标准答案 误差在 10^-5
的结果都被视为正确结果。
示例 1:
输入: arr = [1,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,3]
输出: 2.00000
解释: 删除数组中最大和最小的元素后,所有元素都等于 2,所以平均值为 2 。
示例 2:
输入: arr = [6,2,7,5,1,2,0,3,10,2,5,0,5,5,0,8,7,6,8,0]
输出: 4.00000
示例 3:
输入: 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]
输出: 4.77778
示例 4:
输入: 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]
输出: 5.27778
示例 5:
输入: 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]
输出: 5.29167
提示:
20 <= arr.length <= 1000
arr.length
是20
的倍数0 <= arr[i] <= 10^5
解题思路
首先对数组进行升序排序,使得最小值和最大值出现在数组的两端。
数组中最小的 5% 和最大的 5% 元素分别位于前
left = n * 0.05
和后right = n * 0.05
的位置。遍历数组,仅在索引范围
[left, right)
内的元素累加求和。剩余部分的长度为
n * 0.9
,因此平均值为:sum / (n * 0.9)
复杂度分析
- 时间复杂度:
O(n log n)
,排序:O(n log n)
,遍历剩余部分:O(n)
,总时间复杂度为:O(n log n)
。 - 空间复杂度:
O(1)
,排序使用了原地操作,无额外空间。
代码
/**
* @param {number[]} arr
* @return {number}
*/
var trimMean = function (arr) {
arr.sort((a, b) => a - b);
const n = arr.length;
const left = Math.floor(n * 0.05);
const right = Math.ceil(n * 0.95);
let sum = 0;
for (let i = left; i < right; i++) {
sum += arr[i];
}
return sum / (n * 0.9);
};